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Solve the following.
Given that: A = (1 -1 0; 2 3 4; 0 1 2) and B = (2 2 -4; -4 2 -4; 2 -1 5), find AB. Using this…
Given that: $A = \begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix}$, find $AB$.
Using this result, solve the following system of equation:
$x - y = 3$, $2x + 3y + 4z = 17$ and $y + 2z = 7$
Answer
Answer
Solution: $x = 2$, $y = -1$, $z = 4$
$AB = \begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix}\begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix} = \begin{pmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix} = 6I$
So $A^{-1} = \frac{1}{6}B$.
The system is $AX = C$, where $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ and $C = \begin{pmatrix} 3 \\ 17 \\ 7 \end{pmatrix}$.
$X = A^{-1}C = \frac{1}{6}\begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix}\begin{pmatrix} 3 \\ 17 \\ 7 \end{pmatrix} = \frac{1}{6}\begin{pmatrix} 12 \\ -6 \\ 24 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}$
Hence $x = 2$, $y = -1$, $z = 4$.
Final answer: $AB = 6I$; $x = 2$, $y = -1$, $z = 4$