Solving Linear Equations Using Matrix - Practice Questions & MCQ

Let us consider n linear equations in n unknowns, given as below

$
\begin{aligned}
& \mathrm{a}_{11} \mathrm{x}_1+\mathrm{a}_{12} \mathrm{x}_2+\ldots+\mathrm{a}_{1 \mathrm{n}} \mathrm{x}_{\mathrm{n}}=\mathrm{b}_1 \\
& \mathrm{a}_{21} \mathrm{x}_1+\mathrm{a}_{22} \mathrm{x}_2+\ldots+\mathrm{a}_{2 \mathrm{n}} \mathrm{x}_{\mathrm{n}}=\mathrm{b}_2 \\
& \text {... ... ... ... ... ... } \\
& \text {... ... ... ... ... ... } \\
& \mathrm{a}_{\mathrm{n} 1} \mathrm{x}_1+\mathrm{a}_{\mathrm{n} 2} \mathrm{x}_2+\ldots+\mathrm{a}_{\mathrm{nn}} \mathrm{x}_{\mathrm{n}}=\mathrm{b}_{\mathrm{n}}
\end{aligned}
$
Here $\mathrm{x}_1, \mathrm{x}_2, \ldots \mathrm{x}_{\mathrm{n}}$ are n unknown variables
if $b_1=b_2=\ldots=b_n=0$ then the system of equation is known as homogenous system of equation and if any of $b_1, b_2, \ldots b_n$ is non - zero then it is called non homogenous system of equation

The above system of equations can be written in matrix form as

$
\left[\begin{array}{ccccc}
a_{11} & a_{12} & \ldots & \ldots & a_{1 n} \\
a_{21} & a_{22} & \ldots & \ldots & a_{2 n} \\
\ldots & \ldots & \ldots & \ldots & \ldots \\
\ldots & \ldots & \ldots & \ldots & \ldots \\
a_{n 1} & a_{n 2} & \ldots & \ldots & a_{n n}
\end{array}\right]\left[\begin{array}{c}
x_1 \\
x_2 \\
\ldots \\
\ldots \\
x_n
\end{array}\right]=\left[\begin{array}{c}
b_1 \\
b_2 \\
\ldots \\
\ldots \\
b_n
\end{array}\right]
$

$\Rightarrow \mathrm{AX}=\mathrm{B}$, where

$
\mathrm{A}=\left[\begin{array}{ccccc}
a_{11} & a_{12} & \ldots & \ldots & a_{1 n} \\
a_{21} & a_{22} & \ldots & \ldots & a_{2 n} \\
\ldots & \ldots & \ldots & \ldots & \ldots \\
\ldots & \ldots & \ldots & \ldots & \ldots \\
a_{n 1} & a_{n 2} & \ldots & \ldots & a_{n n}
\end{array}\right], \mathrm{X}=\left[\begin{array}{c}
x_1 \\
x_2 \\
\ldots \\
\ldots \\
x_n
\end{array}\right], \mathrm{B}=\left[\begin{array}{c}
b_1 \\
b_2 \\
\ldots \\
\ldots \\
b_n
\end{array}\right]
$

Premultiplying equation $A X=B$ by $A^{-1}$, we get

$
\begin{aligned}
& A^{-1}(A X)=A^{-1} B \Rightarrow\left(A^{-1} A\right) X=A^{-1} B \\
& \Rightarrow I X=A^{-1} B \\
& \Rightarrow X=A^{-1} B
\end{aligned}
$
$
X=\frac{\operatorname{adj} A}{|A|} B
$
Types of equation:
1. System of equations is non-homogenous:

If $|A| \neq 0$, then the system of equations is consistent and has a unique solution $X=A^{-1} B$
If $|A|=0$ and $(\operatorname{adj} A) \cdot B \neq 0$, then the system of equations is inconsistent and has no solution.
If $|A|=0$ and $(\operatorname{adj} A) \cdot B=0$, then the system of equations is consistent and has infinite number of solutions.
1. System of equations is homogenous:

If $|A| \neq 0$, then the system of equations has only one solution which is the trivial solution.
If $|\mathrm{A}|=0$, then the system of equations has non-trivial solution and it has an infinite number of solutions.