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MATRICES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMATRICES4 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.

C11 = 75,   C12 = 110,   C13 = 75

cofactors are C21 = 150, C22 = -100,  C23 = 0

C31 = 75,   C32 = 30,  C33 = -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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