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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
Using elementary row operations find the inverse of matrix $\text{A} = \begin{pmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{pmatrix}$ and hence solve the following system of equations $3x - 3y + 4z = 21, 2x - 3y + 4z = 20, -y + z = 5.$
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Answer

$\text{A = I.A}$
$\Rightarrow \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A} $
$\text{Applying R}_{1}\rightarrow\text{R}_{1}-\text{R}_{2}$
$\Rightarrow\begin{bmatrix} 1 & 0 & 0 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}= \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A} $
$\text{Applying R}_{2}\rightarrow\text{R}_{2}+\text{2R}_{1}$
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}= \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A} $
$\text{Applying R}_{2}\rightarrow\text{R}_{2}+\text{4R}_{3}$
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix}= \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ 0 & 0 & 1 \end{bmatrix}\text{A} $
$\text{Applying R}_{3}\rightarrow\text{R}_{3}+\text{R}_{2}$
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}= \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}\text{A}$
$\therefore\text{A}^{-1} = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$
The matrix from of given equations
$\begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} \text{x} \\ \text{y}\\ \text{z} \end{bmatrix} = \begin{bmatrix} 21 \\ 20 \\ 5 \end{bmatrix} $
$\Rightarrow\text{AX = B, where X} = \begin{bmatrix} \text{x} \\ \text{y}\\ \text{z} \end{bmatrix}, \text{B}= \begin{bmatrix} 21 \\ 20 \\ 5 \end{bmatrix} $
$\Rightarrow\text{X = A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix} \begin{bmatrix} 21 \\ 20\\ 5 \end{bmatrix} = \begin{bmatrix} 1 \\ -2 \\ 3 \end{bmatrix} $
$\therefore \text{x = 1, y =}{ - 2, \text{z = 3}}$

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