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MATRICES — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMATRICES5 Marks
Question
Using matrix method, solve the following system of equations:
$\frac{2}{\text{x}}+\frac{3}{\text{y}}+\frac{10}{\text{z}}=4,\frac{4}{\text{x}}-\frac{6}{\text{x}}+\frac{5}{\text{z}}=1,\frac{6}{\text{x}}+\frac{9}{\text{y}}-\frac{20}{\text{z}}=2; \text{ x,y,z,}\neq0$
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Answer

Writing the given system of equations as  $\begin{pmatrix} 2 & 3 & 10 \\ 4 & -6 & 5\\ 6 & 9 & -20 \end{pmatrix}\begin{pmatrix} \frac{1}{\text{x}} \\ \frac{1}{\text{y}} \\ \frac{1}{\text{z}} \end{pmatrix}=\begin{pmatrix} 4 \\ 1 \\ 2 \end{pmatrix} \text{or } $
$\text{A}\cdot\text{X}=B |A| = 2(120-45)-3(-80-30)+10(36+36) = 1200,$
$\therefore X = A^{-1} B.C_{11} = 75, C_{12 }= 110, C_{13} = 75$
cofactors are $C_{21} = 150, C_{22} = -100, C_{23} = 0,C_{31} = 75, C_{32} = 30, C_{33} = -24$
$A^{-1} =\frac{1}{1200}\begin{pmatrix} 75 & 150 & 75 \\ 110 & -100 & 30\\ 72 & 0 & -24 \end{pmatrix}$
$\therefore$ $\begin{pmatrix} \frac{1}{\text{x}} \\ \frac{1}{\text{y}} \\ \frac{1}{\text{z}} \end{pmatrix}$$=\frac{1}{1200}\begin{pmatrix} 75 & 150 & 75 \\ 110 & -100 & 30\\ 72 & 0 & -24 \end{pmatrix}$$\begin{pmatrix} 4 \\ 1 \\ 2 \end{pmatrix}$ $=\begin{pmatrix} \frac{1}{\text{2}} \\ \frac{1}{\text{3}} \\ \frac{1}{\text{5}} \end{pmatrix}$
$\therefore x = 2, y = 3, z = 5.$

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