## Precalculus (6th Edition) Blitzer

The determinant is $-407$.
In order to find the given determinant, first evaluate the individual determinant separately. So consider, \begin{align} & \left| \begin{matrix} 5 & 0 \\ 4 & -3 \\ \end{matrix} \right|=5\left( -3 \right)-4\left( 0 \right) \\ & =-15-0 \\ & =-15 \end{align} Next consider, \begin{align} & \left| \begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} \right|=-1\left( -1 \right)-0 \\ & =1-0 \\ & =1 \end{align} Now find, \begin{align} & \left| \begin{matrix} 7 & -5 \\ 4 & 6 \\ \end{matrix} \right|=7\left( 6 \right)-4\left( -5 \right) \\ & =42+20 \\ & =62 \end{align} And finally find, \begin{align} & \left| \begin{matrix} 4 & 1 \\ -3 & 5 \\ \end{matrix} \right|=4\left( 5 \right)-\left( -3 \right)1 \\ & =20+3 \\ & =23 \end{align} Hence, $\left| \begin{matrix} \left| \begin{matrix} 5 & 0 \\ 4 & -3 \\ \end{matrix} \right| & \left| \begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} \right| \\ \left| \begin{matrix} 7 & -5 \\ 4 & 6 \\ \end{matrix} \right| & \left| \begin{matrix} 4 & 1 \\ -3 & 5 \\ \end{matrix} \right| \\ \end{matrix} \right|=\left| \begin{matrix} -15 & 1 \\ 62 & 23 \\ \end{matrix} \right|$ Now calculate, \begin{align} & \left| \begin{matrix} -15 & 1 \\ 62 & 23 \\ \end{matrix} \right|=\left( -15 \right)\left( 23 \right)-62\left( 1 \right) \\ & =-345-62 \\ & =-407 \end{align}