Solve using Augmented Matrix Methods
-4(x1) + 6(x2) = -8
6(x1) - 9(x2) = 12
-4 6 -8
6 -9 12
-->
2 -3 4
3 -3 4
-->
1 0 0 ---- #2 - #1
2 -3 4 --- #1
--->
1 0 0
0 3 -4 -- 2(#1) - #2
1 0 0
0 1 -4/3 --- #2/3
x1=0, x2 = -4/3
(poor choice of method for such a simple system)
To solve the given system of equations using augmented matrix methods, we will first write the equations in matrix form.
Let's represent the system of equations as:
-4x1 + 6x2 = -8 ---- Equation 1
6x1 - 9x2 = 12 ---- Equation 2
To solve it using augmented matrix methods, we'll convert the equations into an augmented matrix, which merges the coefficients of the variables and the constants on the right side of the equal sign.
The augmented matrix for the system is:
[-4 6 | -8]
[ 6 -9 | 12]
Now, we'll perform row operations to transform the augmented matrix into row-echelon form or reduced row-echelon form by using elementary row operations, which include scaling a row, swapping two rows, or adding or subtracting rows.
The goal is to obtain an upper triangular matrix or a diagonal matrix with 1's along the diagonal.
Step 1: Convert the first element in the first column to 1 by performing a row operation.
Multiply the first row by -1/4:
[ 1 -3/2 | 2 ]
[ 6 -9 | 12 ]
Step 2: Use the first row to eliminate the first element in the second row.
Multiply the first row by -6 and add it to the second row:
[ 1 -3/2 | 2 ]
[ 0 0 | 0 ]
The augmented matrix is now in row-echelon form.
To determine the solution, we'll convert the row-echelon form matrix to equations.
From the row-echelon form matrix, we have the equation:
x1 - (3/2)x2 = 2 ---- Equation 3
Notice that x2 is a free variable since it doesn't have a pivot in the coefficient matrix.
Therefore, the solution to the system of equations is:
x1 = 2 + (3/2)x2
x2 = t (where t is any real number)
So the general solution to the system is:
x1 = 2 + (3/2)t
x2 = t
In augmented matrix form, the solution is:
[ x1 ]
[ x2 ] =
[ 2 + (3/2)t ]
[ t ]
Thus, we have solved the system of equations using augmented matrix methods.