Let us first understand why do we prefer using matrices to solve a system of linear equation?

Consider the linear equation in one variable 5x = 2x + 3. We can easily figure out x = 1. Now when we are given a pair of linear equations consisting of two variables, we use simultaneous linear equations concept to find out the value of the unknown variables. Similarly, if we have number of linear equations consisting of number of variables, then the process to find out the value of the unknown variables becomes tedious and complex. To solve such equations, use of inverse matrix is comparatively easier.

So, now we will learn how to use an inverse matrix to solve the system of linear equation.

Consider the system of equations:

\( a_1 x + b_1 y + c_1 z = d_1 \)

\( a_2 x + b_2 y + c_2 z = d_2 \)

\( a_3 x + b_3 y + c_3 z = d_3 \)

where \( d_1, d_2, d_3\)

Let\( A = \begin{bmatrix}

a_1 & b_1 & c_1\cr

a_2 & b_2 & c_2\cr

a_3 & b_3 & c_3

\end{bmatrix}\)

\(B = \begin{bmatrix}

x\cr

y\cr

z

\end{bmatrix}\)

\(C = \begin{bmatrix}

d_1\cr

d_2\cr

d_3

\end{bmatrix}\)

A is known as the coefficient matrix of the system of linear equations.

We can write the given equations in as follows:

\(\begin{bmatrix}

a_1 & b_1 & c_1\cr

a_2 & b_2 & c_2\cr

a_3 & b_3 & c_3

\end{bmatrix}\begin{bmatrix}

x\cr

y\cr

z

\end{bmatrix} \begin{bmatrix}

d_1\cr

d_2\cr

d_3

\end{bmatrix}\)

We can abbreviate the above system of equations as, AX = B , i.e.

Now two cases arise which are explained further,

__Case I: If __A __ is non-singular matrix__

We know that if a matrix is non-singular (i.e. the value of determinant of the matrix is not zero) then the inverse of such a matrix exists.

Since AX = B

Multiplying both sides by A^{-1}

\(\Rightarrow\)^{-1}(AX) = A ^{-1}B

\(\Rightarrow\)^{-1}A)X = A ^{-1}B

(Using associative property)

\(\Rightarrow\)^{-1}B

As inverse of a matrix is unique therefore this matrix equation gives unique solution for the given system of linear equations.

__Case II: If __ __ A is a singular matrix__

If a matrix is singular, then the determinant of A is zero. Then we will have to calculate the value of

\( (adj A)B \)

Now, in a case when\( (adj A)B \neq 0\)

On the other hand, if \((adj A)B = 0\)

Example: Solve the given system of linear equation

\(x + 3y – z = 10\)

\( 2x – y + 6z = 3\)

\( x + y – 2z = 5\)

Solution: We can write the above equations in form of matrix as follows:

\(AX = B\)

\(\begin{bmatrix}

1 & 3 & -2\cr

2 & -1 & 6\cr

1 & 1 & -2

\end{bmatrix}\begin{bmatrix}

x\cr

y\cr

z

\end{bmatrix} = \begin{bmatrix}

10\cr

3\cr

5

\end{bmatrix}\)

To find the matrix \(X\)

X = A^{-1} B

A^{-1} = \(\begin{bmatrix}

\frac{-1}{5} & \frac{1}{5} & \frac{4}{5}\cr

\frac{1}{2} & 0 & \frac{-1}{2}\cr

\frac{3}{20} & \frac{1}{10} & \frac{-7}{20}

\end{bmatrix}\)

\(\Rightarrow \)^{-1}B =\( \begin{bmatrix}

\frac{-1}{5} & \frac{1}{5} & \frac{4}{5}\cr

\frac{1}{2} & 0 & \frac{-1}{2}\cr

\frac{3}{20} & \frac{1}{10} & \frac{-7}{20}

\end{bmatrix}\begin{bmatrix}

10\cr

3\cr

5

\end{bmatrix}\)

\(\Rightarrow\)

x\cr

y\cr

z

\end{bmatrix} \)

\frac{60}{23}\cr

\frac{57}{23}\cr

\frac{1}{23}

\end{bmatrix}\)

Thus the solution of the above equations is:

\(x = \frac{60}{23}\)

\(y = \frac{57}{23}\)

\(z = \frac{1}{23}\)

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