assume a not equal to 0, and that the inverse of A exists. Use row operations to calculate A^−1

Question

Let A = | 1 a 1

c 0 2

1 a 1/c + 1 |

, where c does not equal 0 and a ∈ R.

Now assume a not equal to 0, and that the inverse of A exists. Use row operations to calculate A^−1

Answer 1

This is a hard problem to type into the answer box. I hope that when you see it the columns line up.

I know the answer is correct because when I was done, I verified it.

OK, let's go

To find the inverse by row operations we want to set the identity matrix next to the A matrix and work on both of them at the same time. The set up looks like this: The manipulations will turn the first matrix into the Identity matrix. Then the second matrix will be the inverse.

1 a 1 1 0 0

c 0 2 0 1 0

1 a 1/c + 1 0 0 1

We must create a 1, 0 and 0 in the first column.

For R1, I made no changes. R1 will be our pivot row.

For R2, I multiplied each entry in R1 by (-c) and added R2

For R3, I multiplied each entry in R1 by (-1) and added R3.

Do this all the way across all 6 columns. This is the result:

1 a 1 1 0 0

0 -ac -c+2 -c 1 0

0 0 1/c -1 0 1

Be sure to understand each entry in this step before proceeding.

Next I wanted to create a 1 in the second row, second column.

So I made no changes in R1 and R3.

I multiplied every entry in R2 by (-1/ac).

The resulting matrix is as follows:

1 a 1 1 0 0

0 1 (-c+2)/(-ac) 1/a 1/-ac 0

0 0 1/c -1 0 1

Next we must create a 0 in the 0 in the first row, second column.

We would have to do the same in the third row, second column except that it is already a 0.

So I will multiply each entry in R2 by (-a) and add it to R1.

I will make no changes to R2 and R3.

1 0 2/c 0 1/c 0

0 1 (-c+2)/(-ac) 1/a 1/-ac 0

0 0 1/c -1 0 1

Now for the third column. We want to create a 1 in the third row so we will multiply R3 times c.

R1 and R2 do not change.

1 0 2/c 0 1/c 0

0 1 (-c+2)/(-ac) 1/a 1/-ac 0

0 0 1 -c 0 c

Last step is to get 0's in the third column. This was the hardest one.

R1 becomes (-2/c)R3 + R1

R2 becomes (-c+2)/(ac) R3 + R2

R3 remains the same

1 0 0 2 1/c -2

0 1 0 (c-1)/a 1/-ac (-c+2)/a

0 0 1 -c 0 c

Now we have created the Identity matrix in the first 3 columns and the inverse of A in the last 3 columns.

You can check your answer by multiplying the original matrix A by the answer A^-1.

If you get the Identity matrix for an answer, you know that you have the correct inverse.

If you need help with this, please contact me again on this thread.

Be patient, this can be a real challenge.