Question

Let \(A=\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}\) and \(B=\begin{bmatrix} -1 & -1 \\ 1 & 1 \end{bmatrix}\). If the variables \(x\) and \(y\) satisfy the matrix equation \((A^{-1}+B)\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 0 \\ -2 \end{bmatrix}\), then the value of \(x+3y\) is......

• A. \(8\)
• B. \(10\)
• C. \(12\)
• D. \(14\)

Hint:

Find the inverse of A, add B, and solve the resulting two linear equations.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For a \(2\times2\) matrix, \(A^{-1}=\dfrac{1}{|A|}\,\text{adj}A\). Here \(|A|=4-3=1\).

Step 2: Inverse of A
\(A^{-1}=\begin{bmatrix}2&-1\\-3&2\end{bmatrix}\).

Step 3: Add B
\(A^{-1}+B=\begin{bmatrix}2-1&-1-1\\-3+1&2+1\end{bmatrix}=\begin{bmatrix}1&-2\\-2&3\end{bmatrix}\).

Step 4: Solve the system
\(x-2y=0\) and \(-2x+3y=-2\).
From the first, \(x=2y\). Then \(-4y+3y=-2\), so \(y=2\) and \(x=4\).

Step 5: Required value
\(x+3y=4+6=10\).

Final Answer:
The value of x + 3y is 10. \[ \boxed{\text{(B) }10} \]