Question

Find x, y, z if
\[ \begin{bmatrix} 5 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 & -2 \\ 1 & -2 & 3 \\ -1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x-1 \\ y+1 \\ 2z \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}. \]

Hint:

When multiplying matrices, always check that the inner dimensions match (e.g., \(m \times n\) times \(n \times p\)). The resulting matrix will have the outer dimensions (\(m \times p\)). Proceed with multiplication step-by-step to avoid errors.

Answer 1

Step 1: Multiply the first two matrices. \[ \begin{bmatrix} 5 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix}_{2\times3} \begin{bmatrix} 0 & 1 & -2 \\ 1 & -2 & 3 \\ -1 & 1 & 1 \end{bmatrix}_{3\times3} = \begin{bmatrix} (0+1-0) & (5-2+0) & (-10+3+0) \\ (0+1-1) & (1-2+1) & (-2+3+1) \end{bmatrix} = \begin{bmatrix} 1 & 3 & -7 \\ 0 & 0 & 2 \end{bmatrix} \] Step 2: Multiply the resulting matrix by the column vector. \[ \begin{bmatrix} 1 & 3 & -7 \\ 0 & 0 & 2 \end{bmatrix}_{2\times3} \begin{bmatrix} x-1 \\ y+1 \\ 2z \end{bmatrix}_{3\times1} = \begin{bmatrix} 1(x-1) + 3(y+1) - 7(2z) \\ 0(x-1) + 0(y+1) + 2(2z) \end{bmatrix} = \begin{bmatrix} x - 1 + 3y + 3 - 14z \\ 4z \end{bmatrix} = \begin{bmatrix} x + 3y - 14z + 2 \\ 4z \end{bmatrix} \] Step 3: Set the resulting matrix equal to the right-hand side and solve the system of equations. \[ \begin{bmatrix} x + 3y - 14z + 2 \\ 4z \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \] From the second row: \( 4z = 1 \implies z = \tfrac{1}{4} \). From the first row: \( x + 3y - 14z + 2 = 2 \implies x + 3y - 14z = 0 \). Substitute \( z = \tfrac{1}{4} \): \[ x + 3y - 14\left(\tfrac{1}{4}\right) = 0 \implies x + 3y - \tfrac{7}{2} = 0 \implies x + 3y = \tfrac{7}{2}. \] The problem does not yield a unique solution for \(x\) and \(y\). The solution is \(z = \tfrac{1}{4}\) and any \(x, y\) that satisfy \(x + 3y = \tfrac{7}{2}\).