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Model Paper 2 — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 25 Marks
Question
Solve the system of the following equations: $($Using matrices$):$
$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4 ; \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1 ; \frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$
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Answer

Put $\frac{1}{x}=u, \frac{1}{y}=v$ and $\frac{1}{z}=w$ in the given equations,
$2 u+3 v+10 w=4 ; 4 u-6 v+5 w=1 ; 6 u+9 v-20 w=2$
$\therefore$ The matrix form of given equations is $\left[\begin{array}{ccc}2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20\end{array}\right]\left[\begin{array}{l}x \\ v \\ z\end{array}\right]=\left[\begin{array}{l}4 \\ 1 \\ 2\end{array}\right][ AX = B ]$
Here, $A=\left[\begin{array}{ccc}2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20\end{array}\right], X\left[\begin{array}{l}x \\ v \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 1 \\ 2\end{array}\right]$
$\begin{array}{l}
\therefore|A|=\left|\begin{array}{ccc}
2 & 3 & 10 \\
4 & -6 & 5 \\
6 & 9 & -20
\end{array}\right| \\
=2(120-45)-3(-80-30)+10(36+36) \\
=150+330+750=1200 \neq 0
\end{array}$
$\therefore A^{-1}$ exists and unique solution is $X=A^{-1} B \ldots.(i)$
Now $A_{11}=75, A_{12}=110, A_{13}=72$ and $A_{21}=150, A_{22}=-100, A_{23}=0$ and $A_{31}=75, A_{32}=30, A_{33}=-24$
$\therefore a d j . A=\left[\begin{array}{ccc}
75 & 110 & 72 \\
150 & -100 & 0 \\
75 & 30 & -24
\end{array}\right]^{\prime}=\left[\begin{array}{ccc}
75 & 150 & 75 \\
110 & -100 & 30 \\
72 & 0 & -24
\end{array}\right]$
And $A^{-1}=\frac{a d j . A}{|A|}=\frac{1}{1200}\left[\begin{array}{ccc}75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24\end{array}\right]$
$\therefore$ From eq. $(i),$
$\begin{array}{l}
{\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{1200}\left[\begin{array}{ccc}
75 & 150 & 75 \\
110 & -100 & 30 \\
72 & 0 & -24
\end{array}\right]\left[\begin{array}{l}
4 \\
1 \\
2
\end{array}\right]} \\
\Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{1200}\left[\begin{array}{c}
300+150+150 \\
440-100+60 \\
288+0-48
\end{array}\right]
\end{array}$
$\begin{array}{l}=\frac{1}{1200}\left[\begin{array}{l}600 \\ 400 \\ 240\end{array}\right] \\ =\left[\begin{array}{l}\frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{5}\end{array}\right] \\ \therefore u=\frac{1}{2}, v=\frac{1}{3}, w=\frac{1}{5} \\ \Rightarrow x=\frac{1}{u}=2, y=\frac{1}{v}=3, z=\frac{1}{w}=5\end{array}$

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