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Using Matrix Method, Solve the Following System of Equations: X – 2y = 10, 2x + Y + 3z = 8 and -2y + Z = 7 - Mathematics

Sum

Using matrix method, solve the following system of equations: 
x – 2y = 10, 2x + y + 3z = 8 and -2y + z = 7

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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

`|[ 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

Concept: Inverse of a Matrix - Inverse of a Square Matrix by the Adjoint Method
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