If
\[
P=
\begin{bmatrix}
0 & 1 & 2\\
2 & 3 & 4\\
1 & -1 & 0
\end{bmatrix}
\]
and
\[
Q=
\begin{bmatrix}
2 & -1 & 5\\
-4 & 2 & -4\\
2 & 2 & -4
\end{bmatrix},
\]
find the matrix product
\[
QP.
\]
Hence, solve the following system of linear equations using matrices:
\[
x-y=3,\qquad
2x+3y+4z=17,\qquad
y+2z=7.
\]
Show Hint
Solution and Explanation
Concept: If the product of two matrices is a scalar multiple of the identity matrix, i.e., \[ QP=kI, \] then \[ P^{-1}=\frac{1}{k}Q. \] The given system of linear equations can be written in the matrix form \[ PX=B, \] where \[ X= \begin{bmatrix} x\\ y\\ z \end{bmatrix}. \]
Step 1: Find the matrix product \(QP\). \[ Q= \begin{bmatrix} 2 & -1 & 5\\ -4 & 2 & -4\\ 2 & 2 & -4 \end{bmatrix}, \qquad P= \begin{bmatrix} 0 & 1 & 2\\ 2 & 3 & 4\\ 1 & -1 & 0 \end{bmatrix}. \] Multiplying, \[ QP= \begin{bmatrix} 2 & -1 & 5\\ -4 & 2 & -4\\ 2 & 2 & -4 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2\\ 2 & 3 & 4\\ 1 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 3 & -6 & 0\\ 0 & 6 & 0\\ 0 & 12 & 12 \end{bmatrix}. \] Thus, \[ \boxed{ QP= \begin{bmatrix} 3 & -6 & 0\\ 0 & 6 & 0\\ 0 & 12 & 12 \end{bmatrix} } \] Hence, \[ QP\ne6I. \]
Step 2: Solve the given system. Arrange the equations according to the rows of \(P\): \[ \begin{aligned} y+2z&=7,\\ 2x+3y+4z&=17,\\ x-y&=3. \end{aligned} \] Thus, \[ \begin{bmatrix} 0 & 1 & 2\\ 2 & 3 & 4\\ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 7\\ 17\\ 3 \end{bmatrix}. \] From \[ x-y=3, \] we get \[ x=y+3. \] Also, \[ y+2z=7 \] gives \[ y=7-2z. \] Substituting into the second equation, \[ 2(y+3)+3y+4z=17, \] \[ 5y+4z=11. \] Using \(y=7-2z\), \[ 5(7-2z)+4z=11, \] \[ 35-10z+4z=11, \] \[ -6z=-24, \] \[ z=4. \] Therefore, \[ y=7-2(4)=-1, \] and \[ x=y+3=2. \] Hence, \[ \boxed{x=2,\quad y=-1,\quad z=4.} \]
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