Let \( A=\left[\begin{array}{ccc}0 & 2 & 2 \\ 0 & -2 & -2 \\ 0 & 4 & 4\end{array}\right] \) (a) Find the dimension of ker \( A \) and give \( a \) basis. Dimension:

To find the dimension of the kernel of the matrix \( A = \left[\begin{array}{ccc}0 & 2 & 2 \\ 0 & -2 & -2 \\ 0 & 4 & 4\end{array}\right] \), we first perform row reduction to find its reduced row echelon form (RREF). Starting with \( A \): \[ \begin{bmatrix} 0 & 2 & 2 \\ 0 & -2 & -2 \\ 0 & 4 & 4 \end{bmatrix} \] Swap the second and first rows, and then divide the first row by 2: \[ \begin{bmatrix} 0 & 1 & 1 \\ 0 & -2 & -2 \\ 0 & 4 & 4 \end{bmatrix} \] Next, normalize the second row by multiplying it by -1/2: \[ \begin{bmatrix} 0 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 4 & 4 \end{bmatrix} \] Now eliminate the contributions from row three: \[ \begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \] This matrix indicates that we have 1 leading 1 (i.e., a pivot) in the second column. The number of variables (3) minus the number of pivots (1) gives us the dimension of the kernel. Thus, the dimension of the kernel of \( A \) is \( 3 - 1 = 2 \). To find the basis, set \( x_2 = t \) and \( x_3 = s \). Then from the first row, \[ x_1 + t + s = 0 \implies x_1 = -t - s \] By substituting different values for \( t \) and \( s \), we can write the general solution in vector form: \[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = t \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} + s \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \] Thus, a basis for the kernel of \( A \) is: \[ \left\{ \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \right\} \] To conclude: **Dimension: 2** **Basis: \( \left\{ \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \right\} \)**