Find the inverse of the matrix (if it exists) | vedclass

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(A) Let $A = \left[\begin{array}{lll}1 & 2 & 3 \ 0 & 2 & 4 \ 0 & 0 & 5\end{array}ight]$.First,we find the determinant $|A|$:$|A| = 1(2 	imes 5 -

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$\frac{1}{10}\left[\begin{array}{ccc}10 & -10 & 2 \ 0 & 5 & -4 \ 0 & 0 & 2\end{array}ight]$

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(A) Let $A = \left[\begin{array}{lll}1 & 2 & 3 \ 0 & 2 & 4 \ 0 & 0 & 5\end{array}ight]$.First,we find the determinant $|A|$:$|A| = 1(2 	imes 5 - 4 	imes 0) - 2(0 	imes 5 - 4 	imes 0) + 3(0 	imes 0 - 2 	imes 0) = 1(10) - 2(0) + 3(0) = 10$.Since $|A| 
eq 0$,the inverse $A^{-1}$ exists.Next,we find the cofactor matrix $C = [C_{ij}]$:$C_{11} = +(10-0) = 10, C_{12} = -(0-0) = 0, C_{13} = +(0-0) = 0$$C_{21} = -(10-0) = -10, C_{22} = +(5-0) = 5, C_{23} = -(0-0) = 0$$C_{31} = +(8-6) = 2, C_{32} = -(4-0) = -4, C_{33} = +(2-0) = 2$The adjoint matrix $adj(A)$ is the transpose of the cofactor matrix:$adj(A) = \left[\begin{array}{ccc}10 & -10 & 2 \ 0 & 5 & -4 \ 0 & 0 & 2\end{array}ight]$.Finally,$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{10} \left[\begin{array}{ccc}10 & -10 & 2 \ 0 & 5 & -4 \ 0 & 0 & 2\end{array}ight]$.

Find the inverse of the matrix (if it exists): $\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]$

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