Solve the system of equations using matrices. Use Gaussian elimination with back- substitution.
x+y+z = -5
x-y+3z = -1
4x+y+z = -2
It's difficult to show Gaussian elimination process here. But the answers should be,
x = 1, y = -4 and z = -2
Thank you so much , but I'm supposed to show my work :/
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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To solve the system of equations using matrices and Gaussian elimination with back-substitution, we need to represent the system of equations as an augmented matrix.
The augmented matrix contains the coefficients of the variables on the left side and the constants on the right side. Here's the augmented matrix for the given system:
[ 1 1 1 | -5 ]
[ 1 -1 3 | -1 ]
[ 4 1 1 | -2 ]
Now, let's perform the Gaussian elimination to simplify the augmented matrix:
Step 1: Swap the first row with the second row to start by having a 1 in the pivot position (top left corner).
[ 1 -1 3 | -1 ]
[ 1 1 1 | -5 ]
[ 4 1 1 | -2 ]
Step 2: Subtract the first row from the second row multiplied by -1.
[ 1 -1 3 | -1 ]
[ 0 2 -2 | -4 ]
[ 4 1 1 | -2 ]
Step 3: Subtract the first row from the third row multiplied by 4.
[ 1 -1 3 | -1 ]
[ 0 2 -2 | -4 ]
[ 0 5 -11 | -6 ]
Step 4: Divide the second row by 2 to get a 1 in the second pivot.
[ 1 -1 3 | -1 ]
[ 0 1 -1 | -2 ]
[ 0 5 -11 | -6 ]
Step 5: Subtract the second row from the first row multiplied by 1.
[ 1 0 4 | 1 ]
[ 0 1 -1 | -2 ]
[ 0 5 -11 | -6 ]
Step 6: Subtract the second row from the third row multiplied by 5.
[ 1 0 4 | 1 ]
[ 0 1 -1 | -2 ]
[ 0 0 -6 | 4 ]
Step 7: Divide the third row by -6 to get a 1 in the third pivot.
[ 1 0 4 | 1 ]
[ 0 1 -1 | -2 ]
[ 0 0 1 | -2/3 ]
Now, we need to perform back-substitution to find the values of x, y, and z.
From the third row, we have z = -2/3.
From the second row, we have y - z = -2, substituting z = -2/3, we get y + 2/3 = -2, thus y = -2 - 2/3 = -8/3.
From the first row, we have x + 4z = 1, substituting z = -2/3, we get x + 4(-2/3) = 1, thus x - 8/3 = 1, solving for x, we get x = 1 + 8/3 = 11/3.
Therefore, the solution to the system of equations is x = 11/3, y = -8/3, and z = -2/3.