Question

Let A and B be two square matrices of order 2 such that $A^{-1}B = 2I$ and $A^{-3} + B^{-3} = A$. If $\lambda\bigl(A^{-2}B^{-1} + A^{-1}B^{-2}\bigr) = B$ then find the value of $\lvert \lambda\rvert$.

Answer 1

3

Answer 2

Step 1: Express $B$ in terms of $A$.
From $A^{-1}B = 2I$, we get

$$B = 2A.$$

Step 2: Use the second equation to relate powers of $A$.

$$A^{-3} + B^{-3} = A \quad\Longrightarrow\quad A^{-3} + (2A)^{-3} = A \quad\Longrightarrow\quad A^{-3} + \tfrac{1}{8}A^{-3} = A \quad\Longrightarrow\quad \tfrac{9}{8}A^{-3} = A.$$

Multiply both sides by $A^3$:

$$\frac{9}{8}I = A^4 \quad\Longrightarrow\quad A^4 = \frac{9}{8}I \quad\Longrightarrow\quad A^{-3} = \frac{8}{9}A.$$

Step 3: Compute $A^{-2}B^{-1} + A^{-1}B^{-2}$.

$$A^{-2}B^{-1} = A^{-2}(2A)^{-1} = \tfrac12A^{-3}, \quad A^{-1}B^{-2} = A^{-1}(2A)^{-2} = \tfrac14A^{-3}.$$

So their sum is

$$A^{-2}B^{-1} + A^{-1}B^{-2} = \tfrac12A^{-3} + \tfrac14A^{-3} = \tfrac34\,A^{-3} = \tfrac34\cdot\frac{8}{9}A = \frac{2}{3}A.$$

Step 4: Solve for $\lambda$.
The given equation is

$$\lambda\bigl(A^{-2}B^{-1} + A^{-1}B^{-2}\bigr) = B \;\Longrightarrow\; \lambda\cdot\frac{2}{3}A = 2A.$$

Cancel $A$ (invertible) to get

$$\lambda\cdot\tfrac{2}{3} = 2 \quad\Longrightarrow\quad \lambda = 3.$$

Therefore, $\lvert \lambda\rvert = 3$.