Solving Equations by Matrix Method

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Solving Equations by Matrix Method

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]

Equating the corresponding elements of equal matrices X and A-1B, we get the solution.


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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: 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
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.

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


(■(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))

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?

 

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