Find the inverse of the matrix,if it exists: | vedclass

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(A) Let $A=\left[\begin{array}{ccc}1 & 3 & -2 \ -3 & 0 & -5 \ 2 & 5 & 0\end{array}ight]$.First,calculate the determinant $|A|$:$|A| = 1(0 - (-25))

Vedclass · Accepted Answer

$A^{-1}=\left[\begin{array}{rrr}1 & -\frac{2}{5} & -\frac{3}{5} \ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25}\end{array}ight]$

Vedclass · Answer

(A) Let $A=\left[\begin{array}{ccc}1 & 3 & -2 \ -3 & 0 & -5 \ 2 & 5 & 0\end{array}ight]$.First,calculate the determinant $|A|$:$|A| = 1(0 - (-25)) - 3(0 - (-10)) - 2(-15 - 0)$$|A| = 1(25) - 3(10) - 2(-15) = 25 - 30 + 30 = 25$.Since $|A| 
eq 0$,the inverse exists.Using the formula $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$:Find the matrix of cofactors $C_{ij}$:$C_{11} = +(0 - (-25)) = 25, C_{12} = -(0 - (-10)) = -10, C_{13} = +(-15 - 0) = -15$$C_{21} = -(0 - (-10)) = -10, C_{22} = +(0 - (-4)) = 4, C_{23} = -(5 - 6) = 1$$C_{31} = +(-15 - 0) = -15, C_{32} = -(-5 - 6) = 11, C_{33} = +(0 - (-9)) = 9$$	ext{adj}(A) = \left[\begin{array}{ccc}25 & -10 & -15 \ -10 & 4 & 11 \ -15 & 1 & 9\end{array}ight]^T = \left[\begin{array}{ccc}25 & -10 & -15 \ -10 & 4 & 1 \ -15 & 11 & 9\end{array}ight]$Wait,recalculating cofactors:$C_{23} = -(5 - 6) = 1$. Correct.$C_{32} = -(-5 - 6) = 11$. Correct.$A^{-1} = \frac{1}{25} \left[\begin{array}{ccc}25 & -10 & -15 \ -10 & 4 & 11 \ -15 & 1 & 9\end{array}ight] = \left[\begin{array}{ccc}1 & -\frac{2}{5} & -\frac{3}{5} \ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25}\end{array}ight]$.

Find the inverse of the matrix,if it exists: $\left[\begin{array}{ccc}1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0\end{array}\right]$

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