Respuesta :
Answer:
Step-by-step explanation:
It is given that
[tex]\Delta=\begin{vmatrix}3&0&3\\2 &3&3\\0 &4&-2\end{vmatrix}[/tex]
By cofactor expansion across the first row, we get
[tex]\Delta=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}[/tex]
[tex]\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][/tex]
[tex]\Delta=3\left[-18\right]+0\left[(-1)(-4)\right]+3\left[8\right][/tex]
[tex]\Delta=-54+0+24[/tex]
[tex]\Delta=-30[/tex]
Therefore, the value of determinant is -30.
By cofactor expansion across the second column, we get
[tex]\Delta=a_{12}C_{12}+a_{22}C_{22}+a_{32}C_{32}[/tex]
[tex]\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][/tex]
[tex]\Delta=0\left[(-1)(-4)\right]+3\left[(-6)\right]+4\left[(-1)3\right][/tex]
[tex]\Delta=-18-12[/tex]
[tex]\Delta=-30[/tex]
Therefore, the value of determinant is -30.
