# Find the inverse of a matrix

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• May 17th 2009, 07:28 AM
speckmagoo
Find the inverse of a matrix
Consider the array with matrix A on the left and matrix I on the right. Could i please have some assistance on performing the elementary row operations on the array, so the left hand half of the matrix becomes I and hence A inverse is deduced from the right half of the array.
0 -1 -2 1 0 0
1 1 4 0 1 0
1 3 7 0 0 1
...
...
...
1 0 0 ? ? ?
0 1 0 ? ? ?
0 0 1 ? ? ?
Thankyou in advance for any assistance provided.
• May 17th 2009, 08:06 AM
Opalg
Quote:

Originally Posted by speckmagoo
Consider the array with matrix A on the left and matrix I on the right. Could i please have some assistance on performing the elementary row operations on the array, so the left hand half of the matrix becomes I and hence A inverse is deduced from the right half of the array.
0 -1 -2 1 0 0
1 1 4 0 1 0
1 3 7 0 0 1
...
...
...
1 0 0 ? ? ?
0 1 0 ? ? ?
0 0 1 ? ? ?
Thankyou in advance for any assistance provided.

$\begin{bmatrix} 0 &-1& -2& 1& 0& 0 \\ 1 &1 &4& 0& 1& 0 \\ 1& 3 &7 &0& 0 &1 \end{bmatrix}$

Start by interchanging the top two rows:
$\begin{bmatrix} 1 &1 &4& 0& 1& 0 \\0 &-1& -2& 1& 0& 0 \\ 1& 3 &7 &0& 0 &1\end{bmatrix}$

Now subtract the top row from the bottom row:
$\begin{bmatrix} 1 &1 &4& 0& 1& 0 \\0 &-1& -2& 1& 0& 0 \\ 0& 2 &3 &0& -1 &1\end{bmatrix}$

That gets the first column to be what you want.

Multiply the middle row by –1:
$\begin{bmatrix} 1 &1 &4& 0& 1& 0 \\0 &1& 2& -1& 0& 0 \\ 0& 2 &3 &0& -1 &1\end{bmatrix}$

Next, subtract the middle row from the top row, and subtract twice the middle row from the bottom row. That will get the second column to be what you want. After that, you just have to deal with the third column in a similar way, and you will have the answer.
• May 17th 2009, 08:48 AM
speckmagoo
Reply
Thankyou for your assistance Opalg
I have followed your steps and this is what is what i got
1 0 2 1 1 0
0 1 2 -1 0 0
0 0 -1 2 -1 1

In regards to your suggestion, i keep on finding that i am not able to cancel either of the column 3 2's in rows 2 & 3 without affecting the other entries in the same row. I am unsure whether i am missing something blatently obvious or whether this result has mathematical implications and in particular on the ability to find the determinant of matrix A.
• May 17th 2009, 10:19 AM
Opalg
Quote:

Originally Posted by speckmagoo
Thankyou for your assistance Opalg
I have followed your steps and this is what is what i got
1 0 2 1 1 0
0 1 2 -1 0 0
0 0 -1 2 -1 1

In regards to your suggestion, i keep on finding that i am not able to cancel either of the column 3 2's in rows 2 & 3 without affecting the other entries in the same row. I am unsure whether i am missing something blatently obvious or whether this result has mathematical implications and in particular on the ability to find the determinant of matrix A.

$\begin{bmatrix}1& 0& 2& 1& 1& 0 \\0 &1 &2 &-1& 0& 0 \\0 &0 &-1& 2& -1& 1\end{bmatrix}
$

Now you're nearly there! Multiply the bottom row by –1, then subtract twice it from each of the other two rows, and you're finished.

–Except that you're not quite finished, because it's essential to check that the matrix you have found for $A^{-1}$ really is the inverse of A. So multiply these two matrices together and check that the product is the identity matrix.
• May 17th 2009, 10:19 AM
HallsofIvy
Quote:

Originally Posted by speckmagoo
Thankyou for your assistance Opalg
I have followed your steps and this is what is what i got
1 0 2 1 1 0
0 1 2 -1 0 0
0 0 -1 2 -1 1

In regards to your suggestion, i keep on finding that i am not able to cancel either of the column 3 2's in rows 2 & 3 without affecting the other entries in the same row. I am unsure whether i am missing something blatently obvious or whether this result has mathematical implications and in particular on the ability to find the determinant of matrix A.

Remember that your objective is to is to get 1 in each "pivot" position (on the diagonal) and 0s above and below it.
To get 1 in the last row, multiply the entire third row by -1.
To get 0 in the two first rows, third column, subtract two times that new third row from each.

I am assuming that your original matrix is the first three columns of that 3 by 6 array and the last three columns will become the inverse matrix.

Don't worry about "affecting the other entries in the same row"- you want to! You will not affect entries in the rows before the third column because of those 0s before the pivot position and the rows after the third column are what you want to get.