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

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsSolution of Simultaneous Linear Equations5 Marks
Question
Show that the following system of linear equations is consistent and also find solutions:
$5x +3y + 7z = 4$
$3x + 26y + 2z = 9$
$7x + 2y + 10z = 5$
✓

Answer

This can be written as: $\begin{bmatrix}5&3&7\\ 3&26&2\\ 7&2&10\end{bmatrix}
\begin{bmatrix}\text{X}\\ \text{y}\\ \text{z}\end{bmatrix}
=\begin{bmatrix}4\\ 9\\ 5\end{bmatrix}$ Or $\text{A}\text{X}=\text{B}$
$\text{|A|}=5{(256)}-3{(16)}+7{(6-182)}$ $=0$ So, A is singular. Thus, the given system is either inconsistent or it is consistent with infinitely many solutions according 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 $\text{C}_{11}=256\\ \text{C}_{21}=-16\\ \text{C}_{31}=-176$ $\text{C}_{12}=-16\\ \text{C}_{22}=1\\ \text{C}_{32}=11$ $\text{C}_{13}=-176\\ \text{C}_{23}=11\\ \text{C}_{33}=121$
$\text{adj A}=\begin{bmatrix}256&-16&-176\\ -16&1&11\\ -176&11&121\end{bmatrix}^\text{T}=\begin{bmatrix}256&-16&-176\\ -16&1&11\\ -176&11&121\end{bmatrix}$ $\text{adj}\text{A}\times\text{B}=\begin{bmatrix}256&-16&-176\\ -16&1&11\\ -176&11&121\end{bmatrix}\begin{bmatrix}4\\ 9\\ 5\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$
Thus, AX = B has infinite many solutions. Now, let z = kThen, 5x + 3y = 4 - 7k
3x + 26y = 9 - 2k Which can be written as: $\begin{bmatrix}5&3\\ 3&26\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}4-7\text{k}\\ 9-2\text{k}\end{bmatrix}$
Or $\text{AX = B}$ $\text{|A|}=2$ $\text{adj A}=\begin{bmatrix}26&-3\\ -3&5\end{bmatrix}$ Now, $\text{x}=\text{A}^{-1}\text{B}=\frac{1}{\text{|A|}}\times\text{adj A}\times\text{B}$
$=\frac{1}{121}\begin{bmatrix}26&-3\\ -3&5\end{bmatrix}\begin{bmatrix}4-7\text{k}\\ 9-2\text{k}\end{bmatrix}$ $=\frac{1}{121}\begin{bmatrix}77-176\text{k}\\ 11\text{k}+33\end{bmatrix}$
$\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{7-16\text{k}}{11}\\ \frac{\text{k}+3}{11}\end{bmatrix}$ There values of x, y, z satisfies the third eq. Hence, $\text{x}=\frac{7-16\text{k}}{11},\text{y}=\frac{\text{k}+3}{11},\text{z}=\text{k}$

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