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Question 717092: x+3y-z=0
2x-y+4z=0
x-11y+14z=0
please solve the system of equations using matrix inverse method
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! With all the equations with a zero on the right side, we know that a solution (perhaps the only solution) to this system will be (0, 0, 0). But we will go ahead and find an inverse anyway.
The coefficient matrix for this system is:
1 3 -1
2 -1 4
1 -11 14 (Note: To save time I am not going to put the matrices in brackets. So you'll just have to deal with just the numbers.) To find the inverse of this matrix:- "Append" the identity matrix to the coefficient matrix. This should give you a matrix with the left half being the original coefficient matrix and the right half being the identity matrix.
- Use elementary row operations to change this expanded matrix so the the left half is the identity matrix.
- The right half of this modified, expanded matrix will be the inverse.
Let's see this in action:
1. Append the identity matrix:
1 3 -1 1 0 0
2 -1 4 0 1 0
1 -11 14 0 0 1 2. Use elementary row operations to modify this expanded matrix. (Some notes before I start:- You probably want to check my arithmetic
- The exact steps I use and the order in which I do them are not the only way to accomplish this step. All that is important is that you use proper elementary row operations correctly and that you end up with the right matrix
- Between each matrix I will describe how I am getting from the one above to the one below. In these descriptions I will refer to row 1 as r1, row 2 as r2 and row 3 as r3.
Replace r2 with r2 + -2*r1:
1 3 -1 1 0 0
0 -7 6 -2 1 0
1 -11 14 0 0 1
Replace r3 with r3 + -r1:
1 3 -1 1 0 0
0 -7 6 -2 1 0
0 -14 15 -1 0 1
Replace r3 with r3 + -2*r2:
1 3 -1 1 0 0
0 -7 6 -2 1 0
0 0 3 3 -2 1
Replace r2 with r2 + -2*r3:
1 3 -1 1 0 0
0 -7 0 -8 5 -2
0 0 3 3 -2 1
Replace r2 with (-1/7)*r2 and replace r3 with (1/3)*r3:
1 3 -1 1 0 0
0 1 0 8/7 -5/7 2/7
0 0 1 1 -2/3 1/3
Replace r1 with r1 + -3*r2:
1 0 -1 -17/7 -15/7 -6/7
0 1 0 8/7 -5/7 2/7
0 0 1 1 -2/3 1/3
Replace r1 with r1 + -r3:
1 0 0 -10/7 -59/21 -11/21
0 1 0 8/7 -5/7 2/7
0 0 1 1 -2/3 1/3 The identity matrix has now "moved" from the right half to the left half.
3. The right half is now the inverse of our original coefficient matrix. So:
-10/7 -59/21 -11/21
8/7 -5/7 2/7
1 -2/3 1/3
is the inverse of:
1 3 -1
2 -1 4
1 -11 14 Note: Since the inverse does not have a row of all zeros, then there will be just one solution to this system, the (0, 0, 0) we've know of since the start.
To use this inverse to find a solution to the system, write the system in matrix form and then left-multiply (remember, matrix multiplication is not commutative!) both sides by the inverse matrix. The system, in matrix form is:
1 3 -1 x 0
2 -1 4 * y = 0
1 -11 14 z 0
Left-multiply each side by the inverse:
-10/7 -59/21 -11/21 1 3 -1 x -10/7 -59/21 -11/21 0
8/7 -5/7 2/7 * 2 -1 4 * y = 8/7 -5/7 2/7 * 0
1 -2/3 1/3 1 -11 14 z 1 -2/3 1/3 0On the left side the multiplying first two matrices, if we found the inverse correctly, will result in the identity matrix. And that identity matrix times the variable matrix will result in the variable matrix. On the right side we end up with a 3x1 column matrix with the x value of the solution in the first row, the y value in the second row and the z value in the third row (all zeros as we've known from the start).
P.S. I suggest you multiply the coefficient matrix and its inverse. If you do not get the identity matrix as a result then one or more of the calculations made while we found the inverse must be in error.
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