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Solution of Simultaneous Linear Equations — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSSolution of Simultaneous Linear Equations5 Marks
Question
Show that the following system of linear equations is consistent and also find solution:
$x + y + z = 6$
$x + 2y + 3z = 14$
$x + 4y + 7z = 30$
✓

Answer

This system can be written as:
$\begin{bmatrix}1&1&1\\ 1&2&3\\ 1&4&7\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}6\\ 14\\ 30\end{bmatrix}$
or $\text{AX = B}$
$\text{|A|}=1{(2)}-1{(4)}+1{(2)}$
$=2-4+2$
$=0$
So, $A$ is singular, thus the given system has either no solutions or infinite solutions depending on as
$(\text{Adj A})\times\text{(B)}\neq0$ or $(\text{Adj A})\times\text{(B)}=0$
Let $C_{ij}$ be the co$-$factors of $a_{ij}$ in $A$
$\text{C}_{11}=2\\ \text{C}_{21}=-3\\ \text{C}_{31}=1$
$\text{C}_{12}=-4\\ \text{C}_{22}=6\\ \text{C}_{32}=-2$
$\text{C}_{13}=2\\ \text{C}_{23}=-3\\ \text{C}_{33}=1$
$\text{adj A}=\begin{bmatrix}2&-4&2\\ -3&6&-3\\ 1&-2&1\end{bmatrix}^\text{T}=\begin{bmatrix}2&-3&1\\ -4&6&-2\\ 2&-3&1\end{bmatrix}$
$(\text{adj A})\times\text{B}=\begin{bmatrix}2&-4&2\\ -3&6&-3\\ 1&-2&1\end{bmatrix}\begin{bmatrix}6\\ 14\\ 30\end{bmatrix}\begin{bmatrix}12-42+30\\ -24+84-60\\ 12-42+30\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$
So, $\text{AX = B}$ has infinite solutions.
Now, let $z = k$
So, $x + y = 6 - k$
$x + 2y = 14 - 3k$
which can be written as:
$\begin{bmatrix}1&1\\ 1&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}6-\text{k}\\ 14-\text{3k}\end{bmatrix}$
or $\text{AX = B}$
$\text{|A|}=1\neq0$
$\text{adj A}=\begin{bmatrix}2&-1\\-1&1\end{bmatrix}^\text{T}=\begin{bmatrix}2&-1\\-1&1\end{bmatrix}$
$\text{X = A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{adj A}\times\text{B}$
$\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\frac{1}{1}\begin{bmatrix}2&-1\\ -1&1\end{bmatrix}\begin{bmatrix}6-\text{k}\\ 14-3\text{k}\end{bmatrix}$
$=\begin{bmatrix}12-2\text{k}-14+3\text{k}\\ -6+\text{k}+14-3\text{k}\end{bmatrix}$
$=\begin{bmatrix}12-2\text{k}-14+3\text{k}\\ -6+\text{k}+14-3\text{k}\end{bmatrix}$
$\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}-2+\text{k}\\ 8-2\text{k}\end{bmatrix}$
Hence, $x = k - 2$
$y = 8 - 2k$
$z = k$

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