Using elementary transformations, find the inverse of the matrix A = 8 & 4 & 3 2 & 1 & 1 1 & 2 & 2 and use it to solve the following system of linear equations: 8x + 4y +3z = 19 2x + y + z = 5 x + 2y + 2z = 7

Writing 8 & 4 & 3 2 & 1 & 1 1 & 2 & 2 = 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 A R_ 1 R_ 3 1 & 2 & 2 2 & 1 & 1 8 & 4 & 3 = 0 & 0 & 1 0 & 1 & 0 1 & 0 & 0 A R_ 1 & R_ 1 - & 2R _ 2 R_ 3 & R_ 3 - & 4R_ 2 -3 & 0 & 0 2 & 1 & 1 0 & 0 & -1 = 0 & -2 & 1 0 & 1 & 0 1 & -4 & 0 A R_ 1 & -1 3 R_ 1 R_ 3 & -R_ 3 1 & 0 & 0 2 & 1 & 1 0 & 0 & 1 = 0 & 2 3 & -1 3 0 & 1 & 0 -1 & 4 & 0 A 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 = 0 & 2 3 & -1 3 0 & -1 3 & 2 3 -1 & 4 & 0 A R_ 2 _ 2 - R_ 3 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 = 0 & 2 3 & -1 3 1 & -13 3 & 2 3 -1 & 4 & 0 A ^ -1 = 0 & 2 3 & -1 3 1 & -13 3 & 2 3 -1 & 4 & 0 AX = B = A^ -1 B x y z = 0 & 2 3 & -1 3 1 & -13 3 & 2 3 -1 & 4 & 0 19 5 7 = 1 2 1 x = 1, y = 2, z = 1

Using elementary transformations, find the inverse of the matrix …

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Question

Using elementary transformations, find the inverse of the matrix $\text{A} = \begin{bmatrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{bmatrix}$ and use it to solve the following system of linear equations:
$\text{8x + 4y +3z = 19}$
$\text{2x + y + z = 5} $
$\text{x + 2y + 2z = 7}$

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Answer

$\text{Writing} \begin{bmatrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A} $
$\text{R}_{1}\leftrightarrow \text{R}_{3} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 1 \\ 8 & 4 & 3 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}\text{A} $
$ \begin{matrix} \text{R}_{1}\rightarrow & \text{R}_{1} - & 2\text{R} _{2} \\ \text{R}_{3}\rightarrow& \text{R}_{3} - & 4\text{R}_{2} \\ \end{matrix} $ $ \begin{bmatrix} -3 & 0 & 0 \\ 2 & 1 & 1 \\ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} 0 & -2 & 1 \\ 0 & 1 & 0 \\ 1 & -4 & 0 \end{bmatrix}\text{A} $
$ \begin{matrix} \text{R}_{1}\rightarrow & -\frac{1}{3}\text{R}_{1} \\ \text{R}_{3}\rightarrow & -\text{R}_{3}\\ \end{matrix} $ $ \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & \frac{2}{3} & -\frac{1}{3} \\ 0 & 1 & 0 \\ -1 & 4 & 0 \end{bmatrix}\text{A} $
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & \frac{2}{3} & -\frac{1}{3} \\ 0 & -\frac{1}{3} & \frac{2}{3} \\ -1 & 4 & 0 \end{bmatrix}\text{A}$
$\text{R}_{2}\rightarrow\text{R}_{2} - \text{R}_{3} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & \frac{2}{3} & -\frac{1}{3} \\ 1 & -\frac{13}{3} & \frac{2}{3} \\ -1 & 4 & 0 \end{bmatrix}\text{A} $
$\therefore\text{A}^{-1} = \begin{bmatrix} 0 & \frac{2}{3} & -\frac{1}{3} \\ 1 & -\frac{13}{3} & \frac{2}{3} \\ -1 & 4 & 0 \end{bmatrix} $
$\text{AX = B}\Rightarrow\text{X = A}^{-1}\text{B}$
$\therefore \begin{bmatrix} \text{x} \\ \text{y}\\ \text{z} \end{bmatrix} = \begin{bmatrix} 0 & \frac{2}{3} & -\frac{1}{3} \\ 1 & -\frac{13}{3} & \frac{2}{3} \\ -1 & 4 & 0 \end{bmatrix} \begin{bmatrix} 19 \\ 5 \\ 7 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} $
$\therefore \text{x = 1, y = 2, z = 1}$

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