Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 1 \\ -5 \\ 0 \end{array}\right] , \left[\begin{array}{c} 1 \\ -1 \\ 6 \\ -1 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 1 \\ -5 \\ 0 \end{array}\right] , \left[\begin{array}{c} 1 \\ -1 \\ 6 \\ -1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 2 \\ 1 \\ 0 \\ -2 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ -3 \\ 0 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 3 \, a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ 1 \\ 0 \\ -2 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ -3 \\ 0 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 3 \\ -1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -1 & -1 & 3 \\ -1 & 4 & -7 \\ 0 & -4 & 8 \\ -1 & 1 & -1 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -1 \\ -1 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 4 \\ -4 \\ 1 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -1 \\ -1 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 4 \\ -4 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -3 \\ -3 \\ -3 \\ 4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ -3 \\ 0 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -3 \\ -3 \\ -3 \\ 4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ -3 \\ 0 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 0 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 1 \\ -1 \\ 2 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 1 \\ -1 \\ 2 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 5 & 4 & -2 \\ -4 & -3 & 2 \\ 1 & 4 & 6 \\ 4 & 8 & 8 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 5 \\ -4 \\ 1 \\ 4 \end{array}\right] , \left[\begin{array}{c} 4 \\ -3 \\ 4 \\ 8 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 2 \, a \\ -2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 5 \\ -4 \\ 1 \\ 4 \end{array}\right] , \left[\begin{array}{c} 4 \\ -3 \\ 4 \\ 8 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ -2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & -7 & -8 \\ -1 & 4 & 5 \\ 0 & -2 & -2 \\ 1 & -1 & -2 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} -7 \\ 4 \\ -2 \\ -1 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} -7 \\ 4 \\ -2 \\ -1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} 5 \\ 6 \\ -2 \\ -7 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 0 \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} 5 \\ 6 \\ -2 \\ -7 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -4 & 1 & 6 \\ -5 & 1 & 8 \\ -3 & 0 & 6 \\ -3 & 1 & 4 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -4 \\ -5 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ 0 \\ 1 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 2 \, a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -4 \\ -5 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ 0 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ 2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 2 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -3 \\ 1 \\ 1 \\ 1 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -3 \\ 1 \\ 1 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ -1 \\ 3 \\ -1 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -3 \, a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ -1 \\ 3 \\ -1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -3 \\ 2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 2 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 5 \\ -1 \\ 0 \\ -2 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 3 \, a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 2 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 5 \\ -1 \\ 0 \\ -2 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 3 \\ -1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -4 \\ 1 \\ 5 \\ 4 \end{array}\right] , \left[\begin{array}{c} -3 \\ 3 \\ 5 \\ 5 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -4 \\ 1 \\ 5 \\ 4 \end{array}\right] , \left[\begin{array}{c} -3 \\ 3 \\ 5 \\ 5 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 0 \\ 1 \\ -1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 1 \\ 0 \\ 1 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ -3 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ 1 \\ -1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 1 \\ 0 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -3 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & 4 & -8 \\ -1 & -3 & 6 \\ 0 & 0 & 0 \\ 0 & 4 & -8 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 4 \\ -3 \\ 0 \\ 4 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 0 \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 4 \\ -3 \\ 0 \\ 4 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & -2 & 3 \\ -2 & 5 & -8 \\ -3 & 5 & -7 \\ 0 & 1 & -2 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -2 \\ -3 \\ 0 \end{array}\right] , \left[\begin{array}{c} -2 \\ 5 \\ 5 \\ 1 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -2 \\ -3 \\ 0 \end{array}\right] , \left[\begin{array}{c} -2 \\ 5 \\ 5 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 3 \\ 2 \\ 1 \\ -2 \end{array}\right] , \left[\begin{array}{c} -6 \\ -7 \\ -1 \\ 3 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 0 \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 3 \\ 2 \\ 1 \\ -2 \end{array}\right] , \left[\begin{array}{c} -6 \\ -7 \\ -1 \\ 3 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ -1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -2 \\ 1 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} 5 \\ -3 \\ -1 \\ 0 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -a \\ -2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 1 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} 5 \\ -3 \\ -1 \\ 0 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -1 \\ -2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 7 \\ 2 \\ 5 \\ -5 \end{array}\right] , \left[\begin{array}{c} -4 \\ -1 \\ -6 \\ 2 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 0 \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 7 \\ 2 \\ 5 \\ -5 \end{array}\right] , \left[\begin{array}{c} -4 \\ -1 \\ -6 \\ 2 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -1 \\ 2 \\ 1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 1 \\ -3 \\ 0 \\ -2 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ 3 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -1 \\ 2 \\ 1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 1 \\ -3 \\ 0 \\ -2 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 3 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 0 \\ 2 \\ 1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 2 \\ -1 \\ -3 \\ 1 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 0 \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 2 \\ 1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 2 \\ -1 \\ -3 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 2 \\ 2 \\ 1 \\ -1 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ -2 \\ 5 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ 2 \\ 1 \\ -1 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ -2 \\ 5 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} -5 \\ -4 \\ 5 \\ 5 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} -5 \\ -4 \\ 5 \\ 5 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 4 \\ 5 \\ 4 \\ -3 \end{array}\right] , \left[\begin{array}{c} 4 \\ 1 \\ -4 \\ -4 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 0 \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 4 \\ 5 \\ 4 \\ -3 \end{array}\right] , \left[\begin{array}{c} 4 \\ 1 \\ -4 \\ -4 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -4 \\ -2 \\ -5 \\ -4 \end{array}\right] , \left[\begin{array}{c} 4 \\ 1 \\ 4 \\ 3 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -3 \, a \\ -2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -4 \\ -2 \\ -5 \\ -4 \end{array}\right] , \left[\begin{array}{c} 4 \\ 1 \\ 4 \\ 3 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -3 \\ -2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -2 \\ 0 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 1 \\ -4 \\ 3 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -3 \, a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 0 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 1 \\ -4 \\ 3 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -3 \\ 1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -3 \\ 2 \\ 2 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ -3 \\ -1 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 2 \, a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -3 \\ 2 \\ 2 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ -3 \\ -1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ -1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 1 \\ -1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -7 \\ 3 \\ -3 \\ 3 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 1 \\ -1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -7 \\ 3 \\ -3 \\ 3 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ -1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 3 \\ 1 \\ 1 \\ -2 \end{array}\right] , \left[\begin{array}{c} 0 \\ -1 \\ 1 \\ 7 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 2 \, a \\ 0 \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 3 \\ 1 \\ 1 \\ -2 \end{array}\right] , \left[\begin{array}{c} 0 \\ -1 \\ 1 \\ 7 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ 0 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & -5 & 3 \\ 0 & 1 & -1 \\ -1 & 8 & -6 \\ 1 & -8 & 6 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 1 \\ 8 \\ -8 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 2 \, a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 1 \\ 8 \\ -8 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ 1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -3 \\ 1 \\ 4 \\ -1 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 4 \, a \\ a \\ 0 \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -3 \\ 1 \\ 4 \\ -1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 4 \\ 1 \\ 0 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 4 \\ 1 \\ 1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -4 \\ -3 \\ 0 \\ -2 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 4 \\ 1 \\ 1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -4 \\ -3 \\ 0 \\ -2 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 8 \\ 5 \\ -4 \\ -5 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ 0 \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 8 \\ 5 \\ -4 \\ -5 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 0 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 5 & -8 & -1 \\ 1 & -1 & 1 \\ -4 & 6 & 0 \\ 3 & -4 & 1 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 5 \\ 1 \\ -4 \\ 3 \end{array}\right] , \left[\begin{array}{c} -8 \\ -1 \\ 6 \\ -4 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -3 \, a \\ -2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 5 \\ 1 \\ -4 \\ 3 \end{array}\right] , \left[\begin{array}{c} -8 \\ -1 \\ 6 \\ -4 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -3 \\ -2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & -4 & 2 \\ 1 & -4 & 3 \\ 2 & -8 & 7 \\ 0 & 0 & 4 \end{array}\right] = \left[\begin{array}{ccc} 1 & -4 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 2 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ 3 \\ 7 \\ 4 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 4 \, a \\ a \\ 0 \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 2 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ 3 \\ 7 \\ 4 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 4 \\ 1 \\ 0 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 2 \\ 3 \\ 0 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 2 \\ 0 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ 3 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 2 \\ 3 \\ 0 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 2 \\ 0 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 3 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -1 \\ 0 \\ -1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 0 \\ 1 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -3 \, a \\ 3 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -1 \\ 0 \\ -1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 0 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -3 \\ 3 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 2 \\ 1 \\ 2 \\ 2 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ 1 \\ -1 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 2 \, a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ 1 \\ 2 \\ 2 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ 1 \\ -1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ -1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -2 \\ 0 \\ -1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ 2 \\ -2 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 0 \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 0 \\ -1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ 2 \\ -2 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ -1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 3 \\ -2 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ -2 \\ 6 \\ 1 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 3 \\ -2 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ -2 \\ 6 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -1 \\ -1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -2 \\ -1 \\ -1 \end{array}\right] , \left[\begin{array}{c} 1 \\ -1 \\ 1 \\ 1 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ 3 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -2 \\ -1 \\ -1 \end{array}\right] , \left[\begin{array}{c} 1 \\ -1 \\ 1 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 3 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -1 \\ 1 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 4 \\ 0 \\ 6 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 3 \, a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -1 \\ 1 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 4 \\ 0 \\ 6 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 3 \\ -1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 0 & -1 & 1 \\ 0 & 1 & -1 \\ 1 & -6 & 8 \\ 1 & -3 & 5 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 0 \\ 0 \\ 1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 1 \\ -6 \\ -3 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 0 \\ 1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 1 \\ -6 \\ -3 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 1 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 1 \\ 5 \\ -8 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 4 \, a \\ a \\ 0 \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 1 \\ 5 \\ -8 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 4 \\ 1 \\ 0 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -1 \\ -2 \\ 0 \end{array}\right] , \left[\begin{array}{c} -5 \\ 6 \\ 5 \\ 1 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -3 \, a \\ a \\ 0 \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -1 \\ -2 \\ 0 \end{array}\right] , \left[\begin{array}{c} -5 \\ 6 \\ 5 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -3 \\ 1 \\ 0 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ -1 \end{array}\right] , \left[\begin{array}{c} 3 \\ 1 \\ -1 \\ -1 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ -1 \end{array}\right] , \left[\begin{array}{c} 3 \\ 1 \\ -1 \\ -1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -1 \\ 2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 0 \\ 1 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} -4 \\ -1 \\ 2 \\ 2 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 1 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} -4 \\ -1 \\ 2 \\ 2 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 3 \\ -2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 4 \\ 5 \\ -4 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 3 \\ -2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 4 \\ 5 \\ -4 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

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

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -1 \\ -1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 0 \\ -1 \\ 1 \\ 4 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -1 \\ -1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 0 \\ -1 \\ 1 \\ 4 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & -2 & -5 \\ -1 & 3 & 7 \\ 2 & -3 & -8 \\ 0 & 3 & 6 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 2 \\ 0 \end{array}\right] , \left[\begin{array}{c} -2 \\ 3 \\ -3 \\ 3 \end{array}\right] \right\} \]

   \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ -2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 2 \\ 0 \end{array}\right] , \left[\begin{array}{c} -2 \\ 3 \\ -3 \\ 3 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -2 \\ 1 \end{array}\right] \right\} \)
3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.