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Question
Using matrix method, solve the following system of equations:
x – 2y = 10, 2x + y + 3z = 8 and -2y + z = 7
Solution
Given equation are,
x - 2y = 10 ...(i)
2x + y + 3z = 8 ...(ii)
-2y + z = 7 ...(iii)
Let A = `|[ 1, -2, 0],[2, 1, 3],[0, -2, 1]|, X = |[x], [y], [z]|, B = |[10], [8], [7]|`
`|"A"|` = 1 (1 + 6) + 2(2 - 0) + 0
= 7 + 4 = 11 ≠ 0
∴ A-1 exists
`a_11 = |[1, 3],[-2,1]| = (1 +6) = 7, a_12 = |[2, 3],[0,1]| =-2, a_13 = |[2,1],[0 ,-2]| = -4`
`a_21 = |[-2, 0],[-2,1]| = -(-2) = 2, a_22 = |[1, 0],[0,1]| =1, a_23 = -|[1,-2],[ 0,-2]| = -(-2) = 2`
`a_31 = |[-2, 0],[1,3]| = -6 , a_32 = -|[1, 0],[2,3]| =(-3)= - 3, a_33 = |[1,-2],[ 2,1]| = 1 + 4 = 5`
∴ adj A = `|[7, -2 , -4], [2, 1, 2],[-6, -3, 5]|^T`
= `|[7, 2 , -6], [-2, 1, -3],[-4, 2, 5]|`
∴ A -1 = `(1)/|"A"|. "adj" "A"`
= `(1)/(11) |[7, 2, -6],[-2, 1, -3], [-4, 2, 5]|`
Now. AX = B ⇒ X = A -1 B
`|[ x],[y],[z]| = (1)/(11) |[7, 2, -6],[-2, 1, -3], [-4, 2, 5]| |[10], [8], [7]|`
`|[ x],[y],[z]| = (1)/(11) |[ 70+ 16 -42],[-20 + 8 -21], [-40 + 16 + 35]|`
= `(1)/(11) |[ 44],[-33],[11]|`
Here, x = 4, y = -3, z = 1
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