Question

Find \( x \) and \( y \) if \[ 2 \left[ \begin{array}{cc} 1 & 3 \\ 0 & x \end{array} \right] + \left[ \begin{array}{cc} y & 0 \\ 1 & 2 \end{array} \right] = \left[ \begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array} \right]. \]

Hint:

For solving linear systems of equations involving matrices, perform matrix operations step by step and translate the results into scalar equations.

Answer 1

Step 1: Perform the matrix multiplication and addition: \[ 2 \left[ \begin{array}{cc} 1 & 3 \\ 0 & x \end{array} \right] = \left[ \begin{array}{cc} 2 & 6 \\ 0 & 2x \end{array} \right], \] \[ \left[ \begin{array}{cc} y & 0 \\ 1 & 2 \end{array} \right] = \left[ \begin{array}{cc} y & 0 \\ 1 & 2 \end{array} \right]. \] Now, adding these matrices: \[ \left[ \begin{array}{cc} 2 & 6 \\ 0 & 2x \end{array} \right] + \left[ \begin{array}{cc} y & 0 \\ 1 & 2 \end{array} \right] = \left[ \begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array} \right]. \] This gives the system of equations: \[ 2 + y = 5 \quad \text{(Equation 1)} \] \[ 6 = 6 \quad \text{(Equation 2)} \] \[ 0 + 1 = 1 \quad \text{(Equation 3)} \] \[ 2x + 2 = 8 \quad \text{(Equation 4)}. \] Step 2: Solve Equation 1 for \( y \): \[ y = 5 - 2 = 3. \] Step 3: Solve Equation 4 for \( x \): \[ 2x + 2 = 8 \quad \Rightarrow \quad 2x = 6 \quad \Rightarrow \quad x = 3. \] Thus, the solution is \( x = 3 \) and \( y = 3 \). \bigskip