Question

If \(A=\begin{bmatrix}1&2&3\\2&4&5\\3&5&6\end{bmatrix}\), then find \(A^{-1}\).

Hint:

Compute |A| via cofactor expansion, build the adjugate from all nine cofactors, then divide by |A|.

Answer 1

Step 1: Find \(|A|\) by expanding along the first row:
\[ |A|=1(4\cdot6-5\cdot5)-2(2\cdot6-5\cdot3)+3(2\cdot5-4\cdot3) \]
\[ =1(24-25)-2(12-15)+3(10-12)=-1+6-6=-1 \]

Step 2: Find all nine cofactors:
\(C_{11}=24-25=-1,\ C_{12}=-(12-15)=3,\ C_{13}=10-12=-2\)
\(C_{21}=-(12-15)=3,\ C_{22}=6-9=-3,\ C_{23}=-(5-6)=1\)
\(C_{31}=10-12=-2,\ C_{32}=-(5-6)=1,\ C_{33}=4-4=0\)

Step 3: Build the adjugate (transpose of the cofactor matrix):
\[ \text{adj}(A)=\begin{bmatrix}-1&3&-2\\3&-3&1\\-2&1&0\end{bmatrix} \]

Step 4: Divide by \(|A|=-1\):
\[ A^{-1}=\dfrac1{-1}\begin{bmatrix}-1&3&-2\\3&-3&1\\-2&1&0\end{bmatrix}=\begin{bmatrix}1&-3&2\\-3&3&-1\\2&-1&0\end{bmatrix} \]

Final Answer:
\[ \boxed{A^{-1}=\begin{bmatrix}1&-3&2\\-3&3&-1\\2&-1&0\end{bmatrix}} \]