Use product 1&-1&2 0&2&-3 3&-2&4 -2&0&1 9&2&-3 6&1&-2 to solve th… - Vidyadip

Question

Use product $\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}$ to solve the system of equations $x + 3z = 9, -x + 2y - 2z = 4, 2x - 3y + 4z = -3.$

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Answer

Suppose, $\text{A}=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\text{B}=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}$
$\text{A}\times\text{B}=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}$
$=\begin{bmatrix}-2-9+12&0-2+2&1+3-4\\0+18-18&0+4-3&0-6+6\\-6-18+24&0-4+4&3+6-8\end{bmatrix}$
$=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
Since, $A \times B = I,$
$\therefore$ $B = A^{-1}.....(1)$
Now, the given system of equations is
$x + 3z = 9$
$-x + 2y - 2z = 4$
$2x - 3y + 4z = -3$
This can also be presented as:
$\begin{bmatrix}1&0&3\\-1&2&-2\\2&-3&4\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}9\\4\\-3\end{bmatrix}$
Here, we can observe that $\begin{bmatrix}1&0&3\\-1&2&-2\\2&-3&4\end{bmatrix}=\text{A}^\text{T}$
So, $\text{A}^\text{T}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}9\\4\\-3\end{bmatrix}$
Multiply the above expression by $(A^T)^{-1}.$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=(\text{A}^\text{T})^{-1}\begin{bmatrix}9\\4\\-3\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=(\text{A}^{-1})^\text{T}\begin{bmatrix}9\\4\\-3\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\text{B}^\text{T}\begin{bmatrix}9\\4\\-3\end{bmatrix}$ $[\text{Using}\ (1)]$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}^\text{T}\begin{bmatrix}9\\4\\-3\end{bmatrix}$
$=\begin{bmatrix}-2&9&6\\0&2&1\\1&-3&-2\end{bmatrix}\begin{bmatrix}9\\4\\-3\end{bmatrix}$
$=\begin{bmatrix}-18+36-18\\0+8-3\\9-12+6\end{bmatrix}$
$=\begin{bmatrix}0\\5\\3\end{bmatrix}$
Hence, x = 0, y = 5 and z = 3.

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