The matrix method for solving systems of linear equations is versatile and applicable in many fields such as: Electrical engineering (circuit analysis), Economics (input-output models), Computer graphics (transformations), Optimization problems (linear programming), Network theory (flow problems), Physics (mechanics and structural analysis). In each of these applications, the matrix method provides an efficient way to handle large systems of equations and find solutions to complex real-world problems
Q. Solve the system of linear equations, using matrix method
[ JKBOSE 2023 SET X, Most Imp. ]
$$2 x+y+z=1$$
$$
\begin{array}{r}
x-2 y-z=\frac{3}{2} \\
3 y-5 z=9
\end{array}
$$
Sol. The given equations are
$$
\begin{aligned}
2 x+y+z & =1 \\
x-2 y-z & =\frac{3}{2} \\
3 y-5 z & =9 \text { or } 0 . x+3 y-5 z=9
\end{aligned}
$$
Their matrix form is $$\left[\begin{array}{rrr}2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ \frac{3}{2} \\ 9\end{array}\right]$$
$$
(\Rightarrow \mathrm{AX}=\mathrm{B})
$$
Comparing $$\mathrm{A}=\left[\begin{array}{rrr}2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5\end{array}\right], \mathrm{X}=\left[\begin{array}{c}x \\ y \\ z\end{array}\right]$$ and $$\mathrm{B}=\left[\begin{array}{c}1 \\ \frac{3}{2} \\ 9\end{array}\right]$$
$$
|A|=\left|\begin{array}{rrr}
2 & 1 & 1 \\
1 & -2 & -1 \\
0 & 3 & -5
\end{array}\right|
$$
Expanding along first row, $$=2(10+3)-1(-5-0)+1(3-0)$$ or $$|\mathrm{A}|=2(13)+5+3=26+5+3=34 \neq 0$$
Solution is unique and $$\mathbf{X}=\mathbf{A}^{-1} \mathbf{B}=\frac{1}{|\mathrm{~A}|}$$ (adj. A) B …(i)
Let us find adj. A
$$
\mathrm{A}_{11}=+\left|\begin{array}{rr}
-2 & -1 \\
3 & -5
\end{array}\right|=10+3=13
$$
$$
\begin{aligned}
& A_{12}=-\left|\begin{array}{ll}
1 & -1 \\
0 & -5
\end{array}\right|=-(-5-0)=5, \\
& \mathrm{~A}_{13}=+\left|\begin{array}{rr}
1 & -2 \\
0 & 3
\end{array}\right|=(3-0)=3, \\
& A_{21}=-\left|\begin{array}{rr}
1 & 1 \\
3 & -5
\end{array}\right|=-(-5-3)=8, \\
& \mathrm{~A}_{22}=+\left|\begin{array}{rr}
2 & 1 \\
0 & -5
\end{array}\right|=(-10-0)=-10, \\
& \mathrm{~A}_{23}=-\left|\begin{array}{ll}
2 & 1 \\
0 & 3
\end{array}\right|=-(6-0)=-6, \\
& \mathrm{~A}_{31}=+\left|\begin{array}{rr}
1 & 1 \\
-2 & -1
\end{array}\right|=(-1+2)=1 \text {, } \\
& A_{32}=-\left|\begin{array}{rr}
2 & 1 \\
1 & -1
\end{array}\right|=-(-2-1)=3, \\
& \mathrm{~A}_{33}=+\left|\begin{array}{rr}
2 & 1 \\
1 & -2
\end{array}\right|=-4-1=-5 . \\
& \quad \text { Adj. } A=\left[\begin{array}{rrr}
13 & 5 & 3 \\
8 & -10 & -6 \\
1 & 3 & -5
\end{array}\right]^{\prime}=\left[\begin{array}{rrr}
13 & 8 & 1 \\
5 & -10 & 3 \\
3 & -6 & -5
\end{array}\right]
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{1}{34}\left[\begin{array}{r}
13+12+9 \\
5-15+27 \\
3-9-45
\end{array}\right]=\frac{1}{34}\left[\begin{array}{r}
34 \\
17 \\
-51
\end{array}\right]=\left[\begin{array}{r}
1 \\
\frac{1}{2} \\
\frac{-3}{2}
\end{array}\right]
\end{aligned}
$$
Equating corresponding entries, we have $$x=1$$,
$$
y=\frac{1}{2}, z=-\frac{3}{2} .
$$
