Question

If \(A=\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix}\), then find \(A^{-1}\).

Hint:

Use \(A^{-1}=\dfrac{1}{\det A}\text{adj}(A)\) via cofactors; verify by checking \(AA^{-1}=I\).

Answer 1

Step 1: Understanding the Concept:
\(A^{-1}=\dfrac{1}{\det A}\,\text{adj}(A)\), so first find \(\det A\) (must be nonzero), then all nine cofactors to build the adjugate.

Step 2: Computing the determinant:
Expanding along row 1: \(\det A=2\big(2(-2)-(-4)(1)\big)-(-3)\big(3(-2)-(-4)(1)\big)+5\big(3(1)-2(1)\big)\)\[=2(-4+4)+3(-6+4)+5(3-2)=0-6+5=-1.\]

Step 3: Computing the cofactors:
\(C_{11}=0,\ C_{12}=2,\ C_{13}=1\); \(C_{21}=-1,\ C_{22}=-9,\ C_{23}=-5\); \(C_{31}=2,\ C_{32}=23,\ C_{33}=13\) (each \(C_{ij}\) is \((-1)^{i+j}\) times the \(2\times2\) minor deleting row \(i\), column \(j\)).

Step 4: Building the adjugate (transpose of cofactor matrix):
\(\text{adj}(A)=\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix}\).

Step 5: Dividing by the determinant:
\(A^{-1}=\dfrac{1}{-1}\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix}=\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}\).

Final Answer:
\(A^{-1}=\boxed{\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}}\).