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Answer: \(A^{-1}=\frac{1}{40}\begin{bmatrix}19 & 5 & -27\\-2 & 10 & -14\\-3 & -5 & 19\end{bmatrix}\)

\(|A|=3(35-16)-1(10-8)+5(4-7)=57-2-15=40\neq 0\), so \(A^{-1}\) exists.

Cofactors:
\(A_{11}=19,\ A_{12}=-(10-8)=-2,\ A_{13}=4-7=-3\)
\(A_{21}=-(5-10)=5,\ A_{22}=15-5=10,\ A_{23}=-(6-1)=-5\)
\(A_{31}=8-35=-27,\ A_{32}=-(24-10)=-14,\ A_{33}=21-2=19\)

\(\text{adj}A=\begin{bmatrix}19 & 5 & -27\\-2 & 10 & -14\\-3 & -5 & 19\end{bmatrix}\)

\(A^{-1}=\frac{1}{|A|}\text{adj}A=\frac{1}{40}\begin{bmatrix}19 & 5 & -27\\-2 & 10 & -14\\-3 & -5 & 19\end{bmatrix}\)

Doubtly AI · 29 Oct 2026 AI-verified · Nov 2026 Report
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