Question

Compute the determinant using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column. StartAbsoluteValue Start 3 By 3 Matrix 1st Row 1st Column 3 2nd Column 0 3rd Column 3 2nd Row 1st Column 2 2nd Column 3 3rd Column 3 3rd Row 1st Column 0 2nd Column 4 3rd Column negative 2 EndMatrix EndAbsoluteValue

Answer 1

Answer:

Step-by-step explanation:

It is given that

$\Delta=\begin{vmatrix}3&0&3\\2 &3&3\\0 &4&-2\end{vmatrix}$

By cofactor expansion across the first row, we get

$\Delta=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}$

$\Delta=3\left[(-1)^{1+1}\begin{vmatrix}3&3\\4&-2\end{vmatrix}\right]+0\left[(-1)^{1+2}\begin{vmatrix}2&3\\0&-2\end{vmatrix}\right]+3\left[(-1)^{1+3}\begin{vmatrix}2&3\\0&4\end{vmatrix}\right]$

$\Delta=3\left[-18\right]+0\left[(-1)(-4)\right]+3\left[8\right]$

$\Delta=-54+0+24$

$\Delta=-30$

Therefore, the value of determinant is -30.

By cofactor expansion across the second column, we get

$\Delta=a_{12}C_{12}+a_{22}C_{22}+a_{32}C_{32}$

$\Delta=0\left[(-1)^{2+1}\begin{vmatrix}2&3\\0&-2\end{vmatrix}\right]+3\left[(-1)^{2+2}\begin{vmatrix}3&3\\0&-2\end{vmatrix}\right]+4\left[(-1)^{3+2}\begin{vmatrix}3&3\\2&3\end{vmatrix}\right]$

$\Delta=0\left[(-1)(-4)\right]+3\left[(-6)\right]+4\left[(-1)3\right]$

$\Delta=-18-12$

$\Delta=-30$

Therefore, the value of determinant is -30.