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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:$\frac{2}{\text{x}}+\frac{3}{\text{y}}+\frac{10}{\text{z}}=4,\frac{4}{\text{x}}-\frac{6}{\text{y}}+\frac{5}{\text{z}}=1,\frac{6}{\text{x}}+\frac{9}{\text{y}}-\frac{20}{\text{z}}=2:\text{x},\text{y},\text{z}\neq0$
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Answer

Let 
$\frac{1}{\text{x}}=\text{u},\frac{1}{\text{y}}=\text{v},\frac{1}{\text{z}}=\text{w}$
The above system can be written as:
$\begin{bmatrix}2&3&10\\ 4&-6&5\\ 6&9&-20\end{bmatrix}\begin{bmatrix}\text{u}\\ \text{v}\\ \text{w}\end{bmatrix}=\begin{bmatrix}4\\ 1\\ 2\end{bmatrix}$
Or $AX = B$
$\text{|A|}=2{(75)}-3{(-110)}+10{(72)}=1200\neq0$
So, the above system has a unique solution, given by 
$\text{X}=\text{A}^{-1}\text{B}$
Let $C_{ij}$ be the co-factors of $a_{ij}$ in $A$
$\text{C}_{11}=75\\ \text{C}_{21}=150\\ \text{C}_{31}=75$
$\text{C}_{12}=110\\ \text{C}_{22}=-100\\ \text{C}_{32}=30$
$\text{C}_{13}=72\\ \text{C}_{23}=0\\ \text{C}_{33}=-24$
$\text{adj A}=\begin{bmatrix}75&110&72\\ 150&-100&0\\ 75&30&-24\end{bmatrix}^\text{T}=\begin{bmatrix}75&150&75\\ 110&-100&30\\ 72&0&-24\end{bmatrix}$
Now,
$\text{X}=\text{A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{(Adj A)}\times\text{B}$
$=\frac{1}{1200}\begin{bmatrix}75&150&75\\ 110&-100&30\\ 72&0&-24\end{bmatrix}\begin{bmatrix}4\\ 1\\ 2\end{bmatrix}$
$\begin{bmatrix}\text{u}\\ \text{v}\\ \text{w}\end{bmatrix}=\frac{1}{1200}\begin{bmatrix}600\\ 400\\ 240\end{bmatrix}=\begin{bmatrix}\frac{1}{2}\\ \frac{1}{3}\\ \frac{1}{5}\end{bmatrix}$
Hence, $x = 2, y = 3, z = 5$

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