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 \( (x, y) \).
Show 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} \( (0,0) \).
Concept:
If a matrix \( M \) when multiplied by a vector gives the zero vector,
\[
M \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix},
\]
then one of the following cases must hold:
the determinant of \( M \) is zero, which leads to 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: Calculate the successive powers of matrix \( A \).
\[
A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}
\]
\[
A^2 = \begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix}, \quad
A^3 = \begin{bmatrix} 7 & -12 \\ 3 & -5 \end{bmatrix}
\]
\[
A^4 = \begin{bmatrix} 9 & -16 \\ 4 & -7 \end{bmatrix}, \quad
A^5 = \begin{bmatrix} 11 & -20 \\ 5 & -9 \end{bmatrix}
\]
Step 2: Add matrix \( B \) to \( A^5 \).
\[
A^5 + B
= \begin{bmatrix} 11 & -20 \\ 5 & -9 \end{bmatrix}
+ \begin{bmatrix} 29 & 49 \\ 1 & 2 \end{bmatrix}
= \begin{bmatrix} 40 & 29 \\ 6 & -7 \end{bmatrix}
\]
Step 3: Form the corresponding system of linear equations.
\[
\begin{bmatrix} 40 & 29 \\ 6 & -7 \end{bmatrix}
\begin{bmatrix} x \\ y \end{bmatrix}
=
\begin{bmatrix} 0 \\ 0 \end{bmatrix}
\]
Which gives:
\[
40x + 29y = 0 \quad \cdots (1)
\]
\[
6x - 7y = 0 \quad \cdots (2)
\]
Step 4: Solve the system.
From equation (2),
\[
y = \frac{6}{7}x
\]
Substituting into equation (1),
\[
40x + 29\left(\frac{6}{7}x\right) = 0
\]
\[
x\left(40 + \frac{174}{7}\right) = 0
\]
Since the coefficient of \(x\) is non-zero,
\[
x = 0 \quad \Rightarrow \quad y = 0
\]
