Q. 13

# Solve the followi

Given: - Equations are: –

6x + y – 3z = 5

X + 3y – 2z = 5

2x + y + 4z = 8

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by     and let Dj be the determinant obtained from D after replacing the jth column by Then, provided that D ≠ 0

Now, here we have

6x + y – 3z = 5

x + 3y – 2z = 5

2x + y + 4z = 8

So by comparing with theorem, lets find D , D1 and D2 Solving determinant, expanding along 1st Row

D = 6[(4)(3) – (1)( – 2)] – 1[(4)(1) + 4] – 3[1 – 3(2)]

D = 6[12 + 2] –  – 3[ – 5]

D = 84 – 8 + 15

D = 91

Again, Solve D1 formed by replacing 1st column by B matrices

Here  Solving determinant, expanding along 1st Row

D1 = 5[(4)(3) – ( – 2)(1)] – 1[(5)(4) – ( – 2)(8)] – 3[(5) – (3)(8)]

D1 = 5[12 + 2] – 1[20 + 16] – 3[5 – 24]

D1 = 5 – 36 – 3( – 19)

D1 = 70 – 36 + 57

D1 = 91

Again, Solve D2 formed by replacing 1st column by B matrices

Here  Solving determinant

D2 = 6[20 + 16] – 5[4 – 2( – 2)] + ( – 3)[8 – 10]

D2 = 6 – 5(8) + ( – 3)( – 2)

D2 = 182

And, Solve D3 formed by replacing 1st column by B matrices

Here  Solving determinant, expanding along 1st Row

D3 = 6[24 – 5] – 1[8 – 10] + 5[1 – 6]

D3 = 6 – 1( – 2) + 5( – 5)

D3 = 114 + 2 – 25

D3 = 91

Thus by Cramer’s Rule, we have  x = 1

again,  y = 2

and,  z = 1

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