Question

Let \(A = [\begin{array}{cc}3 & -4 \\ 1 & -1\end{array}]\) and \(B = [\begin{array}{cc}6 & -13 \\ 5 & -10\end{array}]\) be two matrices. If the variables \(x\) and \(y\) satisfy the matrix equation \(((A^{-1})^2+B)[\begin{array}{c}x \\ y\end{array}] = [\begin{array}{c}0 \\ 0\end{array}]\), then the ordered pair \((x,y) =\)

• A. \((3,5)\)
• B. \((10,7)\)
• C. \((4,6)\)
• D. \((5,3)\)

Hint:

Compute the inverse of A, square it, add B and solve the resulting homogeneous system.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We need the matrix \(M = (A^{-1})^2 + B\) and then solve \(M\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\).

Step 2: Key Formula or Approach:
\(|A| = 3(-1) - (-4)(1) = 1\). For a \(2 \times 2\) matrix, \(A^{-1} = \frac{1}{|A|}\begin{pmatrix} -1 & 4 \\ -1 & 3 \end{pmatrix}\).

Step 3: Detailed Explanation:
\(A^{-1} = \begin{pmatrix} -1 & 4 \\ -1 & 3 \end{pmatrix}\).
\((A^{-1})^2 = \begin{pmatrix} 1 - 4 & -4 + 12 \\ 1 - 3 & -4 + 9 \end{pmatrix} = \begin{pmatrix} -3 & 8 \\ -2 & 5 \end{pmatrix}\).
Add \(B\): \(M = \begin{pmatrix} 3 & -5 \\ 3 & -5 \end{pmatrix}\).
Both rows give \(3x - 5y = 0\), so \(\frac{x}{y} = \frac{5}{3}\).
Among the options, \((5, 3)\) satisfies this: \(15 - 15 = 0\).
Checking the others: \((3,5)\) gives \(9 - 25\), \((10, 7)\) gives \(30 - 35\), \((4, 6)\) gives \(12 - 30\), none zero.

Final Answer:
The pair is \((5, 3)\), option (D). \[ \boxed{(5,3)} \]