Solve The Linear Equation In Two Or Three Variables

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.

Linear Equation - Solve The Linear Equation In Two Or Three Variables

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 \( d_1, d_2, d_3\) are not all zero.

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

\(\Rightarrow\) A-1(AX) = A -1B

\(\Rightarrow\) (A-1A)X = A -1B

(Using associative property)

\(\Rightarrow\) X = A-1B

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\( (adj A)B \neq 0\) , then the solution does not exists and the system of equations is known as inconsistent.

On the other hand, if \((adj A)B = 0\), then the given system of linear equations could either be inconsistent or consistent and the system could have either infinitely many solutions or no solution.Let us see an example to understand this concept better.

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 \(X\) we need to find the inverse of matrix  such that
X = A-1 B

A-1 = \(\begin{bmatrix}
\frac{-1}{5} & \frac{1}{5} & \frac{4}{5}\cr
\frac{1}{2} & 0 & \frac{-1}{2}\cr
\frac{3}{20} & \frac{1}{10} & \frac{-7}{20}
\end{bmatrix}\)

\(\Rightarrow \)X = A-1B =\( \begin{bmatrix}
\frac{-1}{5} & \frac{1}{5} & \frac{4}{5}\cr
\frac{1}{2} & 0 & \frac{-1}{2}\cr
\frac{3}{20} & \frac{1}{10} & \frac{-7}{20}
\end{bmatrix}\begin{bmatrix}
10\cr
3\cr
5
\end{bmatrix}\)

\(\Rightarrow\) X = \(\begin{bmatrix}
x\cr
y\cr
z
\end{bmatrix} \)
= \(\begin{bmatrix}
\frac{60}{23}\cr
\frac{57}{23}\cr
\frac{1}{23}
\end{bmatrix}\)

Thus the solution of the above equations is:

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

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

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

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Practise This Question

An electron moves along a straight line as shown inside a charged parallel plate capacitor.  The space between the plates is filled with constant magnetic field of induction B. Time taken for the straight line motion of the electron in the capacitor is