Matrices & Determinants
Matrix Properties and Powers
Grade 12
Question:
<p>Let A and B be two non-singular matrices such that \(A \neq I\), \(B^3 = I\) and \(AB = BA^2\), where I is the identity matrix. Find the least value of k such that \(A^k = I\).</p>
Step-by-Step Solution
Key Concept: Use the given relation AB = BA² to express higher powers and substitute B³ = I to find the power of A that equals the identity matrix.
<p><strong>Given:</strong> $AB = BA^2$</p><p>From $AB = BA^2$, we get $B = A^{-1}BA^2$</p><p>Therefore, $B^3 = (A^{-1}BA)^3 = I$</p><p>Expanding: $(A^{-1}BA)(A^{-1}BA)(A^{-1}BA) = I$</p><p>This gives us: $A^{-1}B(AB)(AB)AA = I$</p><p>Using $AB = BA^2$: $A^{-1}B(BA^2)(BA^2)AA = I$</p><p>Simplifying: $A^{-1}BBA(AB)A^4 = I$</p><p>Using $AB = BA^2$ again: $A^{-1}BBA(BA^2)A^4 = I$</p><p>Further simplification: $A^{-1}BB(AB)A^6 = I$</p><p>$A^{-1}BB(BA^2)A^6 = I$ (using $AB = BA^2$)</p><p>$A^{-1}B^3A^8 = I$</p><p>Since $B^3 = I$: $A^{-1}IA^8 = I$</p><p>$A^{-1}A^8 = I$</p><p>$A^7 = I$</p><p><strong>∴ The least value of k is 7.</strong></p>
Correct Answer: 7