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DETERMINANTS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDETERMINANTS5 Marks
Question
Solve the following system of equations by matrix method:$\frac{2}{\text{x}}-\frac{3}{\text{y}}+\frac{3}{\text{z}}=10$
$\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}=10$
$\frac{3}{\text{x}}-\frac{1}{\text{y}}+\frac{2}{\text{z}}=13$
✓

Answer

Let $\frac{1}{\text{x}}=\text{u},\frac{1}{\text{y}}=\text{v},\frac{1}{\text{z}}=\text{w}$
2u - 3v + 3w = 10
u + v + w = 10
3u - v + 2w = 13
Which can be written as:
$\begin{bmatrix}2&-3&3\\ 1&1&1\\ 3&-2&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}10\\ 10\\ 13\end{bmatrix}$
$\text{|A|}=2{(3)}+3{(-1)}+3{(-4)}$
$=6-3-12=-9\neq0$
Hence, the system has a unique solution, given by 
$\text{X}=\text{A}^{-1}\times\text{B}$
$\text{C}_{11}=3\ \text{C}_{21}=3\ \text{C}_{31}=-6$
$\text{C}_{12}=1\ \text{C}_{22}=-5\ \text{C}_{32}=1$
$\text{C}_{13}=-4\ \text{C}_{23}=-7\ \text{C}_{33}=5$
$\text{X}=\frac{1}{\text{|A|}}\text{(Adj A)}\times\text{(B)}$
$=\frac{1}{-9}\begin{bmatrix}3&3&-6\\ 1&-5&1\\ -4&-7&5\end{bmatrix}\begin{bmatrix}10\\ 10\\ 13 \end{bmatrix}$
$\frac{1}{-9}\begin{bmatrix}30+30-78\\ 10-50+13\\ -40-70+65\end{bmatrix}$
$\begin{bmatrix}\text{u}\\ \text{v}\\ \text{w}\end{bmatrix}=\frac{-1}{9}\begin{bmatrix}-18\\ -27\\ -45\end{bmatrix}=\begin{bmatrix}2\\ 3\\ 5\end{bmatrix}$
Hence, $\text{x}=\frac{1}{2},\text{y}=\frac{1}{3},\text{z}=\frac{1}{5}$

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