Use row operations on augmented matrices to solve the given systems of linear equations.
x + y − z = 0
x + 2y + 4z = −3
3x − y − 22z = 19
1 1 -1 0
1 2 4 -3
3 -1 -22 19
1 1 -1 0
0 1 5 -3 --- I did #2 - #1
0 -4 -19 19 -- I took #3 - 4x#2
1 1 -1 0
0 1 5 -3
0 0 1 7 --- I took #3 - 3x#2
so z = 7
in #2: y + 35 = -3
y = -38
in #1
x - 38 - 7 = 0
x = 45
x=45 , y=-38 , z=7
To solve the system of linear equations using row operations on augmented matrices, let's represent the system of equations with an augmented matrix:
[[1, 1, -1, 0],
[1, 2, 4, -3],
[3, -1, -22, 19]]
Now, let's perform row operations to eliminate the variables below the leading coefficients.
R2 = R2 - R1
R3 = R3 - 3*R1
The updated matrix becomes:
[[1, 1, -1, 0],
[0, 1, 5, -3],
[0, -4, -19, 19]]
Next, let's eliminate the variable below the leading coefficient in the third row.
R3 = 4*R2 + R3
The updated matrix becomes:
[[1, 1, -1, 0],
[0, 1, 5, -3],
[0, 0, 1, 1]]
Now, we'll back-substitute to solve for the variables.
From the third row, we have z = 1.
Substituting z = 1 into the second row equation:
y + 5(1) = -3
y + 5 = -3
y = -8
Substituting z = 1 and y = -8 into the first row equation:
x + (-8) - 1 = 0
x - 9 = 0
x = 9
Therefore, the solution to the system of equations is x = 9, y = -8, and z = 1.
To solve the system of linear equations using row operations on augmented matrices, we need to follow these steps:
Step 1: Write the augmented matrix:
Start by writing the coefficients of the variables and the constants on the right-hand side of each equation as an augmented matrix:
[ 1 1 -1 | 0 ]
[ 1 2 4 | -3 ]
[ 3 -1 -22 | 19 ]
Step 2: Perform row operations to transform the augmented matrix into row-echelon form.
The goal is to have a triangular matrix with leading coefficients of 1 in the leftmost column. We can achieve this by performing row operations. Below, I'll show you the row operations one by one:
R2 = R2 - R1
R3 = R3 - 3R1
The new matrix after these row operations:
[ 1 1 -1 | 0 ]
[ 0 1 5 | -3 ]
[ 0 -4 -19 | 19 ]
Step 3: Continue performing row operations to bring the augmented matrix to reduced row-echelon form.
Now, we want to get 0's below the leading coefficient of 1s in the second column. To achieve this, we'll perform the following row operation:
R3 = R3 + 4R2
The new matrix after this row operation:
[ 1 1 -1 | 0 ]
[ 0 1 5 | -3 ]
[ 0 0 -1 | 7 ]
Step 4: Back substitution.
Now that we have the matrix in reduced row-echelon form, we can solve for the variables. Start from the bottom row and work your way up:
-1z = 7 --> z = -7
y + 5z = -3
y + 5(-7) = -3
y - 35 = -3
y = 32
x + y - z = 0
x + 32 - (-7) = 0
x + 32 + 7 = 0
x + 39 = 0
x = -39
Therefore, the solution to the system of linear equations is:
x = -39, y = 32, z = -7.