Problem Statement
જો શ્રેણિક $A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}$ હોય, તો સાબિત કરો કે $A^2 - 6A + 11I = O$ અને તે પરથી $A^{-1}$ મેળવો.
Verified Solution & Marking Scheme
A² ની ગણતરી (Compute A²)
$A^2 = A \times A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 4 - 3 & -2 - 4 \\ 6 + 12 & -3 + 16 \end{bmatrix} = \begin{bmatrix} 1 & -6 \\ 18 & 13 \end{bmatrix}$
A² - 6A + 11I = O ચકાસણી (Verify matrix polynomial)
$6A = \begin{bmatrix} 12 & -6 \\ 18 & 24 \end{bmatrix}, \quad 11I = \begin{bmatrix} 11 & 0 \\ 0 & 11 \end{bmatrix}$
$A^2 - 6A + 11I = \begin{bmatrix} 1 & -6 \\ 18 & 13 \end{bmatrix} - \begin{bmatrix} 12 & -6 \\ 18 & 24 \end{bmatrix} + \begin{bmatrix} 11 & 0 \\ 0 & 11 \end{bmatrix} = \begin{bmatrix} 1 - 12 + 11 & -6 - (-6) + 0 \\ 18 - 18 + 0 & 13 - 24 + 11 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$
A⁻¹ મેળવવો (Compute A⁻¹ from polynomial)
$A^2 - 6A + 11I = O \implies 11I = 6A - A^2$
બંને બાજુ $A^{-1}$ વડે ગુણતાં:
$11 A^{-1} = 6(A A^{-1}) - A^2 A^{-1} = 6I - A$
$11 A^{-1} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix}$
$A^{-1} = \frac{1}{11} \begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix}$