Problem Statement
If $A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$, verify that $A^2 - 6A + 11I = O$. Hence, find $A^{-1}$ and compute $A^3$.
Verified Solution & Marking Scheme
Compute A² via Matrix Multiplication
$A^2 = A \cdot A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 2(2) + (-1)(3) & 2(-1) + (-1)(4) \\ 3(2) + 4(3) & 3(-1) + 4(4) \end{pmatrix} = \begin{pmatrix} 1 & -6 \\ 18 & 13 \end{pmatrix}$
Verify A² - 6A + 11I = O
$6A = \begin{pmatrix} 12 & -6 \\ 18 & 24 \end{pmatrix}, \quad 11I = \begin{pmatrix} 11 & 0 \\ 0 & 11 \end{pmatrix}$
$A^2 - 6A + 11I = \begin{pmatrix} 1 & -6 \\ 18 & 13 \end{pmatrix} - \begin{pmatrix} 12 & -6 \\ 18 & 24 \end{pmatrix} + \begin{pmatrix} 11 & 0 \\ 0 & 11 \end{pmatrix}$
$= \begin{pmatrix} 1 - 12 + 11 & -6 - (-6) + 0 \\ 18 - 18 + 0 & 13 - 24 + 11 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = O$
Hence verified.
Determine A⁻¹ Using the Matrix Equation
Multiplying the equation $A^2 - 6A + 11I = O$ by $A^{-1}$:
$A^{-1} A^2 - 6 A^{-1} A + 11 A^{-1} I = O$
$A - 6I + 11 A^{-1} = O \implies 11 A^{-1} = 6I - A$
$A^{-1} = \frac{1}{11} (6I - A) = \frac{1}{11} \left[ \begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix} - \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} \right] = \frac{1}{11} \begin{pmatrix} 4 & 1 \\ -3 & 2 \end{pmatrix}$
Compute A³ Using A² = 6A - 11I
Multiplying $A^2 = 6A - 11I$ by $A$:
$A^3 = 6A^2 - 11A = 6(6A - 11I) - 11A = 36A - 66I - 11A = 25A - 66I$
$A^3 = 25 \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} - 66 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 50 - 66 & -25 - 0 \\ 75 - 0 & 100 - 66 \end{pmatrix} = \begin{pmatrix} -16 & -25 \\ 75 & 34 \end{pmatrix}$