Question

Let \( A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \) and \( B = \begin{bmatrix} 29 & 49 \\1 & 2 \end{bmatrix} \). If \( (A^5 + B)\begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \), then find \( (x, y) \).

Hint:

If a square matrix multiplying a vector gives the zero vector and the matrix is non-singular, the only possible solution is the \textbf{trivial solution} \( (0,0) \).

Answer 1

Correct Answer: 0

Concept: If a matrix \( M \) multiplies a vector and gives the zero vector, \[ M \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}, \] then either:
the determinant of \( M \) is zero (infinitely many solutions), or
the determinant of \( M \) is non-zero, in which case the only solution is the trivial solution \( x = 0, y = 0 \).
Step 1: Compute successive powers of \( A \). \[ A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \] \[ A^2 = \begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix}, \quad A^3 = \begin{bmatrix} 7 & -12 \\ 3 & -5 \end{bmatrix} \] \[ A^4 = \begin{bmatrix} 9 & -16 \\ 4 & -7 \end{bmatrix}, \quad A^5 = \begin{bmatrix} 11 & -20 \\ 5 & -9 \end{bmatrix} \]
Step 2: Add matrix \( B \). \[ A^5 + B = \begin{bmatrix} 11 & -20 \\ 5 & -9 \end{bmatrix} + \begin{bmatrix} 29 & 49 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 40 & 29 \\ 6 & -7 \end{bmatrix} \]
Step 3: Form the system of linear equations. \[ \begin{bmatrix} 40 & 29 \\ 6 & -7 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] This gives: \[ 40x + 29y = 0 \quad \cdots (1) \] \[ 6x - 7y = 0 \quad \cdots (2) \]
Step 4: Solve the equations. From equation (2): \[ y = \frac{6}{7}x \] Substitute into equation (1): \[ 40x + 29\left(\frac{6}{7}x\right) = 0 \] \[ x\left(40 + \frac{174}{7}\right) = 0 \] Since the coefficient is non-zero, \[ x = 0 \quad \Rightarrow \quad y = 0 \]