Answer:
Step-by-step explanation:
Solve this using Cramer's Rule. Find the determinant of the original matrix X, then find the determinant of X1, then X2, then X3, and divide each one of these by the determinant of X to get the solution.
The original matrix is
1 2 -1
1 2 1
-1 -1 2
to find the determinant, pick up the first two columns and stick them on the end and do the difference of the sums of the major axes multiplied and the sums of the minor axes multiplied. That looks like this:
1 2 -1 1 2
1 2 1 1 2
-1 -1 2 -1 -1
Multiplying the numbers on the major axes and then adding the products and subtracting from the same process using the minor axes:
[1*2*2)+(2*1*1)+(-1*1*-1)]-[(-1*2*-1)+(-1*1*1)+(2*1*2)]=3 - 5 = -2
The determinant of X is -2. Now do the same for the determinant of X1, but replace the X1 column with the solution column and expand:
-4 2 -1 -4 2
2 2 1 2 2
6 -1 2 6 -1
[(-4*2*2)+(2*1*6)+(-1*2*-1)]-[(6*2*-1)+(-1*1*-4)+(2*2*2)] = -2 - 0 = -2
The determinant of X1 is also -2. Now do the same for the determinant of X2, but replace the X2 column with the solution column and expand:
1 -4 -1 1 -4
1 2 1 1 2
-1 6 2 -1 6
[(1*2*2)+(-4*1*-1)+(-1*1*6)]-[(-1*2*-1)+(6*1*1)+(2*1*-4)] = 2 - 0 = 2
Now for the determinant of X3:
1 2 -4 1 2
1 2 2 1 2
-1 -1 6 -1 -1
[(1*2*6)+(2*2*-1)+(-4*1*-1)]-[(-1*2*-4)+(-1*2*1)+(6*1*2)] = 12 - 18 = -6
To find X1:
[tex]\frac{X1}{X}=\frac{-2}{-2}=1[/tex]
To find X2:
[tex]\frac{X2}{X}=\frac{2}{-2}=-1[/tex]
To find X3:
[tex]\frac{X3}{X}=\frac{-6}{-2}=3[/tex]
The solution is (1, -1, 3)
