Question

If \(X = A^{-1}B,\) Where \(A = \left[ \begin{array}{cc}1 & -1 2 & 1 \end{array} \right],B = \left[ \begin{array}{c}3 6 \end{array} \right]\) and \(X = \left[ \begin{array}{c}x_{1} x_{2} \end{array} \right],\) then \(x_{1} + x_{2} =\)

• A. 3
• B. 4
• C. 5
• D. 6
• E. 7

Hint:

For \(2\times2\) matrix, \(A^{-1} = \frac{1}{ad-bc}\left[\begin{array}{cc}d & -b -c & a\end{array}\right]\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Solve \(AX = B\). \(\det A = 1\cdot1 - (-1)\cdot2 = 1+2=3\). \(A^{-1} = \frac{1}{3}\left[\begin{array}{cc}1 & 1 -2 & 1\end{array}\right]\). Then \(X = A^{-1}B = \frac{1}{3}\left[\begin{array}{cc}1 & 1 -2 & 1\end{array}\right]\left[\begin{array}{c}36\end{array}\right] = \frac{1}{3}\left[\begin{array}{c}3+6 -6+6\end{array}\right] = \left[\begin{array}{c}30\end{array}\right]\). So \(x_1+x_2=3\).

Step 2: Detailed Explanation:
Alternatively solve system: \(x_1 - x_2 = 3\), \(2x_1 + x_2 = 6\). Add: \(3x_1=9 \implies x_1=3, x_2=0\). Sum=3.

Step 3: Final Answer:
Option (A).