Question-34

Find \(\displaystyle \text{det}( A)\)

\[ A=\begin{bmatrix} 1 & 0 & 1 & 0 & 0 & 0\\ 0 & -1 & 2 & 0 & 0 & 0\\ 0 & 0 & -3 & 0 & 0 & 0\\ 1 & -1 & 2 & 1 & 2 & -1\\ 2 & 1 & 3 & 0 & 2 & 3\\ 0 & 1 & 2 & 0 & 0 & 3 \end{bmatrix} \]

Answer

\(18\)

Solution

We can partition this matrix as follows:

\[ A=\begin{bmatrix} 1 & 0 & 1 & 0 & 0 & 0\\ 0 & -1 & 2 & 0 & 0 & 0\\ 0 & 0 & -3 & 0 & 0 & 0\\ 1 & -1 & 2 & 1 & 2 & -1\\ 2 & 1 & 3 & 0 & 2 & 3\\ 0 & 1 & 2 & 0 & 0 & 3 \end{bmatrix} =\begin{bmatrix} B & 0\\ D & E \end{bmatrix} \]

Now, we use the fact the determinant of \(\displaystyle A\) is \(\displaystyle \text{det}( B)\text{det}( E)\) when one of the off-diagonal blocks is zero. This turns out to be \(\displaystyle 18\).