Question

Consider two matrices \(A=\begin{bmatrix}3 & -4 1 & -1\end{bmatrix}\) and \(B=\begin{bmatrix}6 & -13 5 & -10\end{bmatrix}\). If the following matrix equation holds true: \[ \left((A^{-1})^2 + B\right)\begin{bmatrix}x y\end{bmatrix} = \begin{bmatrix}0 0\end{bmatrix} \] Find the values of \(x\) and \(y\).

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

Hint:

When a matrix equation gives \(M\mathbf{x}=0\), the vector solution must satisfy the linear equations formed by matrix \(M\). Compute matrix operations carefully (inverse, square, addition) before solving the resulting system.

Answer 1

The Correct Option is D

Concept: If a matrix equation is of the form \[ M \begin{bmatrix}xy\end{bmatrix} = \begin{bmatrix}00\end{bmatrix} \] then a non-trivial solution exists when the matrix \(M\) becomes the zero matrix or produces a dependent system of equations. Here, \[ M = (A^{-1})^2 + B \] Hence we first compute \(A^{-1}\), then square it, and finally add matrix \(B\).

Step 1: Find \(A^{-1}\). For matrix \[ A= \begin{bmatrix} 3 & -4 1 & -1 \end{bmatrix} \] Determinant: \[ |A| = 3(-1) - (-4)(1) \] \[ |A| = -3 + 4 = 1 \] Thus, \[ A^{-1}= \begin{bmatrix} -1 & 4 -1 & 3 \end{bmatrix} \]

Step 2: Compute \((A^{-1})^2\). \[ (A^{-1})^2 = \begin{bmatrix} -1 & 4 -1 & 3 \end{bmatrix} \begin{bmatrix} -1 & 4 -1 & 3 \end{bmatrix} \] Multiplying matrices: \[ (A^{-1})^2= \begin{bmatrix} -3 & 8 -2 & 5 \end{bmatrix} \]

Step 3: Add matrix \(B\). \[ B= \begin{bmatrix} 6 & -13 5 & -10 \end{bmatrix} \] \[ (A^{-1})^2 + B = \begin{bmatrix} -3 & 8 -2 & 5 \end{bmatrix} + \begin{bmatrix} 6 & -13 5 & -10 \end{bmatrix} \] \[ = \begin{bmatrix} 3 & -5 3 & -5 \end{bmatrix} \]

Step 4: Form the linear equations. \[ \begin{bmatrix} 3 & -5 3 & -5 \end{bmatrix} \begin{bmatrix} xy \end{bmatrix} = \begin{bmatrix} 00 \end{bmatrix} \] This gives \[ 3x - 5y = 0 \]

Step 5: Solve the relation. \[ 3x = 5y \] \[ \frac{x}{y} = \frac{5}{3} \] Thus, \[ x=5,\qquad y=3 \] \[ \boxed{(x,y)=(5,3)} \]