## Precalculus (6th Edition) Blitzer

Expand along row 1, $\left| \begin{matrix} 2 & 0 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 4 \\ \end{matrix} \right|=2\left| \begin{matrix} 3 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 4 \\ \end{matrix} \right|\text{ (as other terms are zero)}$ Again, expand along row 1 $2\left| \begin{matrix} 3 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 4 \\ \end{matrix} \right|\text{= 2}\cdot \text{3}\left| \begin{matrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \\ \end{matrix}\text{ } \right|\text{(as other terms are zero)}$ Again, expand along row 1 \begin{align} & \text{2}\cdot \text{3}\left| \begin{matrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \\ \end{matrix}\text{ } \right|\text{=2}\cdot 3\cdot 2\left| \begin{matrix} 1 & 0 \\ 0 & 4 \\ \end{matrix} \right|\text{(as other terms are zero)} \\ & \text{ =2}\cdot 3\cdot 2(1\cdot 4-0) \\ & \text{ =2}\cdot 3\cdot 2\cdot 4 \\ & \text{ =48} \\ \end{align} So the solution of given determinant is 48.