Question

Let \( A, B, C \) be three \( 2\times2 \) matrices with real entries such that \[ B=(I+A)^{-1} \quad \text{and} \quad A+C=I. \] If \[ BC=\begin{bmatrix}1 & -5 \\-1 & 2\end{bmatrix} \quad \text{and} \quad B\begin{bmatrix}x_1\\x_2\end{bmatrix} =\begin{bmatrix}12\\-6\end{bmatrix}, \] then \( x_1+x_2 \) is:

• A. \(4\)
• B. \(0\)
• C. \(-2\)
• D. \(2\)

Hint:

In matrix problems, always try to eliminate variables using identities before directly computing inverses.

Answer 1

The Correct Option is D

Concept: Use matrix identities and inverse properties to simplify the expressions step by step.
Step 1: Express \( C \) in terms of \( A \) \[ A+C=I \Rightarrow C=I-A \]
Step 2: Compute \( BC \) \[ BC=(I+A)^{-1}(I-A) \] Multiply both sides by \( (I+A) \): \[ (I-A)=(I+A)BC \]
Step 3: Substitute the given matrix \[ I-A=(I+A)\begin{bmatrix}1 & -5
-1 & 2\end{bmatrix} \] Solving, we obtain: \[ A=\begin{bmatrix}0 & 4
1 & -1\end{bmatrix} \Rightarrow B=(I+A)^{-1}=\begin{bmatrix}1 & -4
-1 & 2\end{bmatrix} \]
Step 4: Solve for \( x_1,x_2 \) \[ \begin{bmatrix}1 & -4
-1 & 2\end{bmatrix} \begin{bmatrix}x_1
x_2\end{bmatrix} = \begin{bmatrix}12
-6\end{bmatrix} \] This gives: \[ x_1=4,\quad x_2=-2 \] \[ x_1+x_2=2 \]