Solving Equations by Matrix Method
The simultaneous equations in two
variables can be solved by different methods: substitution method, cross
multiplication method, etc. Here, we deal with the method of solving linear
equations of two variables using matrices.
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Let us consider the following linear
equations,
a1x + b1y = c1
………………. (i)
a2x + b2y = c2
………………. (ii)
Writing equations (i) and (ii) in the
matrix form, we get
![(■(a_1&b_1@a_2&b_2 ))(■(x@y)) = (■(c_1@c_2 )) i.e. AX = B ………………….. (iii) Where A = (■(a_1&b_1@a_2&b_2 )), X = (■(x@y)) and B = (■(c_1@c_2 )) Pre-multiplying both sides of equation (iii) by inverse of A i.e. A-1, we get A-1(AX) = A-1B Or, (A-1A)X = A-1B Or, IX = A-1B [∵ A-1A = I] ∴ X = A-1B [∵ IX = X]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg07tLVOfCA3tAPsc9pUc-UrWwML-hjh_WU6Mb0f3WKwPjpec6HDiLfiJZ1nMpG1Jyeq3c-qD4nY75K-_l0cI38L0TUlIHcIAEOZ-tsnHOaRpbCOrn2c20HO3dlx5g9OEg8728hyphenhyphenmEg7_Wf/s16000/simultaneous+equation+in+matrix+form.png "equations (i) and (ii) in the matrix form")
Equating the corresponding elements of equal matrices X and A-1B, we get the solution.
Note: If the determinant |A| = 0, the system of simultaneous equations have no unique
solution, therefore when |A| ≠ 0, the equations have unique solution.
Worked Out Examples
Example 3: If the cost of 17 kg sugar and 4 kg tea is Rs. 1110 and
the cost of 8 kg sugar and 2 kg tea is Rs. 540. Find the cost of sugar per kg and
tea per kg by matrix method.
Solution: Let,
Cost of sugar per kg = Rs. x
Cost of tea per kg = Rs. y
Then according to questions,
1st case, 17x + 4y = 1110 …………… (i)
2nd case, 8x + 2y = 540 ……………… (ii)
Writing the given equations in the matrix form, we get
Comparing the corresponding elements, we get
x = 30, y = 150
Hence, The cost of
sugar per kg = Rs. 30
The cost of tea per
kg = Rs. 150
Do you have any questions regarding the matrix method for solving equations?
You can ask your questions or problems here, in the comment section below.
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