Given the matrix below, we can solve the equation A x = b by computing A^-1 b, provided we can find the matrix A^-1.

In order to find A^-1, we reduce the first matrix shown below, consisting of the matrix A and the identity matrix I side-by-side, to the second form, where the identity matrix I is on the left.

• We reduce A to I, and let the identity matrix on the right follow along.
• Whatever we end up with on the right will be the inverse matrix A^-1.

We will proceed as usual, obtaining 0's in the first column for the second in third rows, then obtaining 0 in the second column of the third row.

• We begin by multiplying the first row of the matrix by 1/2. This step is not necessary, but it is convenient have 1 at the head of the first column.
• We then subtract 5 times the first row from the second, which will accomplish our goal of having 0 as the first number in the second row.
• We then subtract 2 times the first row from the third, obtaining 0 as the first number in the third row.

The second figure below shows how we subtract 5 times the second row from the first.

The figure below shows how we might work with the circled numbers in the second matrix when we subtract twice the first row from the second.

• We thus subtract twice the 3/2 from the -1, obtaining -4.  This result appears circled in the third matrix.

We also see how we treat the numbers in the rectangular boxes.

• We subtract twice the -2 from the -3, obtaining 1.  This result is in a box in the third matrix.

Video File #01

We proceed to obtain the desired 0 in the second column of the third row.

• We first multiply the second row by -2 in order to obtain a 1 in the second column.
• We then add 4 times the second row to the third.
• Since we have 0's the in the first column of both rows, these sums will maintain 0's in those positions.
• The circled and boxed numbers show how we multiply the number in the second row of the second matrix by 4 then added it to the number in the third to obtain the corresponding number in the third row of the third matrix.

We now complete the reduction of A to the identity matrix.

• We obtain 1 as the third diagonal by dividing the third row by -127.
• Adding 32 times the third row to the second and 2 times the third row to the first we obtain 0's above the 1 in the third row.

• Since the first two elements of the third row are both 0, this does not disrupt what we have accomplished in the first two rows.

• The numbers in the right half of the matrix follow along as usual.
• The second figure shows how we add the 1/2 in the first row of the first matrix to double the -19/127 in the third and obtain the 51/254 in the second matrix.

We finally subtract 3/2 the second row from the first to obtain our final off-diagonal 0.

• At this point the matrix on the right is A^-1.

We see that A^-1, which is the inverse of the original matrix, is as indicated.

• You should multiply A by A^-1 to verify that A * A^-1 = I.