Question

Prove that the matrix \[ A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \] satisfies the equation \[ A^{2} - 4A + I_{2} = 0, \] where $I_{2}$ is the $2 \times 2$ identity matrix and $0$ is the $2 \times 2$ zero matrix. Also, find $A^{-1}$ with the help of this.

Hint:

Cayley-Hamilton theorem provides a direct method to compute $A^{-1}$ without using the adjoint and determinant.

Answer 1

Step 1: Compute $A^{2}$. \[ A^{2} = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 4+3 & 6+6 \\ 2+2 & 3+4 \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} \] Step 2: Compute $A^{2} - 4A + I_{2}$. \[ A^{2} - 4A + I_{2} = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} - 4 \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} - \begin{bmatrix} 8 & 12 \\ 4 & 8 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \] Hence proved. ✅ Step 3: Find $A^{-1}$. From Cayley-Hamilton theorem: \[ A^{2} - 4A + I = 0 \] \[ \implies A^{2} - 4A = -I \] \[ \implies A(A - 4I) = -I \] \[ \implies A^{-1} = -(A - 4I) \] \[ = -\left(\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} - \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}\right) = -\begin{bmatrix} -2 & 3 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix} \]

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