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: Verifying by direct multiplication instead of re-deriving:
Rather than re-doing the cofactor expansion, verify the claimed inverse \(B=\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}\) by checking \(AB=I\).

Step 2: Computing row 1 of AB:
Row1(A)\(\cdot\)Col1(B): \(2(0)+(-3)(-2)+5(-1)=0+6-5=1\). Row1\(\cdot\)Col2: \(2(1)+(-3)(9)+5(5)=2-27+25=0\). Row1\(\cdot\)Col3: \(2(-2)+(-3)(-23)+5(-13)=-4+69-65=0\). So row 1 of \(AB\) is \((1,0,0)\), as required.

Step 3: Computing rows 2 and 3 of AB similarly:
Row2\(\cdot\)Col1: \(3(0)+2(-2)+(-4)(-1)=0-4+4=0\); Row2\(\cdot\)Col2: \(3(1)+2(9)+(-4)(5)=3+18-20=1\); Row2\(\cdot\)Col3: \(3(-2)+2(-23)+(-4)(-13)=-6-46+52=0\). Row3\(\cdot\)Col1: \(1(0)+1(-2)+(-2)(-1)=-2+2=0\); Row3\(\cdot\)Col2: \(1(1)+1(9)+(-2)(5)=1+9-10=0\); Row3\(\cdot\)Col3: \(1(-2)+1(-23)+(-2)(-13)=-2-23+26=1\). So \(AB=I\).

Final Answer:
Since \(AB=I\), \(B\) is confirmed to be \(A^{-1}=\boxed{\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}}\).