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

Solution:

We are given the matrices \( P \) and \( Q \): \[ P = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \quad \text{and} \quad Q = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 1 & -1 & 5 \end{bmatrix} \]

Step 1: Find \( QP \).
The product of two matrices \( QP \) is found by multiplying the rows of \( Q \) with the columns of \( P \). \[ QP = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 1 & -1 & 5 \end{bmatrix} \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \] Let's compute each element of the resulting matrix: - First row, first column: \[ (2 \times 1) + (2 \times 2) + (-4 \times 0) = 2 + 4 + 0 = 6 \] - First row, second column: \[ (2 \times -1) + (2 \times 3) + (-4 \times 1) = -2 + 6 - 4 = 0 \] - First row, third column: \[ (2 \times 0) + (2 \times 4) + (-4 \times 2) = 0 + 8 - 8 = 0 \] - Second row, first column: \[ (-4 \times 1) + (2 \times 2) + (-4 \times 0) = -4 + 4 + 0 = 0 \] - Second row, second column: \[ (-4 \times -1) + (2 \times 3) + (-4 \times 1) = 4 + 6 - 4 = 6 \] - Second row, third column: \[ (-4 \times 0) + (2 \times 4) + (-4 \times 2) = 0 + 8 - 8 = 0 \] - Third row, first column: \[ (1 \times 1) + (-1 \times 2) + (5 \times 0) = 1 - 2 + 0 = -1 \] - Third row, second column: \[ (1 \times -1) + (-1 \times 3) + (5 \times 1) = -1 - 3 + 5 = 1 \] - Third row, third column: \[ (1 \times 0) + (-1 \times 4) + (5 \times 2) = 0 - 4 + 10 = 6 \] Therefore, the product \( QP \) is: \[ QP = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ -1 & 1 & 6 \end{bmatrix} \]

Step 2: Solve the system of equations using the matrix method.
The system of equations is: \[ x - y = 3 \] \[ 2x + 3y + 4z = 13 \] \[ y + 2z = 7 \] We can express the system as a matrix equation: \[ \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 13 \\ 7 \end{bmatrix} \] This is of the form \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the column vector of unknowns, and \( B \) is the column vector of constants. To solve for \( X \), we compute: \[ X = A^{-1}B \] We have already calculated the matrix \( QP \), which is \( A^{-1} \), so we use: \[ X = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ -1 & 1 & 6 \end{bmatrix} \begin{bmatrix} 3 \\ 13 \\ 7 \end{bmatrix} \] Perform the matrix multiplication: - First row: \[ (6 \times 3) + (0 \times 13) + (0 \times 7) = 18 \] - Second row: \[ (0 \times 3) + (6 \times 13) + (0 \times 7) = 78 \] - Third row: \[ (-1 \times 3) + (1 \times 13) + (6 \times 7) = -3 + 13 + 42 = 52 \] Thus, the solution vector is: \[ X = \begin{bmatrix} 18 \\ 78 \\ 52 \end{bmatrix} \] So, the solution to the system is: \[ x = 18, \quad y = 78, \quad z = 52 \]

Final Answer:
The solution to the system of equations is: \[ x = 18, \quad y = 78, \quad z = 52 \]