Question

If \[ P = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \quad \text{and} \quad Q = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 1 & -1 & 5 \end{bmatrix} \] find \( QP \) and hence solve the following system of equations using matrix method:
\[ x - y = 3,\quad 2x + 3y + 4z = 13,\quad y + 2z = 7 \]

Hint:

Matrix multiplication can be used to simplify systems of equations. If a matrix product produces a diagonal matrix, solving the system becomes very easy.

Answer 1

Step 1: Write matrices clearly.
\[ P = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \] \[ Q = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 1 & -1 & 5 \end{bmatrix} \] Step 2: Find the matrix product \( QP \).
Multiply rows of \( Q \) with columns of \( P \).
First row calculations:
\[ (2)(1) + (2)(2) + (-4)(0) = 6 \] \[ (2)(-1) + (2)(3) + (-4)(1) = 0 \] \[ (2)(0) + (2)(4) + (-4)(2) = 0 \] Second row calculations:
\[ (-4)(1) + (2)(2) + (-4)(0) = 0 \] \[ (-4)(-1) + (2)(3) + (-4)(1) = 6 \] \[ (-4)(0) + (2)(4) + (-4)(2) = 0 \] Third row calculations:
\[ (1)(1) + (-1)(2) + (5)(0) = -1 \] \[ (1)(-1) + (-1)(3) + (5)(1) = 1 \] \[ (1)(0) + (-1)(4) + (5)(2) = 6 \] Thus
\[ QP = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ -1 & 1 & 6 \end{bmatrix} \] Step 3: Write the system in matrix form.
\[ P \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 13 \\ 7 \end{bmatrix} \] Multiply both sides by \( Q \):
\[ QP \begin{bmatrix} x \\ y \\ z \end{bmatrix} = Q \begin{bmatrix} 3 \\ 13 \\ 7 \end{bmatrix} \] Step 4: Evaluate the right side.
\[ Q \begin{bmatrix} 3 \\ 13 \\ 7 \end{bmatrix} = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 1 & -1 & 5 \end{bmatrix} \begin{bmatrix} 3 \\ 13 \\ 7 \end{bmatrix} \] \[ = \begin{bmatrix} 6 + 26 - 28 \\ -12 + 26 - 28 \\ 3 - 13 + 35 \end{bmatrix} \] \[ = \begin{bmatrix} 4 \\ -14 \\ 25 \end{bmatrix} \] Step 5: Solve the resulting system.
\[ \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ -1 & 1 & 6 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 4 \\ -14 \\ 25 \end{bmatrix} \] From first equation:
\[ 6x = 4 \] \[ x = \frac{2}{3} \] From second equation:
\[ 6y = -14 \] \[ y = -\frac{7}{3} \] From third equation:
\[ -x + y + 6z = 25 \] Substitute \( x = \frac{2}{3}, y = -\frac{7}{3} \)
\[ -\frac{2}{3} - \frac{7}{3} + 6z = 25 \] \[ -\frac{9}{3} + 6z = 25 \] \[ -3 + 6z = 25 \] \[ 6z = 28 \] \[ z = \frac{14}{3} \] Step 6: Final solution.
\[ x = \frac{2}{3},\quad y = -\frac{7}{3},\quad z = \frac{14}{3} \] Final Answer:
\[ (x, y, z) = \left( \frac{2}{3}, -\frac{7}{3}, \frac{14}{3} \right) \]