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: Direct verification of the Cayley–Hamilton-type identity:
For a 2×2 matrix \(A\), the characteristic equation is \(A^2-(\text{tr }A)A+(\det A)I=O\). Here \(\text{tr }A=2+4=6\) and \(\det A=(2)(4)-(-1)(3)=8+3=11\), so the characteristic equation is exactly \(A^2-6A+11I=O\) — this is guaranteed by the Cayley–Hamilton theorem, and matches what's asked to prove.

Step 2: Confirming by direct computation (cross-check):
As computed directly, \(A^2=\begin{bmatrix}1&-6\\18&13\end{bmatrix}\), and \(A^2-6A+11I\) evaluates entrywise to the zero matrix, confirming the theorem's prediction.

Step 3: Deriving A⁻¹ from the identity algebraically:
Multiply \(A^2-6A+11I=O\) on the left (or right) by \(A^{-1}\): \(A-6I+11A^{-1}=O\Rightarrow 11A^{-1}=6I-A\Rightarrow A^{-1}=\dfrac{1}{11}(6I-A)\).

Step 4: Computing explicitly:
\(6I-A=\begin{bmatrix}4&1\\-3&2\end{bmatrix}\), so \(A^{-1}=\dfrac1{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}\).

Step 5: Verifying A·A⁻¹ = I:
\(A\cdot\dfrac1{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}=\dfrac1{11}\begin{bmatrix}8+3&2-2\\12-12&3+8\end{bmatrix}=\dfrac1{11}\begin{bmatrix}11&0\\0&11\end{bmatrix}=I\). Confirmed.

Final Answer:
\[ \boxed{A^{-1}=\dfrac1{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}} \]