\[ \begin{array}{l} A=\left[\begin{array}{ccc} 2 & -3 & 2 \\ 2 & 3 & -4 \\ -2 & -6 & 7 \end{array}\right] \\ W=\operatorname{Col}(4) \\ \vec{v}=\left[\begin{array}{l} 4 \\ 5 \\ 0 \end{array}\right] \end{array} \] 1) Is A invertible? How do you know?

Step 1 ：A square matrix is invertible if and only if its determinant is not zero. So, to determine if matrix A is invertible, we need to calculate its determinant. If the determinant is not zero, then the matrix is invertible. If the determinant is zero, then the matrix is not invertible.

Step 2 ：Matrix A is given by \[ A = \begin{bmatrix} 2 & -3 & 2 \\ 2 & 3 & -4 \\ -2 & -6 & 7 \end{bmatrix} \]

Step 3 ：Calculate the determinant of matrix A, \(\text{det}_A\)

Step 4 ：The determinant of matrix A is approximately zero, which means that matrix A is not invertible.

Step 5 ：\(\boxed{\text{Matrix A is not invertible because its determinant is approximately zero}}\)