Introduction
By notating the linear equations in matrix form and using the Inverse of Matrices, we can find the solution of the system of linear equations.
Linear Equation: Any equation in the form of ax + by = c is called linear equation. Some examples of linear equations are:
$6\text{x} + 5\text{y} = 1$
$9\text{x} – 3\text{y} = 4$
How to represent a linear equation in matrix form?
By using multiplication of matrices and property of equality of matrices, we can represent two linear equations in matrix form as follows:
Let us consider the following two linear equations:
i) $a_1x + b_1y = c_1$
ii) $a_2x + b_2y = c_2$
Matrix form of equations (i) and (ii) would be:
$\left ( \displaylines{a_1 & b_1 \\ a_2 & b_2} \right ) \left ( \displaylines{x \\ y} \right ) = \left ( \displaylines{ c_1 \\ c_2} \right )$
Question: Use matrix multiplication and law of equality of matrices to expand the matrix $\left ( \displaylines{a_1 & b_1 \\ a_2 & b_2} \right ) \left ( \displaylines{x \\ y} \right ) = \left ( \displaylines{ c_1 \\ c_2} \right )$ and receive linear equations mentioned in equation (i) and equation (ii).
Given,
$\left (\displaylines{ a_1 & b_1 \\ a_2 & b_2 } \right ) \left ( \displaylines{ x \\ y} \right ) = \left ( \displaylines{ c_1 \\ c_2} \right )$ – (1)
Multiplying two matrices on the left hand side, we get,
$= \left (\displaylines{ a_1 & b_1 \\ a_2 & b_2 } \right ) \left ( \displaylines{ x \\ y} \right )$
$= \left ( \displaylines{ a_1 x + b_1 y \\ a_2 x + b_2 y} \right )$
Now, put value of above multiplication in equation (1), we get,
$\left ( \displaylines{ a_1 x + b_1 y \\ a_2 x + b_2 y} \right ) = \left ( \displaylines{ c_1 \\ c_2} \right )$
By law of equality of matrices, we get,
$a_1x + b_1y = c_1$ – (2) and $a_2x + b_2y = c_2$ – (3)
Equations (2) and (3) are the same equation we had in above equations (i) and (ii).
Solution of System of Linear Equations
Let us consider a set of linear equations (same as above):
i) $a_1x + b_1y = c_1$
ii) $a_2x + b_2y = c_2$
The matrix form of the above two equations would be:
$\left ( \displaylines{a_1 & b_1 \\ a_2 & b_2} \right ) \left ( \displaylines{x \\ y} \right ) = \left ( \displaylines{ c_1 \\ c_2} \right )$
Let the first matrix $\left ( \displaylines{a_1 & b_1 \\ a_2 & b_2} \right )$ be represented by A. (This matrix contains the coefficient of variables x and y.)
Let the second matrix $\left ( \displaylines{ x \\ y} \right )$ be represented by X.
Let the third matrix $\left ( \displaylines{ c_1 \\ c_2} \right )$ be represented by B.
Then, we have,
$\text{AX} = \text{B}$
$\text{or, A}^{-1}\text{AX} = \text{A}^{-1}\text{B}$
$\text{or, I x X} = \text{A}^{-1}\text{B}$
$\therefore X = \text{A}^{-1}\text{B}$ – (iii)
Thus, with the help of the equation (iii), we can solve a given set of linear equations if the solutions exist.
How do I know if a given system of linear equations has unique solutions or not?
You can know if the given system of linear equations have unique solutions by finding the determinant of matrix A. Matrix A is the matrix consisting the coefficients of variables x and y, as shown earlier.
We know,
$X = \text{A}^{-1}\text{B}$
Above equation helps us to find the values of x and y. But for this purpose, the matrix $\text{A}^{-1}$ must be defined.
For $\text{A}^{-1}$ to be defined, we have the following conditions:
- Matrix A must be a square matrix.
- It must be a non-singular matrix. Or, the value of determinant of matrix A must not be zero.
Solved Examples
Example 1: Check whether the following system of equations contain unique solutions or not: $3x + 2y = 5 \text{ and } 6x + 4y = 10$.
Solution:
Arranging in matrix form,
$\left ( \displaylines { 3&6\\2&4}\right ) \left ( \displaylines{x \\y} \right ) = \left ( \displaylines{5 \\10} \right )
$i.e. AX = B
We have,
A = $\left ( \displaylines{ 3&6\\2&4} \right )$
$\text{or,} |A| = \left | \displaylines{3&6\\2&4} \right )$
$= 3(4) – 6(2)$
$= 12 – 12$
$= 0$
Since |A| = 0, the system of solutions does not have unique solutions.
Example 2: Find $\left ( \displaylines{x\\y}\right )$ when: $\left ( \displaylines{-2&1\\-3&1} \right ) \left ( \displaylines{ x \\y} \right )\left ( \displaylines{11\\1} \right )$
Solution:
$\left ( \displaylines{-2&1\\-3&1} \right ) \left ( \displaylines{x \\y} \right ) \left ( \displaylines{ 11\\1} \right )$
We have,
AX = B
So,A = $\left ( \displaylines{ -2&1\\-3&1} \right )$
Now,
|A| = $\left | \displaylines{ -2&1\\-3&1} \right |$
$= 2(1) – (-3)1$
$= 2 + 3$
$= 5$
Since |A| ≠ 0, the system of equations have unique solutions.
And,
A$^{-1}$ = $\dfrac{1}{|A|} (adjoint of A)$
$= \dfrac{1}{5} \left ( \displaylines{ 1&-1\\3&2} \right )$
Also,
X = $A^{-1}B$
$or, \left ( \displaylines{ x\\y} \right )= \dfrac{1}{5} \left ( \displaylines{1&-1\\3&2} \right ) \left ( \displaylines{11\\1} \right )$
$= \dfrac{1}{5} \left ( \displaylines{ 1(11) -1(1)\\3(11)+2(1) } \right )$
$= \dfrac{1}{5} \left ( \displaylines{11-1\\33+2} \right )$
$= \dfrac{1}{5} \left ( \displaylines{10\\35} \right )$
$= \left ( \displaylines{ \frac{10}{5} \\ \frac{35}{5} } \right )$
$= \left ( \displaylines{ 2 \\7} \right )$
Hence, x = 2 and y = 7.
Example 3: Solve by using Matrix method: $3x + 4y = 5$ and $x – y = -3$
Arranging the given equations in matrix form, we get,
$\left ( \displaylines{3&4\\1&-1} \right ) \left ( \displaylines{x \\y} \right ) = \left ( \displaylines{5\\-3} \right )$
i.e. AX = B
So, A = $\left ( \displaylines{3&4\\1&-1} \right )$
|A| = $\left | \displaylines{3&4\\1&-1} \right )$
$= 3(-1) – 4(1)$
$= -3 – 4$
$= -7$
Since |A| ≠ 0, the system of linear equations have unique solutions.
We know,X = A$^{-1}$B
$or, \left ( \displaylines{x\\y} \right ) = \dfrac{1}{|A|} \left ( Adjoint of A \right ) \left ( \displaylines{5\\-3} \right )$
$= \dfrac{1}{-7} \left ( \displaylines{-1&-4\\-1&3} \right ) \left ( \displaylines{5\\-3} \right)$
$= \dfrac{1}{-7} \left ( \displaylines{-1(5)-4(-3)\\-1(5)+3(-3)} \right )$
$= \dfrac{1}{-7} \left ( \displaylines{ -5+12\\-5-9} \right )$
$= \dfrac{1}{-7} \left ( \displaylines{7\\-14} \right )$
$= \left ( \displaylines{\frac{7}{-7} \\ \frac{-14}{-7} } \right )$
$= \left ( \displaylines{-1\\2} \right )$
Hence, the required values of x and y are -1 and 2, respectively.
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