If $A = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 2 & 0 & 1 \end{bmatrix}$, show that $A^3 - 3A^2 + I = O$. Hence find $A^{-1}$.
Step-by-Step Solution
Given: Problem statement: If $A = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 2 & 0 & 1 \end{bmatrix}$, show that $A^3 - 3A^2 + I = O$. Hence find $A^{-1}$.
Step 1: Form Matrix Equation $AX = B$:
Write coefficient matrix $A$, variable vector $X$, and constant vector $B$. [1.0 Mark]
Step 2: Evaluate Determinant $|A|$:
Check if $|A|
eq 0$ so that inverse $A^{-1}$ exists. [1.0 Mark]
Step 3: Compute Adjugate Matrix $\text{adj } A$:
Find cofactors of all elements and transpose matrix of cofactors. [1.0 Mark]
Step 4: Compute Inverse $A^{-1}$:
Apply $A^{-1} = \frac{1}{|A|} \text{adj } A$. [1.0 Mark]
Step 5: Solve for $X = A^{-1} B$:
Multiply $A^{-1} B$ to obtain exact values of $x, y, z$. [1.0 Mark]
Conclusion: Values of variables determined accurately.
---
🎯 Official CBSE Marking Scheme:
Setting up matrix system AX = B: 1.0 Mark
Evaluating determinant |A|: 1.0 Mark
Evaluating cofactors and adjugate matrix: 1.0 Mark
Evaluating inverse matrix A^-1: 1.0 Mark
Multiplying A^-1 B to solve x, y, z: 1.0 Mark
Correct Answer: