If A = 2&-3&5 3&2&-4 1&1&-2 , find A^ -1 . Using A^ -1 solve the … - Vidyadip

Question

If $\text{A} = \begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix},$ find $A^{-1}.$ Using $A^{-1}$ solve the system of equations:

$2x - 3y + 5z = 11$

$3x + 2y - 4z = -5$

$x + y - 2z = -3$

✓

Answer

$\text{Given:}\ \text{Matrix A}=\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix}$ $\therefore\ \text{|A|}=\begin{vmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{vmatrix}$
$\Rightarrow\ \text{|A|}=2(-4+4)-3(-3)(-6+4)+5(3-2)=0-6+5=-1\neq0$
$\therefore\ \text{A}^{-1}\ \text{exists and A}^{-1}=\frac{1}{\text{|A|}}\text{(adj. A)} \dots\dots(1)$
Now, $A_{11}= 0, A_{12} = 2, A_{13} = 1$ and $A_{21} = -1, A_{22} = -9, A_{23} = -5$ and $A_{31} = 2, A_{32} = 23, A_{13} = 13$
$\therefore\ \text{adj.A}=\begin{bmatrix}0&2&1\\-1&-9&-5\\2&23&13\end{bmatrix}=\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix}$ $\therefore$ From eq. (1),
$\text{A}^{-1}=\frac{1}{-1}\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix}=\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}$
Now, Matrix form of given equations is AX = B
$ \Rightarrow\ \begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}11\\-5\\-3\end{bmatrix}$
$\text{Here}\ \text{A}=\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix},\ \text{X}=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{and B}=\begin{bmatrix}11\\-5\\-3\end{bmatrix}$ Therefore, solution is unique and
$X = A^{-1}B$ $\Rightarrow\ \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}\begin{bmatrix}11\\-5\\-3\end{bmatrix}$
$=\begin{bmatrix}0-5+6\\-22-45+69\\-11-25+39\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}$
Therefore, $x = 1, y = 2$ and $ z = 3$

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