![Solving Equations by Matrix Method Solving Equations by Matrix Method](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgff0MdZIQzkLhI9v3xdnxlBn8AHVFomDrkNrx-DJWdG_vUPGywle0J68QM_tCstvM8te527KzHcGaGprV-FkQO5IeH8ix6mFjtKmqsBfFBE_-9g11vJsgUyw9rPxkPOEdhSrBcEQX1lTvC/s16000/Solving+Equations+by+Matrix+Method.png)
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
![equations (i) and (ii) in the matrix form (■(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)
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 1 Example 1: Solve uning matrices: x – 2y = -7 and 3x + 7y = 5 Solution: Here, x – 2y = -7 …………. (i) 3x + 7y = 5 ………… (ii) Writing equations (i) and (ii) in matrix form, we get (■(1&-2@3&7))(■(x@y)) = (■(-7@5)) i.e. AX = B Where, A = (■(1&-2@3&7)), X = (■(x@y)) and B = (■(-7@5)) ∴ |A| = |■(1&-2@3&7)| = 7 + 6 = 13 ∴ A-1 = 1/(|A|) (■(d&-b@-c&a)) = 1/13 (■(7&2@-3&1)) = (■(7/13&2/13@(-3)/13&1/13)) Now, by using formula X = A-1B, we have X = A-1B i.e. (■(x@y)) = (■(7/13&2/13@(-3)/13&1/13))(■(-7@5)) Or, (■(x@y)) = (■(7/13×-7+2/13×5@(-3)/13×-7+1/13×5)) Or, (■(x@y)) = (■(-3@2)) Comparing the corresponding elements, x = -3, y = 2](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiK8oTSIrwnxoRUmQScjeCQJhvXodpQflebL_dsbQQ7u0v2jSoL-aZIFK7GwATrOT1SHvvq_9aVfr346y1wW1LGEtTmlIav974zsx5QbfDQGZYkEUfjjj3gSAzbHr6zPjv7dTGxa5IvIjix/s16000/example+1.png)
![Example 2 Example 2: Solve by matrix method 4/x + 3/y = 7 and 3/x + 2/y = 4. Solution: Here, 4/x + 3/y = 7 ………………….. (i) 3/x + 2/y = 4 …………………. (ii) Writing equations (i) and (ii) in matrix form, we get (■(4&3@3&2))(■(1/x@1/y)) = (■(7@4)) i.e. AX = B Where, A = (■(4&3@3&2)), X = (■(1/x@1/y)) and B = (■(7@4)) ∴|A| = |■(4&3@3&2)| = 8 – 9 = -1 ∴ A-1 = 1/(|A|) (■(d&-b@-c&a)) = 1/(-1) (■(2&-3@-3&4)) = (■(-2&3@3&-4)) Now, by using formula X = A-1B, we have X = A-1B i.e. (■(1/x@1/y)) = (■(-2&3@3&-4))(■(7@4)) = (■(-14+12@21-16)) = (■(-2@5)) Comparing the corresponding elements, 1/x = -2 or, x = (-1)/2 1/y = 5 or, y = 1/5 ∴ The values of x and y are (-1)/2 and 1/5.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg0-QTA75cfYXjg85WSUVhzN2-SVg6Y6y1nLD88eF2Kpd9mVEpdD5HYXPtCKrAhJtG2vc_tbDyLV3DNsqYB7mbSVfT69x8PxQkka1dr8qq1XACNMUFYoCdSA5VAl6i0HCS3DSfQOAZJ19yv/s16000/example+2.png)
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
![Example 3 (■(17&4@8&2))(■(x@y)) = (■(1110@540)) i.e. AX = B Where, A = (■(17&4@8&2)), X = (■(x@y)) and B = (■(1110@540)) ∴ |A| = |■(17&4@8&2)| = 34 – 32 = 2 ≠ 0 So, the given equations have unique solution. Now, A-1 = 1/(|A|) (■(d&-b@-c&a)) = 1/2 (■(2&-4@-8&17)) Now, X = A-1B i.e. (■(x@y)) = 1/2 (■(2&-4@-8&17))(■(1110@540)) Or, (■(x@y)) = 1/2 (■(2×1110-4×540@-8×1110+17×540)) Or, (■(x@y)) = 1/2 (■(60@300)) = (■(30@150))](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiN1_zGuwtFelOYpK2awgIx75Ugx8HAzL1Hyv9B7wS1gXWAm6hGapW11e2EuAQp9Yu6BMiEdM5b3TNLV0Q4Huc3wif3lb0tgPaIF0sMyz6GdiH3OaxErxAyMgdWB0iQ9skhZ_VMA38FJOXd/s16000/example+3.png)
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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