EXAMPLE 6 We can use Gaussian elimination to find a right inverse of an m xn matrix A, so long as the rank of A is equal to m. The fact that we have free variables when m < n will give many choices of right inverse. For example, taking 1. A = -1 2 -1 we apply Gaussian elimination to the augmented matrix 1 -1 1 1 -1 1 1 0. 2 -1 0. 0. 0. 1 -2 -2 1 0- 1 -1 1 -2 -2 []- From this we see that the general solution of Ax = is -1 X = +s is and the general solution of Ax = X = + If we take s =t = 0, we get the right inverse |-1 B = -2 1 0. 0.

Advanced Engineering Mathematics
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Chapter2: Second-order Linear Odes
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I really cannot see how this problem is going from the reduced matrix (via Gaussian elimination) to the general solution. Can anyone please explain the steps to arrive to the general solution and then give another right inverse of matrix A.

EXAMPLE 6
We can use Gaussian elimination to find a right inverse of an m xn matrix A, so long as
the rank of A is equal to m. The fact that we have free variables when m < n will give
many choices of right inverse. For example, taking
1.
A =
-1
2 -1
we apply Gaussian elimination to the augmented matrix
-1
1
1
-1
1
1
0.
-1
0.
0.
1
-2
-2
1
1
-1
1 -2
-2
[]-
From this we see that the general solution of Ax =
is
-1
X =
+s
is
and the general solution of Ax =
X =
+1
If we take s =t = 0, we get the right inverse
-1
B =
-2
1
0.
0.
Transcribed Image Text:EXAMPLE 6 We can use Gaussian elimination to find a right inverse of an m xn matrix A, so long as the rank of A is equal to m. The fact that we have free variables when m < n will give many choices of right inverse. For example, taking 1. A = -1 2 -1 we apply Gaussian elimination to the augmented matrix -1 1 1 -1 1 1 0. -1 0. 0. 1 -2 -2 1 1 -1 1 -2 -2 []- From this we see that the general solution of Ax = is -1 X = +s is and the general solution of Ax = X = +1 If we take s =t = 0, we get the right inverse -1 B = -2 1 0. 0.
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