Question

Prove that the matrix \(A=\begin{bmatrix}2&-1\\3&4\end{bmatrix}\) satisfies the equation \(A^2-6A+11I=O\), where \(I\) is the identity matrix of order \(2\times2\) and \(O\) is the zero matrix of order \(2\times2\). With the help of it, find \(A^{-1}\).

Hint:

Compute A², combine with 6A and 11I to get O, then solve A(6I−A)=11I for A⁻¹.

Answer 1

Step 1: Computing A²:
\(A^2=\begin{bmatrix}2&-1\\3&4\end{bmatrix}\begin{bmatrix}2&-1\\3&4\end{bmatrix}=\begin{bmatrix}2(2)+(-1)(3)&2(-1)+(-1)(4)\\3(2)+4(3)&3(-1)+4(4)\end{bmatrix}=\begin{bmatrix}1&-6\\18&13\end{bmatrix}\).

Step 2: Computing 6A and 11I:
\(6A=\begin{bmatrix}12&-6\\18&24\end{bmatrix}\), \(11I=\begin{bmatrix}11&0\\0&11\end{bmatrix}\).

Step 3: Combining:
\(A^2-6A+11I=\begin{bmatrix}1-12+11&-6+6+0\\18-18+0&13-24+11\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O\). Verified.

Step 4: Using the identity to find A⁻¹:
From \(A^2-6A+11I=O\), rearrange: \(11I=6A-A^2=A(6I-A)\), so \(A\cdot\dfrac{1}{11}(6I-A)=I\), meaning \(A^{-1}=\dfrac{1}{11}(6I-A)\).

Step 5: Computing 6I − A:
\(6I-A=\begin{bmatrix}6-2&0-(-1)\\0-3&6-4\end{bmatrix}=\begin{bmatrix}4&1\\-3&2\end{bmatrix}\).

Final Answer:
\[ \boxed{A^{-1}=\dfrac{1}{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}} \]