Solve the system of equations using matrices. Use Gaussian elimination with back- substitution.

Question

x+y+z = -5
x-y+3z = -1
4x+y+z = -2

Answer 1

It's difficult to show Gaussian elimination process here. But the answers should be,
x = 1, y = -4 and z = -2

Answer 2

Thank you so much , but I'm supposed to show my work :/

Answer 3

x +y + z = -5...i

x - y + 3z = -1...ii

4x + y + z = -2 ...iii

frm (i)

y = -5-x-z ....(iv)

frm (ii)

y = x+3z+1....(v)

frm (iii)

y = -2-4x-z .....(vi)

equate (iv) = (v)

-5-x-z = x + 3z + 1

x+x+3z+z = -5-1

2x+4z = -6 .....(VII)

equate (iv) = (vi)

-5-x-z = -2-4x-z

4x-x-z+z = 5-2

3x = 3

x = 3/3

x = 1

to get z let subt value of x = 1 in (VII) , we have

2x + 4z = -6

2(1) + 4z = -6

4z = -6-2

4z = -8

z = -8/4

z = -2

to get y .. Let subtitute da value of x nd z in equatn (i) , we have

x + y + z = -5

(1) + y + (-2) = -5

1 + y - 2 = -5

y = -5 + 2 - 1

y = -4

so values are :

x = 1
y = -4
z = -2

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