inverse matrix

[mathstresser]

Show that the matrix is nonsingular.

. . .\(\displaystyle \L A\, =\, \left[\, \begin{array}{rrrr}.5&.5&0&0\\ 0&0&.5&.5\\ .5&-.5&0&0\\ 0&0&.5&-.5 \end{array}\, \right]\)

I augmented this matrix, and I tried to find its inverse.

I reduced it, but I couldn’t reduce it completely. I got

. . .\(\displaystyle \L A\, =\, \left[\, \begin{array}{rrrr}1&0&0&0\\ 0&1&0&1\\ 0&0&1&2\\ 0&0&-1&1 \end{array}\, \right]\)

I don’t think it matters but the right side of the augmented matrix is

. . .\(\displaystyle \L A\, =\, \left[\, \begin{array}{rrrr}1&0&1&0\\ 0&1&-2&-1\\ 0&-1&4&3\\ 0&0&0&-2 \end{array}\, \right]\)

I can’t reduce the last few digits to make them 0. So what do I do?

 

[arthur ohlsten]

The matrix is non singular, and reduces to a 3x4 matrix, and thus has no inverse
.5 .5 0 0
0 0 .5 .5
.5 -.5 0 0
0 0 .5 -.5
multiply 1st row by -1 and add to 3rd

.5 .5 0 0
0 0 .5 .5
0 -1 0 0
0 0 .5 .5
interchange 2nd and 3rd rows

.5 .5 0 0
0 -1 .5 .5
0 0 .5 .5
0 0 .5 .5
multiply 3rd row by -1 and add to row 2 and row 4

.5 .5 0 0
0 -1 0 0
0 0 .5 .5
0 0 0 0
multiply 2nd row by .5 and add to 1st row
multiply 2nd row by -1
multiply 3rd row by 2
multiply 1st row by 2

1 0 0 0
0 1 0 0
0 0 1 1
0 0 0 0

only square matricies have inverses

Arthur