Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & -2 & -5 \\ 3 & 3 & -2 \\ -2 & 0 & -2 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ 3 \\ -2 \end{array}\right] , \left[\begin{array}{c} -2 \\ 3 \\ 0 \end{array}\right] , \left[\begin{array}{c} -5 \\ -2 \\ -2 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^3\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ 3 \\ -2 \end{array}\right] , \left[\begin{array}{c} -2 \\ 3 \\ 0 \end{array}\right] , \left[\begin{array}{c} -5 \\ -2 \\ -2 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^3\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ 3 \\ -2 \end{array}\right] , \left[\begin{array}{c} -2 \\ 3 \\ 0 \end{array}\right] , \left[\begin{array}{c} -5 \\ -2 \\ -2 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^3\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -4 & -2 & 1 \\ 3 & 4 & 4 \\ -1 & -1 & -3 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ 3 \\ -1 \end{array}\right] , \left[\begin{array}{c} -2 \\ 4 \\ -1 \end{array}\right] , \left[\begin{array}{c} 1 \\ 4 \\ -3 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^3\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ 3 \\ -1 \end{array}\right] , \left[\begin{array}{c} -2 \\ 4 \\ -1 \end{array}\right] , \left[\begin{array}{c} 1 \\ 4 \\ -3 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^3\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ 3 \\ -1 \end{array}\right] , \left[\begin{array}{c} -2 \\ 4 \\ -1 \end{array}\right] , \left[\begin{array}{c} 1 \\ 4 \\ -3 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^3\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} 2 & 2 & 3 & -1 \\ 0 & -4 & 4 & -3 \\ 3 & 1 & -3 & -5 \\ -1 & -4 & -5 & 2 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 2 \\ 0 \\ 3 \\ -1 \end{array}\right] , \left[\begin{array}{c} 2 \\ -4 \\ 1 \\ -4 \end{array}\right] , \left[\begin{array}{c} 3 \\ 4 \\ -3 \\ -5 \end{array}\right] , \left[\begin{array}{c} -1 \\ -3 \\ -5 \\ 2 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 2 \\ 0 \\ 3 \\ -1 \end{array}\right] , \left[\begin{array}{c} 2 \\ -4 \\ 1 \\ -4 \end{array}\right] , \left[\begin{array}{c} 3 \\ 4 \\ -3 \\ -5 \end{array}\right] , \left[\begin{array}{c} -1 \\ -3 \\ -5 \\ 2 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^4\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 2 \\ 0 \\ 3 \\ -1 \end{array}\right] , \left[\begin{array}{c} 2 \\ -4 \\ 1 \\ -4 \end{array}\right] , \left[\begin{array}{c} 3 \\ 4 \\ -3 \\ -5 \end{array}\right] , \left[\begin{array}{c} -1 \\ -3 \\ -5 \\ 2 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} -4 & -1 & -4 & -2 & -1 \\ -5 & -5 & -1 & -2 & 4 \\ 3 & -4 & 3 & 0 & 2 \\ 0 & -2 & 3 & -3 & -2 \\ 1 & -5 & -3 & -2 & 0 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ -5 \\ 3 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} -1 \\ -5 \\ -4 \\ -2 \\ -5 \end{array}\right] , \left[\begin{array}{c} -4 \\ -1 \\ 3 \\ 3 \\ -3 \end{array}\right] , \left[\begin{array}{c} -2 \\ -2 \\ 0 \\ -3 \\ -2 \end{array}\right] , \left[\begin{array}{c} -1 \\ 4 \\ 2 \\ -2 \\ 0 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ -5 \\ 3 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} -1 \\ -5 \\ -4 \\ -2 \\ -5 \end{array}\right] , \left[\begin{array}{c} -4 \\ -1 \\ 3 \\ 3 \\ -3 \end{array}\right] , \left[\begin{array}{c} -2 \\ -2 \\ 0 \\ -3 \\ -2 \end{array}\right] , \left[\begin{array}{c} -1 \\ 4 \\ 2 \\ -2 \\ 0 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^5\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ -5 \\ 3 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} -1 \\ -5 \\ -4 \\ -2 \\ -5 \end{array}\right] , \left[\begin{array}{c} -4 \\ -1 \\ 3 \\ 3 \\ -3 \end{array}\right] , \left[\begin{array}{c} -2 \\ -2 \\ 0 \\ -3 \\ -2 \end{array}\right] , \left[\begin{array}{c} -1 \\ 4 \\ 2 \\ -2 \\ 0 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} -4 & -1 & 4 & 2 & 0 \\ -5 & -2 & 2 & -4 & -5 \\ -1 & 0 & -2 & 0 & -2 \\ 0 & -3 & 2 & -1 & 4 \\ 4 & 4 & -5 & -5 & -1 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ -5 \\ -1 \\ 0 \\ 4 \end{array}\right] , \left[\begin{array}{c} -1 \\ -2 \\ 0 \\ -3 \\ 4 \end{array}\right] , \left[\begin{array}{c} 4 \\ 2 \\ -2 \\ 2 \\ -5 \end{array}\right] , \left[\begin{array}{c} 2 \\ -4 \\ 0 \\ -1 \\ -5 \end{array}\right] , \left[\begin{array}{c} 0 \\ -5 \\ -2 \\ 4 \\ -1 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ -5 \\ -1 \\ 0 \\ 4 \end{array}\right] , \left[\begin{array}{c} -1 \\ -2 \\ 0 \\ -3 \\ 4 \end{array}\right] , \left[\begin{array}{c} 4 \\ 2 \\ -2 \\ 2 \\ -5 \end{array}\right] , \left[\begin{array}{c} 2 \\ -4 \\ 0 \\ -1 \\ -5 \end{array}\right] , \left[\begin{array}{c} 0 \\ -5 \\ -2 \\ 4 \\ -1 \end{array}\right] \right\} \)both spans \(\mathbb{R}^5\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ -5 \\ -1 \\ 0 \\ 4 \end{array}\right] , \left[\begin{array}{c} -1 \\ -2 \\ 0 \\ -3 \\ 4 \end{array}\right] , \left[\begin{array}{c} 4 \\ 2 \\ -2 \\ 2 \\ -5 \end{array}\right] , \left[\begin{array}{c} 2 \\ -4 \\ 0 \\ -1 \\ -5 \end{array}\right] , \left[\begin{array}{c} 0 \\ -5 \\ -2 \\ 4 \\ -1 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} 0 & 4 & 1 & 1 \\ 1 & 0 & -1 & -5 \\ 0 & 2 & -4 & -1 \\ -4 & -2 & -5 & 3 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 0 \\ 1 \\ 0 \\ -4 \end{array}\right] , \left[\begin{array}{c} 4 \\ 0 \\ 2 \\ -2 \end{array}\right] , \left[\begin{array}{c} 1 \\ -1 \\ -4 \\ -5 \end{array}\right] , \left[\begin{array}{c} 1 \\ -5 \\ -1 \\ 3 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 0 \\ 1 \\ 0 \\ -4 \end{array}\right] , \left[\begin{array}{c} 4 \\ 0 \\ 2 \\ -2 \end{array}\right] , \left[\begin{array}{c} 1 \\ -1 \\ -4 \\ -5 \end{array}\right] , \left[\begin{array}{c} 1 \\ -5 \\ -1 \\ 3 \end{array}\right] \right\} \)both spans \(\mathbb{R}^4\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 0 \\ 1 \\ 0 \\ -4 \end{array}\right] , \left[\begin{array}{c} 4 \\ 0 \\ 2 \\ -2 \end{array}\right] , \left[\begin{array}{c} 1 \\ -1 \\ -4 \\ -5 \end{array}\right] , \left[\begin{array}{c} 1 \\ -5 \\ -1 \\ 3 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} -3 & 1 & -4 & -2 \\ 4 & 4 & 3 & 1 \\ -3 & -4 & 3 & -2 \\ -3 & -1 & -4 & 1 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -3 \\ 4 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 1 \\ 4 \\ -4 \\ -1 \end{array}\right] , \left[\begin{array}{c} -4 \\ 3 \\ 3 \\ -4 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ -2 \\ 1 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -3 \\ 4 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 1 \\ 4 \\ -4 \\ -1 \end{array}\right] , \left[\begin{array}{c} -4 \\ 3 \\ 3 \\ -4 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ -2 \\ 1 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^4\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -3 \\ 4 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 1 \\ 4 \\ -4 \\ -1 \end{array}\right] , \left[\begin{array}{c} -4 \\ 3 \\ 3 \\ -4 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ -2 \\ 1 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} -1 & 0 & -4 & -7 \\ -3 & 2 & 2 & 3 \\ -5 & 4 & -1 & -5 \\ 0 & 0 & -2 & -4 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -1 \\ -3 \\ -5 \\ 0 \end{array}\right] , \left[\begin{array}{c} 0 \\ 2 \\ 4 \\ 0 \end{array}\right] , \left[\begin{array}{c} -4 \\ 2 \\ -1 \\ -2 \end{array}\right] , \left[\begin{array}{c} -7 \\ 3 \\ -5 \\ -4 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -1 \\ -3 \\ -5 \\ 0 \end{array}\right] , \left[\begin{array}{c} 0 \\ 2 \\ 4 \\ 0 \end{array}\right] , \left[\begin{array}{c} -4 \\ 2 \\ -1 \\ -2 \end{array}\right] , \left[\begin{array}{c} -7 \\ 3 \\ -5 \\ -4 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^4\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -1 \\ -3 \\ -5 \\ 0 \end{array}\right] , \left[\begin{array}{c} 0 \\ 2 \\ 4 \\ 0 \end{array}\right] , \left[\begin{array}{c} -4 \\ 2 \\ -1 \\ -2 \end{array}\right] , \left[\begin{array}{c} -7 \\ 3 \\ -5 \\ -4 \end{array}\right] \right\} \) is not a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} -2 & 4 & 2 & -2 \\ -2 & -4 & -5 & -5 \\ 2 & 4 & -5 & 1 \\ -3 & -2 & -5 & -4 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -2 \\ -2 \\ 2 \\ -3 \end{array}\right] , \left[\begin{array}{c} 4 \\ -4 \\ 4 \\ -2 \end{array}\right] , \left[\begin{array}{c} 2 \\ -5 \\ -5 \\ -5 \end{array}\right] , \left[\begin{array}{c} -2 \\ -5 \\ 1 \\ -4 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -2 \\ -2 \\ 2 \\ -3 \end{array}\right] , \left[\begin{array}{c} 4 \\ -4 \\ 4 \\ -2 \end{array}\right] , \left[\begin{array}{c} 2 \\ -5 \\ -5 \\ -5 \end{array}\right] , \left[\begin{array}{c} -2 \\ -5 \\ 1 \\ -4 \end{array}\right] \right\} \)both spans \(\mathbb{R}^4\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -2 \\ -2 \\ 2 \\ -3 \end{array}\right] , \left[\begin{array}{c} 4 \\ -4 \\ 4 \\ -2 \end{array}\right] , \left[\begin{array}{c} 2 \\ -5 \\ -5 \\ -5 \end{array}\right] , \left[\begin{array}{c} -2 \\ -5 \\ 1 \\ -4 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} 0 & 1 & -3 & 1 \\ 4 & 2 & 0 & 3 \\ -2 & 0 & 1 & 2 \\ -2 & -3 & 4 & 1 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 0 \\ 4 \\ -2 \\ -2 \end{array}\right] , \left[\begin{array}{c} 1 \\ 2 \\ 0 \\ -3 \end{array}\right] , \left[\begin{array}{c} -3 \\ 0 \\ 1 \\ 4 \end{array}\right] , \left[\begin{array}{c} 1 \\ 3 \\ 2 \\ 1 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 0 \\ 4 \\ -2 \\ -2 \end{array}\right] , \left[\begin{array}{c} 1 \\ 2 \\ 0 \\ -3 \end{array}\right] , \left[\begin{array}{c} -3 \\ 0 \\ 1 \\ 4 \end{array}\right] , \left[\begin{array}{c} 1 \\ 3 \\ 2 \\ 1 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^4\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 0 \\ 4 \\ -2 \\ -2 \end{array}\right] , \left[\begin{array}{c} 1 \\ 2 \\ 0 \\ -3 \end{array}\right] , \left[\begin{array}{c} -3 \\ 0 \\ 1 \\ 4 \end{array}\right] , \left[\begin{array}{c} 1 \\ 3 \\ 2 \\ 1 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} 0 & 0 & -3 & 4 \\ 0 & 2 & -2 & 3 \\ 4 & -1 & 0 & 3 \\ 2 & 3 & 1 & 2 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 0 \\ 0 \\ 4 \\ 2 \end{array}\right] , \left[\begin{array}{c} 0 \\ 2 \\ -1 \\ 3 \end{array}\right] , \left[\begin{array}{c} -3 \\ -2 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} 4 \\ 3 \\ 3 \\ 2 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 0 \\ 0 \\ 4 \\ 2 \end{array}\right] , \left[\begin{array}{c} 0 \\ 2 \\ -1 \\ 3 \end{array}\right] , \left[\begin{array}{c} -3 \\ -2 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} 4 \\ 3 \\ 3 \\ 2 \end{array}\right] \right\} \)both spans \(\mathbb{R}^4\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 0 \\ 0 \\ 4 \\ 2 \end{array}\right] , \left[\begin{array}{c} 0 \\ 2 \\ -1 \\ 3 \end{array}\right] , \left[\begin{array}{c} -3 \\ -2 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} 4 \\ 3 \\ 3 \\ 2 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} -1 & -1 & -5 & -2 & 4 \\ -1 & 3 & -5 & -3 & -5 \\ 2 & 4 & -2 & -2 & 3 \\ 3 & 2 & 3 & -4 & 1 \\ 3 & -1 & -5 & 1 & 3 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -1 \\ -1 \\ 2 \\ 3 \\ 3 \end{array}\right] , \left[\begin{array}{c} -1 \\ 3 \\ 4 \\ 2 \\ -1 \end{array}\right] , \left[\begin{array}{c} -5 \\ -5 \\ -2 \\ 3 \\ -5 \end{array}\right] , \left[\begin{array}{c} -2 \\ -3 \\ -2 \\ -4 \\ 1 \end{array}\right] , \left[\begin{array}{c} 4 \\ -5 \\ 3 \\ 1 \\ 3 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -1 \\ -1 \\ 2 \\ 3 \\ 3 \end{array}\right] , \left[\begin{array}{c} -1 \\ 3 \\ 4 \\ 2 \\ -1 \end{array}\right] , \left[\begin{array}{c} -5 \\ -5 \\ -2 \\ 3 \\ -5 \end{array}\right] , \left[\begin{array}{c} -2 \\ -3 \\ -2 \\ -4 \\ 1 \end{array}\right] , \left[\begin{array}{c} 4 \\ -5 \\ 3 \\ 1 \\ 3 \end{array}\right] \right\} \)both spans \(\mathbb{R}^5\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -1 \\ -1 \\ 2 \\ 3 \\ 3 \end{array}\right] , \left[\begin{array}{c} -1 \\ 3 \\ 4 \\ 2 \\ -1 \end{array}\right] , \left[\begin{array}{c} -5 \\ -5 \\ -2 \\ 3 \\ -5 \end{array}\right] , \left[\begin{array}{c} -2 \\ -3 \\ -2 \\ -4 \\ 1 \end{array}\right] , \left[\begin{array}{c} 4 \\ -5 \\ 3 \\ 1 \\ 3 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} 1 & 3 & -1 & -1 \\ -2 & 0 & -5 & -5 \\ 2 & -1 & 0 & 3 \\ -2 & 1 & -1 & -1 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ -2 \\ 2 \\ -2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -1 \\ -5 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} -1 \\ -5 \\ 3 \\ -1 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ -2 \\ 2 \\ -2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -1 \\ -5 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} -1 \\ -5 \\ 3 \\ -1 \end{array}\right] \right\} \)both spans \(\mathbb{R}^4\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ -2 \\ 2 \\ -2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -1 \\ -5 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} -1 \\ -5 \\ 3 \\ -1 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} 2 & 3 & -5 & -5 & 0 \\ -4 & -5 & -3 & -4 & 39 \\ -5 & 0 & 0 & -1 & 18 \\ -2 & -3 & 0 & -5 & 30 \\ -3 & -1 & 0 & -3 & 21 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 0 & 0 & -3 \\ 0 & 1 & 0 & 0 & -3 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & -3 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 2 \\ -4 \\ -5 \\ -2 \\ -3 \end{array}\right] , \left[\begin{array}{c} 3 \\ -5 \\ 0 \\ -3 \\ -1 \end{array}\right] , \left[\begin{array}{c} -5 \\ -3 \\ 0 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} -5 \\ -4 \\ -1 \\ -5 \\ -3 \end{array}\right] , \left[\begin{array}{c} 0 \\ 39 \\ 18 \\ 30 \\ 21 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 2 \\ -4 \\ -5 \\ -2 \\ -3 \end{array}\right] , \left[\begin{array}{c} 3 \\ -5 \\ 0 \\ -3 \\ -1 \end{array}\right] , \left[\begin{array}{c} -5 \\ -3 \\ 0 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} -5 \\ -4 \\ -1 \\ -5 \\ -3 \end{array}\right] , \left[\begin{array}{c} 0 \\ 39 \\ 18 \\ 30 \\ 21 \end{array}\right] \right\} \)both spans \(\mathbb{R}^5\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 2 \\ -4 \\ -5 \\ -2 \\ -3 \end{array}\right] , \left[\begin{array}{c} 3 \\ -5 \\ 0 \\ -3 \\ -1 \end{array}\right] , \left[\begin{array}{c} -5 \\ -3 \\ 0 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} -5 \\ -4 \\ -1 \\ -5 \\ -3 \end{array}\right] , \left[\begin{array}{c} 0 \\ 39 \\ 18 \\ 30 \\ 21 \end{array}\right] \right\} \) is not a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} 3 & 0 & 2 & 0 \\ -1 & 0 & 4 & 2 \\ 4 & 1 & -3 & 4 \\ 0 & 4 & -4 & -1 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 3 \\ -1 \\ 4 \\ 0 \end{array}\right] , \left[\begin{array}{c} 0 \\ 0 \\ 1 \\ 4 \end{array}\right] , \left[\begin{array}{c} 2 \\ 4 \\ -3 \\ -4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 2 \\ 4 \\ -1 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 3 \\ -1 \\ 4 \\ 0 \end{array}\right] , \left[\begin{array}{c} 0 \\ 0 \\ 1 \\ 4 \end{array}\right] , \left[\begin{array}{c} 2 \\ 4 \\ -3 \\ -4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 2 \\ 4 \\ -1 \end{array}\right] \right\} \)both spans \(\mathbb{R}^4\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 3 \\ -1 \\ 4 \\ 0 \end{array}\right] , \left[\begin{array}{c} 0 \\ 0 \\ 1 \\ 4 \end{array}\right] , \left[\begin{array}{c} 2 \\ 4 \\ -3 \\ -4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 2 \\ 4 \\ -1 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} 3 & 0 & -3 & -12 & 12 \\ 3 & 1 & -2 & -11 & 8 \\ 3 & 4 & 1 & -8 & -4 \\ 2 & 2 & 0 & -6 & 0 \\ 3 & -1 & -4 & -13 & 16 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & -1 & -4 & 4 \\ 0 & 1 & 1 & 1 & -4 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 3 \\ 3 \\ 3 \\ 2 \\ 3 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 4 \\ 2 \\ -1 \end{array}\right] , \left[\begin{array}{c} -3 \\ -2 \\ 1 \\ 0 \\ -4 \end{array}\right] , \left[\begin{array}{c} -12 \\ -11 \\ -8 \\ -6 \\ -13 \end{array}\right] , \left[\begin{array}{c} 12 \\ 8 \\ -4 \\ 0 \\ 16 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 3 \\ 3 \\ 3 \\ 2 \\ 3 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 4 \\ 2 \\ -1 \end{array}\right] , \left[\begin{array}{c} -3 \\ -2 \\ 1 \\ 0 \\ -4 \end{array}\right] , \left[\begin{array}{c} -12 \\ -11 \\ -8 \\ -6 \\ -13 \end{array}\right] , \left[\begin{array}{c} 12 \\ 8 \\ -4 \\ 0 \\ 16 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^5\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 3 \\ 3 \\ 3 \\ 2 \\ 3 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 4 \\ 2 \\ -1 \end{array}\right] , \left[\begin{array}{c} -3 \\ -2 \\ 1 \\ 0 \\ -4 \end{array}\right] , \left[\begin{array}{c} -12 \\ -11 \\ -8 \\ -6 \\ -13 \end{array}\right] , \left[\begin{array}{c} 12 \\ 8 \\ -4 \\ 0 \\ 16 \end{array}\right] \right\} \) is not a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -2 & -5 & -2 \\ -3 & 1 & -5 \\ -3 & -4 & -4 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -2 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} -5 \\ 1 \\ -4 \end{array}\right] , \left[\begin{array}{c} -2 \\ -5 \\ -4 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^3\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -2 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} -5 \\ 1 \\ -4 \end{array}\right] , \left[\begin{array}{c} -2 \\ -5 \\ -4 \end{array}\right] \right\} \)both spans \(\mathbb{R}^3\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -2 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} -5 \\ 1 \\ -4 \end{array}\right] , \left[\begin{array}{c} -2 \\ -5 \\ -4 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^3\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -3 & 2 & -6 \\ 3 & -2 & 6 \\ 0 & 1 & -3 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & -3 \\ 0 & 0 & 0 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -3 \\ 3 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} -6 \\ 6 \\ -3 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^3\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -3 \\ 3 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} -6 \\ 6 \\ -3 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^3\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -3 \\ 3 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} -6 \\ 6 \\ -3 \end{array}\right] \right\} \) is not a basis of \(\mathbb{R}^3\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} 4 & -1 & -1 & -1 \\ -5 & 2 & 3 & -4 \\ -3 & -5 & 2 & -1 \\ -1 & 2 & -5 & 3 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 4 \\ -5 \\ -3 \\ -1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 2 \\ -5 \\ 2 \end{array}\right] , \left[\begin{array}{c} -1 \\ 3 \\ 2 \\ -5 \end{array}\right] , \left[\begin{array}{c} -1 \\ -4 \\ -1 \\ 3 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 4 \\ -5 \\ -3 \\ -1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 2 \\ -5 \\ 2 \end{array}\right] , \left[\begin{array}{c} -1 \\ 3 \\ 2 \\ -5 \end{array}\right] , \left[\begin{array}{c} -1 \\ -4 \\ -1 \\ 3 \end{array}\right] \right\} \)both spans \(\mathbb{R}^4\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 4 \\ -5 \\ -3 \\ -1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 2 \\ -5 \\ 2 \end{array}\right] , \left[\begin{array}{c} -1 \\ 3 \\ 2 \\ -5 \end{array}\right] , \left[\begin{array}{c} -1 \\ -4 \\ -1 \\ 3 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} -5 & -1 & -4 & -3 \\ -1 & -2 & -4 & 3 \\ -4 & -2 & 4 & 4 \\ -4 & -4 & -4 & -3 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -5 \\ -1 \\ -4 \\ -4 \end{array}\right] , \left[\begin{array}{c} -1 \\ -2 \\ -2 \\ -4 \end{array}\right] , \left[\begin{array}{c} -4 \\ -4 \\ 4 \\ -4 \end{array}\right] , \left[\begin{array}{c} -3 \\ 3 \\ 4 \\ -3 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -5 \\ -1 \\ -4 \\ -4 \end{array}\right] , \left[\begin{array}{c} -1 \\ -2 \\ -2 \\ -4 \end{array}\right] , \left[\begin{array}{c} -4 \\ -4 \\ 4 \\ -4 \end{array}\right] , \left[\begin{array}{c} -3 \\ 3 \\ 4 \\ -3 \end{array}\right] \right\} \)both spans \(\mathbb{R}^4\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -5 \\ -1 \\ -4 \\ -4 \end{array}\right] , \left[\begin{array}{c} -1 \\ -2 \\ -2 \\ -4 \end{array}\right] , \left[\begin{array}{c} -4 \\ -4 \\ 4 \\ -4 \end{array}\right] , \left[\begin{array}{c} -3 \\ 3 \\ 4 \\ -3 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} 0 & 0 & -2 & -5 \\ 4 & -4 & 3 & -1 \\ 4 & -2 & 2 & -5 \\ -3 & 3 & 4 & -1 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 0 \\ 4 \\ 4 \\ -3 \end{array}\right] , \left[\begin{array}{c} 0 \\ -4 \\ -2 \\ 3 \end{array}\right] , \left[\begin{array}{c} -2 \\ 3 \\ 2 \\ 4 \end{array}\right] , \left[\begin{array}{c} -5 \\ -1 \\ -5 \\ -1 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 0 \\ 4 \\ 4 \\ -3 \end{array}\right] , \left[\begin{array}{c} 0 \\ -4 \\ -2 \\ 3 \end{array}\right] , \left[\begin{array}{c} -2 \\ 3 \\ 2 \\ 4 \end{array}\right] , \left[\begin{array}{c} -5 \\ -1 \\ -5 \\ -1 \end{array}\right] \right\} \)both spans \(\mathbb{R}^4\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 0 \\ 4 \\ 4 \\ -3 \end{array}\right] , \left[\begin{array}{c} 0 \\ -4 \\ -2 \\ 3 \end{array}\right] , \left[\begin{array}{c} -2 \\ 3 \\ 2 \\ 4 \end{array}\right] , \left[\begin{array}{c} -5 \\ -1 \\ -5 \\ -1 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -3 & 2 & -4 \\ -4 & -1 & 2 \\ 0 & 2 & -4 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -3 \\ -4 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ -1 \\ 2 \end{array}\right] , \left[\begin{array}{c} -4 \\ 2 \\ -4 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^3\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -3 \\ -4 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ -1 \\ 2 \end{array}\right] , \left[\begin{array}{c} -4 \\ 2 \\ -4 \end{array}\right] \right\} \)both spans \(\mathbb{R}^3\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -3 \\ -4 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ -1 \\ 2 \end{array}\right] , \left[\begin{array}{c} -4 \\ 2 \\ -4 \end{array}\right] \right\} \) is not a basis of \(\mathbb{R}^3\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} -4 & -1 & -3 & 1 & 0 \\ -4 & 3 & -2 & -2 & -3 \\ -2 & -1 & -3 & -3 & -4 \\ -2 & 2 & 2 & -3 & 2 \\ -2 & 3 & 1 & 2 & -3 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ -4 \\ -2 \\ -2 \\ -2 \end{array}\right] , \left[\begin{array}{c} -1 \\ 3 \\ -1 \\ 2 \\ 3 \end{array}\right] , \left[\begin{array}{c} -3 \\ -2 \\ -3 \\ 2 \\ 1 \end{array}\right] , \left[\begin{array}{c} 1 \\ -2 \\ -3 \\ -3 \\ 2 \end{array}\right] , \left[\begin{array}{c} 0 \\ -3 \\ -4 \\ 2 \\ -3 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ -4 \\ -2 \\ -2 \\ -2 \end{array}\right] , \left[\begin{array}{c} -1 \\ 3 \\ -1 \\ 2 \\ 3 \end{array}\right] , \left[\begin{array}{c} -3 \\ -2 \\ -3 \\ 2 \\ 1 \end{array}\right] , \left[\begin{array}{c} 1 \\ -2 \\ -3 \\ -3 \\ 2 \end{array}\right] , \left[\begin{array}{c} 0 \\ -3 \\ -4 \\ 2 \\ -3 \end{array}\right] \right\} \)both spans \(\mathbb{R}^5\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ -4 \\ -2 \\ -2 \\ -2 \end{array}\right] , \left[\begin{array}{c} -1 \\ 3 \\ -1 \\ 2 \\ 3 \end{array}\right] , \left[\begin{array}{c} -3 \\ -2 \\ -3 \\ 2 \\ 1 \end{array}\right] , \left[\begin{array}{c} 1 \\ -2 \\ -3 \\ -3 \\ 2 \end{array}\right] , \left[\begin{array}{c} 0 \\ -3 \\ -4 \\ 2 \\ -3 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} 0 & 2 & -2 & 1 \\ 0 & -2 & 1 & -5 \\ -1 & -5 & -4 & -3 \\ 0 & 3 & -4 & -4 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 0 \\ 0 \\ -1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ -2 \\ -5 \\ 3 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ -4 \\ -4 \end{array}\right] , \left[\begin{array}{c} 1 \\ -5 \\ -3 \\ -4 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 0 \\ 0 \\ -1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ -2 \\ -5 \\ 3 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ -4 \\ -4 \end{array}\right] , \left[\begin{array}{c} 1 \\ -5 \\ -3 \\ -4 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^4\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 0 \\ 0 \\ -1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ -2 \\ -5 \\ 3 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ -4 \\ -4 \end{array}\right] , \left[\begin{array}{c} 1 \\ -5 \\ -3 \\ -4 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} 2 & -4 & -2 & -1 & 4 \\ 3 & -1 & 0 & -2 & -1 \\ 1 & -1 & -5 & 4 & 1 \\ -1 & 0 & 4 & 0 & -4 \\ 4 & -2 & -3 & 0 & -2 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 2 \\ 3 \\ 1 \\ -1 \\ 4 \end{array}\right] , \left[\begin{array}{c} -4 \\ -1 \\ -1 \\ 0 \\ -2 \end{array}\right] , \left[\begin{array}{c} -2 \\ 0 \\ -5 \\ 4 \\ -3 \end{array}\right] , \left[\begin{array}{c} -1 \\ -2 \\ 4 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 4 \\ -1 \\ 1 \\ -4 \\ -2 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 2 \\ 3 \\ 1 \\ -1 \\ 4 \end{array}\right] , \left[\begin{array}{c} -4 \\ -1 \\ -1 \\ 0 \\ -2 \end{array}\right] , \left[\begin{array}{c} -2 \\ 0 \\ -5 \\ 4 \\ -3 \end{array}\right] , \left[\begin{array}{c} -1 \\ -2 \\ 4 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 4 \\ -1 \\ 1 \\ -4 \\ -2 \end{array}\right] \right\} \)both spans \(\mathbb{R}^5\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 2 \\ 3 \\ 1 \\ -1 \\ 4 \end{array}\right] , \left[\begin{array}{c} -4 \\ -1 \\ -1 \\ 0 \\ -2 \end{array}\right] , \left[\begin{array}{c} -2 \\ 0 \\ -5 \\ 4 \\ -3 \end{array}\right] , \left[\begin{array}{c} -1 \\ -2 \\ 4 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 4 \\ -1 \\ 1 \\ -4 \\ -2 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 4 & -2 & 2 \\ 2 & -4 & -5 \\ -2 & 1 & 4 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 4 \\ 2 \\ -2 \end{array}\right] , \left[\begin{array}{c} -2 \\ -4 \\ 1 \end{array}\right] , \left[\begin{array}{c} 2 \\ -5 \\ 4 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^3\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 4 \\ 2 \\ -2 \end{array}\right] , \left[\begin{array}{c} -2 \\ -4 \\ 1 \end{array}\right] , \left[\begin{array}{c} 2 \\ -5 \\ 4 \end{array}\right] \right\} \)both spans \(\mathbb{R}^3\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 4 \\ 2 \\ -2 \end{array}\right] , \left[\begin{array}{c} -2 \\ -4 \\ 1 \end{array}\right] , \left[\begin{array}{c} 2 \\ -5 \\ 4 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^3\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} -5 & 0 & 2 & 2 & -4 \\ 1 & 0 & -1 & -5 & 2 \\ 1 & -5 & 3 & 0 & 3 \\ 4 & -4 & 0 & -1 & -4 \\ -5 & -4 & -2 & 3 & 0 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -5 \\ 1 \\ 1 \\ 4 \\ -5 \end{array}\right] , \left[\begin{array}{c} 0 \\ 0 \\ -5 \\ -4 \\ -4 \end{array}\right] , \left[\begin{array}{c} 2 \\ -1 \\ 3 \\ 0 \\ -2 \end{array}\right] , \left[\begin{array}{c} 2 \\ -5 \\ 0 \\ -1 \\ 3 \end{array}\right] , \left[\begin{array}{c} -4 \\ 2 \\ 3 \\ -4 \\ 0 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -5 \\ 1 \\ 1 \\ 4 \\ -5 \end{array}\right] , \left[\begin{array}{c} 0 \\ 0 \\ -5 \\ -4 \\ -4 \end{array}\right] , \left[\begin{array}{c} 2 \\ -1 \\ 3 \\ 0 \\ -2 \end{array}\right] , \left[\begin{array}{c} 2 \\ -5 \\ 0 \\ -1 \\ 3 \end{array}\right] , \left[\begin{array}{c} -4 \\ 2 \\ 3 \\ -4 \\ 0 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^5\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -5 \\ 1 \\ 1 \\ 4 \\ -5 \end{array}\right] , \left[\begin{array}{c} 0 \\ 0 \\ -5 \\ -4 \\ -4 \end{array}\right] , \left[\begin{array}{c} 2 \\ -1 \\ 3 \\ 0 \\ -2 \end{array}\right] , \left[\begin{array}{c} 2 \\ -5 \\ 0 \\ -1 \\ 3 \end{array}\right] , \left[\begin{array}{c} -4 \\ 2 \\ 3 \\ -4 \\ 0 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} 1 & 2 & 4 & 1 & 1 \\ 1 & 2 & -2 & -5 & -5 \\ 1 & -5 & -4 & -5 & -2 \\ -1 & -5 & -4 & -3 & -4 \\ -5 & 4 & -1 & -3 & -5 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 1 \\ -1 \\ -5 \end{array}\right] , \left[\begin{array}{c} 2 \\ 2 \\ -5 \\ -5 \\ 4 \end{array}\right] , \left[\begin{array}{c} 4 \\ -2 \\ -4 \\ -4 \\ -1 \end{array}\right] , \left[\begin{array}{c} 1 \\ -5 \\ -5 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 1 \\ -5 \\ -2 \\ -4 \\ -5 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 1 \\ -1 \\ -5 \end{array}\right] , \left[\begin{array}{c} 2 \\ 2 \\ -5 \\ -5 \\ 4 \end{array}\right] , \left[\begin{array}{c} 4 \\ -2 \\ -4 \\ -4 \\ -1 \end{array}\right] , \left[\begin{array}{c} 1 \\ -5 \\ -5 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 1 \\ -5 \\ -2 \\ -4 \\ -5 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^5\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 1 \\ -1 \\ -5 \end{array}\right] , \left[\begin{array}{c} 2 \\ 2 \\ -5 \\ -5 \\ 4 \end{array}\right] , \left[\begin{array}{c} 4 \\ -2 \\ -4 \\ -4 \\ -1 \end{array}\right] , \left[\begin{array}{c} 1 \\ -5 \\ -5 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 1 \\ -5 \\ -2 \\ -4 \\ -5 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} -1 & 3 & -2 & -4 & -7 \\ -2 & -4 & 6 & 0 & 6 \\ 3 & 0 & -3 & -4 & 3 \\ -4 & 1 & 3 & -1 & -6 \\ -5 & -2 & 7 & 3 & -1 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & -1 & 0 & 1 \\ 0 & 1 & -1 & 0 & -2 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -1 \\ -2 \\ 3 \\ -4 \\ -5 \end{array}\right] , \left[\begin{array}{c} 3 \\ -4 \\ 0 \\ 1 \\ -2 \end{array}\right] , \left[\begin{array}{c} -2 \\ 6 \\ -3 \\ 3 \\ 7 \end{array}\right] , \left[\begin{array}{c} -4 \\ 0 \\ -4 \\ -1 \\ 3 \end{array}\right] , \left[\begin{array}{c} -7 \\ 6 \\ 3 \\ -6 \\ -1 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -1 \\ -2 \\ 3 \\ -4 \\ -5 \end{array}\right] , \left[\begin{array}{c} 3 \\ -4 \\ 0 \\ 1 \\ -2 \end{array}\right] , \left[\begin{array}{c} -2 \\ 6 \\ -3 \\ 3 \\ 7 \end{array}\right] , \left[\begin{array}{c} -4 \\ 0 \\ -4 \\ -1 \\ 3 \end{array}\right] , \left[\begin{array}{c} -7 \\ 6 \\ 3 \\ -6 \\ -1 \end{array}\right] \right\} \)both spans \(\mathbb{R}^5\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -1 \\ -2 \\ 3 \\ -4 \\ -5 \end{array}\right] , \left[\begin{array}{c} 3 \\ -4 \\ 0 \\ 1 \\ -2 \end{array}\right] , \left[\begin{array}{c} -2 \\ 6 \\ -3 \\ 3 \\ 7 \end{array}\right] , \left[\begin{array}{c} -4 \\ 0 \\ -4 \\ -1 \\ 3 \end{array}\right] , \left[\begin{array}{c} -7 \\ 6 \\ 3 \\ -6 \\ -1 \end{array}\right] \right\} \) is not a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} -4 & 4 & -3 & -3 & 2 \\ -2 & 2 & 3 & -4 & 4 \\ -2 & 3 & 1 & 0 & 4 \\ -3 & 1 & 1 & 1 & 3 \\ -3 & -5 & 4 & 1 & 1 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ -2 \\ -2 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 4 \\ 2 \\ 3 \\ 1 \\ -5 \end{array}\right] , \left[\begin{array}{c} -3 \\ 3 \\ 1 \\ 1 \\ 4 \end{array}\right] , \left[\begin{array}{c} -3 \\ -4 \\ 0 \\ 1 \\ 1 \end{array}\right] , \left[\begin{array}{c} 2 \\ 4 \\ 4 \\ 3 \\ 1 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ -2 \\ -2 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 4 \\ 2 \\ 3 \\ 1 \\ -5 \end{array}\right] , \left[\begin{array}{c} -3 \\ 3 \\ 1 \\ 1 \\ 4 \end{array}\right] , \left[\begin{array}{c} -3 \\ -4 \\ 0 \\ 1 \\ 1 \end{array}\right] , \left[\begin{array}{c} 2 \\ 4 \\ 4 \\ 3 \\ 1 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^5\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ -2 \\ -2 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 4 \\ 2 \\ 3 \\ 1 \\ -5 \end{array}\right] , \left[\begin{array}{c} -3 \\ 3 \\ 1 \\ 1 \\ 4 \end{array}\right] , \left[\begin{array}{c} -3 \\ -4 \\ 0 \\ 1 \\ 1 \end{array}\right] , \left[\begin{array}{c} 2 \\ 4 \\ 4 \\ 3 \\ 1 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} -2 & 1 & -2 & -4 \\ 3 & -4 & -2 & -5 \\ -2 & 4 & -3 & 3 \\ 1 & -3 & -4 & 1 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -2 \\ 3 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} 1 \\ -4 \\ 4 \\ -3 \end{array}\right] , \left[\begin{array}{c} -2 \\ -2 \\ -3 \\ -4 \end{array}\right] , \left[\begin{array}{c} -4 \\ -5 \\ 3 \\ 1 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -2 \\ 3 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} 1 \\ -4 \\ 4 \\ -3 \end{array}\right] , \left[\begin{array}{c} -2 \\ -2 \\ -3 \\ -4 \end{array}\right] , \left[\begin{array}{c} -4 \\ -5 \\ 3 \\ 1 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^4\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -2 \\ 3 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} 1 \\ -4 \\ 4 \\ -3 \end{array}\right] , \left[\begin{array}{c} -2 \\ -2 \\ -3 \\ -4 \end{array}\right] , \left[\begin{array}{c} -4 \\ -5 \\ 3 \\ 1 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} -5 & -1 & 3 & -15 & -2 \\ 4 & -3 & 1 & 9 & -3 \\ -1 & 4 & 2 & -10 & 1 \\ -4 & -3 & -4 & 3 & 0 \\ -4 & 2 & -2 & -6 & 3 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 0 & 2 & 0 \\ 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -5 \\ 4 \\ -1 \\ -4 \\ -4 \end{array}\right] , \left[\begin{array}{c} -1 \\ -3 \\ 4 \\ -3 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 1 \\ 2 \\ -4 \\ -2 \end{array}\right] , \left[\begin{array}{c} -15 \\ 9 \\ -10 \\ 3 \\ -6 \end{array}\right] , \left[\begin{array}{c} -2 \\ -3 \\ 1 \\ 0 \\ 3 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -5 \\ 4 \\ -1 \\ -4 \\ -4 \end{array}\right] , \left[\begin{array}{c} -1 \\ -3 \\ 4 \\ -3 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 1 \\ 2 \\ -4 \\ -2 \end{array}\right] , \left[\begin{array}{c} -15 \\ 9 \\ -10 \\ 3 \\ -6 \end{array}\right] , \left[\begin{array}{c} -2 \\ -3 \\ 1 \\ 0 \\ 3 \end{array}\right] \right\} \)both spans \(\mathbb{R}^5\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -5 \\ 4 \\ -1 \\ -4 \\ -4 \end{array}\right] , \left[\begin{array}{c} -1 \\ -3 \\ 4 \\ -3 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 1 \\ 2 \\ -4 \\ -2 \end{array}\right] , \left[\begin{array}{c} -15 \\ 9 \\ -10 \\ 3 \\ -6 \end{array}\right] , \left[\begin{array}{c} -2 \\ -3 \\ 1 \\ 0 \\ 3 \end{array}\right] \right\} \) is not a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} 2 & 3 & 1 & -2 & -4 \\ -2 & -3 & -3 & 2 & -2 \\ 4 & -2 & 3 & -2 & 3 \\ -3 & 1 & -1 & -1 & -2 \\ -3 & 2 & 2 & 2 & 0 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 2 \\ -2 \\ 4 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 3 \\ -3 \\ -2 \\ 1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 1 \\ -3 \\ 3 \\ -1 \\ 2 \end{array}\right] , \left[\begin{array}{c} -2 \\ 2 \\ -2 \\ -1 \\ 2 \end{array}\right] , \left[\begin{array}{c} -4 \\ -2 \\ 3 \\ -2 \\ 0 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 2 \\ -2 \\ 4 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 3 \\ -3 \\ -2 \\ 1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 1 \\ -3 \\ 3 \\ -1 \\ 2 \end{array}\right] , \left[\begin{array}{c} -2 \\ 2 \\ -2 \\ -1 \\ 2 \end{array}\right] , \left[\begin{array}{c} -4 \\ -2 \\ 3 \\ -2 \\ 0 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^5\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 2 \\ -2 \\ 4 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 3 \\ -3 \\ -2 \\ 1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 1 \\ -3 \\ 3 \\ -1 \\ 2 \end{array}\right] , \left[\begin{array}{c} -2 \\ 2 \\ -2 \\ -1 \\ 2 \end{array}\right] , \left[\begin{array}{c} -4 \\ -2 \\ 3 \\ -2 \\ 0 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} -5 & 1 & -5 & -1 & 4 \\ -5 & 4 & 4 & -1 & -3 \\ 3 & 2 & -2 & -2 & -4 \\ -4 & -5 & 2 & -5 & -5 \\ -5 & -1 & -2 & 2 & -3 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -5 \\ -5 \\ 3 \\ -4 \\ -5 \end{array}\right] , \left[\begin{array}{c} 1 \\ 4 \\ 2 \\ -5 \\ -1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 4 \\ -2 \\ 2 \\ -2 \end{array}\right] , \left[\begin{array}{c} -1 \\ -1 \\ -2 \\ -5 \\ 2 \end{array}\right] , \left[\begin{array}{c} 4 \\ -3 \\ -4 \\ -5 \\ -3 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -5 \\ -5 \\ 3 \\ -4 \\ -5 \end{array}\right] , \left[\begin{array}{c} 1 \\ 4 \\ 2 \\ -5 \\ -1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 4 \\ -2 \\ 2 \\ -2 \end{array}\right] , \left[\begin{array}{c} -1 \\ -1 \\ -2 \\ -5 \\ 2 \end{array}\right] , \left[\begin{array}{c} 4 \\ -3 \\ -4 \\ -5 \\ -3 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^5\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -5 \\ -5 \\ 3 \\ -4 \\ -5 \end{array}\right] , \left[\begin{array}{c} 1 \\ 4 \\ 2 \\ -5 \\ -1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 4 \\ -2 \\ 2 \\ -2 \end{array}\right] , \left[\begin{array}{c} -1 \\ -1 \\ -2 \\ -5 \\ 2 \end{array}\right] , \left[\begin{array}{c} 4 \\ -3 \\ -4 \\ -5 \\ -3 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 4 & 3 & -1 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ 4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 0 \\ 3 \end{array}\right] , \left[\begin{array}{c} 0 \\ -1 \\ -1 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^3\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ 4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 0 \\ 3 \end{array}\right] , \left[\begin{array}{c} 0 \\ -1 \\ -1 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^3\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ 4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 0 \\ 3 \end{array}\right] , \left[\begin{array}{c} 0 \\ -1 \\ -1 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^3\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} 1 & -2 & 0 & 3 & -1 \\ 2 & 1 & -5 & -3 & -5 \\ -2 & 3 & -4 & 0 & -3 \\ -3 & -4 & -5 & -3 & 4 \\ -3 & 1 & -1 & 3 & -1 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ 2 \\ -2 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ 3 \\ -4 \\ 1 \end{array}\right] , \left[\begin{array}{c} 0 \\ -5 \\ -4 \\ -5 \\ -1 \end{array}\right] , \left[\begin{array}{c} 3 \\ -3 \\ 0 \\ -3 \\ 3 \end{array}\right] , \left[\begin{array}{c} -1 \\ -5 \\ -3 \\ 4 \\ -1 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ 2 \\ -2 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ 3 \\ -4 \\ 1 \end{array}\right] , \left[\begin{array}{c} 0 \\ -5 \\ -4 \\ -5 \\ -1 \end{array}\right] , \left[\begin{array}{c} 3 \\ -3 \\ 0 \\ -3 \\ 3 \end{array}\right] , \left[\begin{array}{c} -1 \\ -5 \\ -3 \\ 4 \\ -1 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^5\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ 2 \\ -2 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ 3 \\ -4 \\ 1 \end{array}\right] , \left[\begin{array}{c} 0 \\ -5 \\ -4 \\ -5 \\ -1 \end{array}\right] , \left[\begin{array}{c} 3 \\ -3 \\ 0 \\ -3 \\ 3 \end{array}\right] , \left[\begin{array}{c} -1 \\ -5 \\ -3 \\ 4 \\ -1 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} 4 & 2 & 4 & -2 \\ 2 & 1 & -5 & -5 \\ -5 & -2 & 0 & -4 \\ 0 & 0 & 2 & -3 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 4 \\ 2 \\ -5 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ 1 \\ -2 \\ 0 \end{array}\right] , \left[\begin{array}{c} 4 \\ -5 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} -2 \\ -5 \\ -4 \\ -3 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 4 \\ 2 \\ -5 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ 1 \\ -2 \\ 0 \end{array}\right] , \left[\begin{array}{c} 4 \\ -5 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} -2 \\ -5 \\ -4 \\ -3 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^4\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 4 \\ 2 \\ -5 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ 1 \\ -2 \\ 0 \end{array}\right] , \left[\begin{array}{c} 4 \\ -5 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} -2 \\ -5 \\ -4 \\ -3 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -2 & 1 & -3 \\ 0 & -3 & -3 \\ 2 & 3 & -1 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -2 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 1 \\ -3 \\ 3 \end{array}\right] , \left[\begin{array}{c} -3 \\ -3 \\ -1 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^3\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -2 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 1 \\ -3 \\ 3 \end{array}\right] , \left[\begin{array}{c} -3 \\ -3 \\ -1 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^3\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -2 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 1 \\ -3 \\ 3 \end{array}\right] , \left[\begin{array}{c} -3 \\ -3 \\ -1 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^3\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} 2 & 4 & -2 & 2 \\ 2 & 0 & 1 & 4 \\ -3 & 1 & 3 & -2 \\ 3 & 3 & -2 & 3 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 2 \\ 2 \\ -3 \\ 3 \end{array}\right] , \left[\begin{array}{c} 4 \\ 0 \\ 1 \\ 3 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ 3 \\ -2 \end{array}\right] , \left[\begin{array}{c} 2 \\ 4 \\ -2 \\ 3 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 2 \\ 2 \\ -3 \\ 3 \end{array}\right] , \left[\begin{array}{c} 4 \\ 0 \\ 1 \\ 3 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ 3 \\ -2 \end{array}\right] , \left[\begin{array}{c} 2 \\ 4 \\ -2 \\ 3 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^4\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 2 \\ 2 \\ -3 \\ 3 \end{array}\right] , \left[\begin{array}{c} 4 \\ 0 \\ 1 \\ 3 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ 3 \\ -2 \end{array}\right] , \left[\begin{array}{c} 2 \\ 4 \\ -2 \\ 3 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} -3 & -5 & 7 & -2 \\ 2 & 3 & -4 & 1 \\ 2 & 3 & -4 & 1 \\ -5 & 4 & -13 & 9 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 1 & -1 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -3 \\ 2 \\ 2 \\ -5 \end{array}\right] , \left[\begin{array}{c} -5 \\ 3 \\ 3 \\ 4 \end{array}\right] , \left[\begin{array}{c} 7 \\ -4 \\ -4 \\ -13 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ 1 \\ 9 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -3 \\ 2 \\ 2 \\ -5 \end{array}\right] , \left[\begin{array}{c} -5 \\ 3 \\ 3 \\ 4 \end{array}\right] , \left[\begin{array}{c} 7 \\ -4 \\ -4 \\ -13 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ 1 \\ 9 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^4\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -3 \\ 2 \\ 2 \\ -5 \end{array}\right] , \left[\begin{array}{c} -5 \\ 3 \\ 3 \\ 4 \end{array}\right] , \left[\begin{array}{c} 7 \\ -4 \\ -4 \\ -13 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ 1 \\ 9 \end{array}\right] \right\} \) is not a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} -1 & -3 & -1 & -4 & 1 \\ 3 & -4 & 4 & -4 & 0 \\ -1 & -3 & -5 & 0 & -5 \\ -4 & 3 & 2 & 4 & -5 \\ -5 & 2 & 1 & -1 & 4 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -1 \\ 3 \\ -1 \\ -4 \\ -5 \end{array}\right] , \left[\begin{array}{c} -3 \\ -4 \\ -3 \\ 3 \\ 2 \end{array}\right] , \left[\begin{array}{c} -1 \\ 4 \\ -5 \\ 2 \\ 1 \end{array}\right] , \left[\begin{array}{c} -4 \\ -4 \\ 0 \\ 4 \\ -1 \end{array}\right] , \left[\begin{array}{c} 1 \\ 0 \\ -5 \\ -5 \\ 4 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -1 \\ 3 \\ -1 \\ -4 \\ -5 \end{array}\right] , \left[\begin{array}{c} -3 \\ -4 \\ -3 \\ 3 \\ 2 \end{array}\right] , \left[\begin{array}{c} -1 \\ 4 \\ -5 \\ 2 \\ 1 \end{array}\right] , \left[\begin{array}{c} -4 \\ -4 \\ 0 \\ 4 \\ -1 \end{array}\right] , \left[\begin{array}{c} 1 \\ 0 \\ -5 \\ -5 \\ 4 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^5\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -1 \\ 3 \\ -1 \\ -4 \\ -5 \end{array}\right] , \left[\begin{array}{c} -3 \\ -4 \\ -3 \\ 3 \\ 2 \end{array}\right] , \left[\begin{array}{c} -1 \\ 4 \\ -5 \\ 2 \\ 1 \end{array}\right] , \left[\begin{array}{c} -4 \\ -4 \\ 0 \\ 4 \\ -1 \end{array}\right] , \left[\begin{array}{c} 1 \\ 0 \\ -5 \\ -5 \\ 4 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} -5 & 3 & -7 & -3 \\ 0 & 2 & 2 & -3 \\ 0 & -3 & -3 & -1 \\ -4 & 4 & -4 & -4 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -5 \\ 0 \\ 0 \\ -4 \end{array}\right] , \left[\begin{array}{c} 3 \\ 2 \\ -3 \\ 4 \end{array}\right] , \left[\begin{array}{c} -7 \\ 2 \\ -3 \\ -4 \end{array}\right] , \left[\begin{array}{c} -3 \\ -3 \\ -1 \\ -4 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -5 \\ 0 \\ 0 \\ -4 \end{array}\right] , \left[\begin{array}{c} 3 \\ 2 \\ -3 \\ 4 \end{array}\right] , \left[\begin{array}{c} -7 \\ 2 \\ -3 \\ -4 \end{array}\right] , \left[\begin{array}{c} -3 \\ -3 \\ -1 \\ -4 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^4\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -5 \\ 0 \\ 0 \\ -4 \end{array}\right] , \left[\begin{array}{c} 3 \\ 2 \\ -3 \\ 4 \end{array}\right] , \left[\begin{array}{c} -7 \\ 2 \\ -3 \\ -4 \end{array}\right] , \left[\begin{array}{c} -3 \\ -3 \\ -1 \\ -4 \end{array}\right] \right\} \) is not a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 3 & -4 & 6 \\ 1 & 4 & -14 \\ 0 & -3 & 9 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & -3 \\ 0 & 0 & 0 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 3 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -4 \\ 4 \\ -3 \end{array}\right] , \left[\begin{array}{c} 6 \\ -14 \\ 9 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^3\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 3 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -4 \\ 4 \\ -3 \end{array}\right] , \left[\begin{array}{c} 6 \\ -14 \\ 9 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^3\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 3 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -4 \\ 4 \\ -3 \end{array}\right] , \left[\begin{array}{c} 6 \\ -14 \\ 9 \end{array}\right] \right\} \) is not a basis of \(\mathbb{R}^3\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} -5 & 3 & 11 & -41 & 83 \\ -2 & -1 & 0 & -1 & -2 \\ -3 & 4 & 11 & -40 & 85 \\ -3 & -2 & -1 & 2 & -11 \\ -3 & 1 & 5 & -19 & 37 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & -1 & 4 & -7 \\ 0 & 1 & 2 & -7 & 16 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -5 \\ -2 \\ -3 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 3 \\ -1 \\ 4 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} 11 \\ 0 \\ 11 \\ -1 \\ 5 \end{array}\right] , \left[\begin{array}{c} -41 \\ -1 \\ -40 \\ 2 \\ -19 \end{array}\right] , \left[\begin{array}{c} 83 \\ -2 \\ 85 \\ -11 \\ 37 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -5 \\ -2 \\ -3 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 3 \\ -1 \\ 4 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} 11 \\ 0 \\ 11 \\ -1 \\ 5 \end{array}\right] , \left[\begin{array}{c} -41 \\ -1 \\ -40 \\ 2 \\ -19 \end{array}\right] , \left[\begin{array}{c} 83 \\ -2 \\ 85 \\ -11 \\ 37 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^5\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -5 \\ -2 \\ -3 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 3 \\ -1 \\ 4 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} 11 \\ 0 \\ 11 \\ -1 \\ 5 \end{array}\right] , \left[\begin{array}{c} -41 \\ -1 \\ -40 \\ 2 \\ -19 \end{array}\right] , \left[\begin{array}{c} 83 \\ -2 \\ 85 \\ -11 \\ 37 \end{array}\right] \right\} \) is not a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} 2 & -1 & 5 & 6 & -2 \\ 0 & -2 & 2 & 4 & 0 \\ 3 & -3 & 9 & 12 & -3 \\ -2 & -3 & -1 & 2 & 2 \\ -2 & -3 & -1 & 2 & 2 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 2 & 2 & -1 \\ 0 & 1 & -1 & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 2 \\ 0 \\ 3 \\ -2 \\ -2 \end{array}\right] , \left[\begin{array}{c} -1 \\ -2 \\ -3 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 5 \\ 2 \\ 9 \\ -1 \\ -1 \end{array}\right] , \left[\begin{array}{c} 6 \\ 4 \\ 12 \\ 2 \\ 2 \end{array}\right] , \left[\begin{array}{c} -2 \\ 0 \\ -3 \\ 2 \\ 2 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 2 \\ 0 \\ 3 \\ -2 \\ -2 \end{array}\right] , \left[\begin{array}{c} -1 \\ -2 \\ -3 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 5 \\ 2 \\ 9 \\ -1 \\ -1 \end{array}\right] , \left[\begin{array}{c} 6 \\ 4 \\ 12 \\ 2 \\ 2 \end{array}\right] , \left[\begin{array}{c} -2 \\ 0 \\ -3 \\ 2 \\ 2 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^5\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 2 \\ 0 \\ 3 \\ -2 \\ -2 \end{array}\right] , \left[\begin{array}{c} -1 \\ -2 \\ -3 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 5 \\ 2 \\ 9 \\ -1 \\ -1 \end{array}\right] , \left[\begin{array}{c} 6 \\ 4 \\ 12 \\ 2 \\ 2 \end{array}\right] , \left[\begin{array}{c} -2 \\ 0 \\ -3 \\ 2 \\ 2 \end{array}\right] \right\} \) is not a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 3 & 0 & 3 \\ 2 & 1 & -2 \\ 4 & 3 & -1 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 3 \\ 2 \\ 4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 3 \end{array}\right] , \left[\begin{array}{c} 3 \\ -2 \\ -1 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^3\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 3 \\ 2 \\ 4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 3 \end{array}\right] , \left[\begin{array}{c} 3 \\ -2 \\ -1 \end{array}\right] \right\} \)both spans \(\mathbb{R}^3\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 3 \\ 2 \\ 4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 3 \end{array}\right] , \left[\begin{array}{c} 3 \\ -2 \\ -1 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^3\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} -4 & -1 & -2 & 3 \\ -3 & -5 & 7 & -3 \\ 4 & 4 & -4 & 0 \\ -4 & 4 & -12 & 1 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ -3 \\ 4 \\ -4 \end{array}\right] , \left[\begin{array}{c} -1 \\ -5 \\ 4 \\ 4 \end{array}\right] , \left[\begin{array}{c} -2 \\ 7 \\ -4 \\ -12 \end{array}\right] , \left[\begin{array}{c} 3 \\ -3 \\ 0 \\ 1 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ -3 \\ 4 \\ -4 \end{array}\right] , \left[\begin{array}{c} -1 \\ -5 \\ 4 \\ 4 \end{array}\right] , \left[\begin{array}{c} -2 \\ 7 \\ -4 \\ -12 \end{array}\right] , \left[\begin{array}{c} 3 \\ -3 \\ 0 \\ 1 \end{array}\right] \right\} \)both spans \(\mathbb{R}^4\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ -3 \\ 4 \\ -4 \end{array}\right] , \left[\begin{array}{c} -1 \\ -5 \\ 4 \\ 4 \end{array}\right] , \left[\begin{array}{c} -2 \\ 7 \\ -4 \\ -12 \end{array}\right] , \left[\begin{array}{c} 3 \\ -3 \\ 0 \\ 1 \end{array}\right] \right\} \) is not a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc} -4 & 0 & -8 & 0 \\ -4 & -5 & -18 & 5 \\ -2 & 3 & 2 & -3 \\ -2 & 0 & -4 & 0 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 2 & 0 \\ 0 & 1 & 2 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ -4 \\ -2 \\ -2 \end{array}\right] , \left[\begin{array}{c} 0 \\ -5 \\ 3 \\ 0 \end{array}\right] , \left[\begin{array}{c} -8 \\ -18 \\ 2 \\ -4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 5 \\ -3 \\ 0 \end{array}\right] \right\} \)is not a basis of \(\mathbb{R}^4\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ -4 \\ -2 \\ -2 \end{array}\right] , \left[\begin{array}{c} 0 \\ -5 \\ 3 \\ 0 \end{array}\right] , \left[\begin{array}{c} -8 \\ -18 \\ 2 \\ -4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 5 \\ -3 \\ 0 \end{array}\right] \right\} \)either doesn't span \(\mathbb{R}^4\) or is linearly dependent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} -4 \\ -4 \\ -2 \\ -2 \end{array}\right] , \left[\begin{array}{c} 0 \\ -5 \\ 3 \\ 0 \end{array}\right] , \left[\begin{array}{c} -8 \\ -18 \\ 2 \\ -4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 5 \\ -3 \\ 0 \end{array}\right] \right\} \) is not a basis of \(\mathbb{R}^4\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} 1 & -4 & 3 & -5 & -3 \\ 4 & -4 & 1 & 2 & -3 \\ -2 & -1 & 2 & 2 & 2 \\ 0 & 1 & -2 & -3 & 4 \\ -2 & 2 & -4 & 2 & -3 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ 4 \\ -2 \\ 0 \\ -2 \end{array}\right] , \left[\begin{array}{c} -4 \\ -4 \\ -1 \\ 1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 1 \\ 2 \\ -2 \\ -4 \end{array}\right] , \left[\begin{array}{c} -5 \\ 2 \\ 2 \\ -3 \\ 2 \end{array}\right] , \left[\begin{array}{c} -3 \\ -3 \\ 2 \\ 4 \\ -3 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ 4 \\ -2 \\ 0 \\ -2 \end{array}\right] , \left[\begin{array}{c} -4 \\ -4 \\ -1 \\ 1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 1 \\ 2 \\ -2 \\ -4 \end{array}\right] , \left[\begin{array}{c} -5 \\ 2 \\ 2 \\ -3 \\ 2 \end{array}\right] , \left[\begin{array}{c} -3 \\ -3 \\ 2 \\ 4 \\ -3 \end{array}\right] \right\} \)both spans \(\mathbb{R}^5\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ 4 \\ -2 \\ 0 \\ -2 \end{array}\right] , \left[\begin{array}{c} -4 \\ -4 \\ -1 \\ 1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 1 \\ 2 \\ -2 \\ -4 \end{array}\right] , \left[\begin{array}{c} -5 \\ 2 \\ 2 \\ -3 \\ 2 \end{array}\right] , \left[\begin{array}{c} -3 \\ -3 \\ 2 \\ 4 \\ -3 \end{array}\right] \right\} \) is a basis of \(\mathbb{R}^5\)

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc} 1 & 3 & 2 & -3 & -5 \\ 0 & 1 & 1 & -1 & -3 \\ -5 & -3 & -5 & -4 & -5 \\ -1 & -2 & -1 & 2 & 0 \\ -5 & 1 & 3 & -4 & -3 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & -2 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

1. The statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -5 \\ -1 \\ -5 \end{array}\right] , \left[\begin{array}{c} 3 \\ 1 \\ -3 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} 2 \\ 1 \\ -5 \\ -1 \\ 3 \end{array}\right] , \left[\begin{array}{c} -3 \\ -1 \\ -4 \\ 2 \\ -4 \end{array}\right] , \left[\begin{array}{c} -5 \\ -3 \\ -5 \\ 0 \\ -3 \end{array}\right] \right\} \)is a basis of \(\mathbb{R}^5\).

   is equivalent to the statement

   The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -5 \\ -1 \\ -5 \end{array}\right] , \left[\begin{array}{c} 3 \\ 1 \\ -3 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} 2 \\ 1 \\ -5 \\ -1 \\ 3 \end{array}\right] , \left[\begin{array}{c} -3 \\ -1 \\ -4 \\ 2 \\ -4 \end{array}\right] , \left[\begin{array}{c} -5 \\ -3 \\ -5 \\ 0 \\ -3 \end{array}\right] \right\} \)both spans \(\mathbb{R}^5\) and is linearly independent.

2. The set of vectors \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -5 \\ -1 \\ -5 \end{array}\right] , \left[\begin{array}{c} 3 \\ 1 \\ -3 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} 2 \\ 1 \\ -5 \\ -1 \\ 3 \end{array}\right] , \left[\begin{array}{c} -3 \\ -1 \\ -4 \\ 2 \\ -4 \end{array}\right] , \left[\begin{array}{c} -5 \\ -3 \\ -5 \\ 0 \\ -3 \end{array}\right] \right\} \) is not a basis of \(\mathbb{R}^5\)