Question 1.9.1: Compute the determinant of the matrix

By definition,

det \begin{bmatrix} 2 & -1 &1 \\ 1 & 1&2 \\ -3&0&-3 \end{bmatrix}=2 det \begin{bmatrix} 1 & 2 \\ 0 & -3 \end{bmatrix} −(−1) det \begin{bmatrix} 1 & 2 \\ -3 & -3 \end{bmatrix} +1det \begin{bmatrix} 1 & 1 \\ -3 & 0 \end{bmatrix}

= 2(−3−0)+1(−3−(−6))+1(0−(−3))
=−6+3+3
= 0

Next, to determine whether or not A is invertible, we row-veduce A to see if Ahas a pivot position in every row. Doing so, we find that

\begin{bmatrix} 2 & -1 &1 \\ 1 & 1&2 \\ -3&0&-3 \end{bmatrix} →\begin{bmatrix} 1 & 0 &1 \\ 0 & 1&1 \\ 0&0&0 \end{bmatrix}

Thus, we see that A does not have a pivot in every row, and therefore A is not invertible.