Linear Algebra

2.1.4 Finding The Inverse Of A Matrix

A little later a formula is given for the inverse of a matrix. However, it is not a good way to find the inverse for a matrix. There is a much easier way and it is this which is presented here. It is also important to note that not all matrices have inverses.

One might think A would have an inverse because it does not equal zero. However,

( ) ( ) ( ) 1 1 − 1 0 1 1 1 = 0

and if A⁻¹ existed, this could not happen because you could multiply on the left by the inverse A and conclude the vector

^T = ^T. Thus the answer is that A does not have an inverse.

Suppose you want to find B such that AB = I. Let

( ) B = b1 ⋅⋅⋅ bn

Also the i^th column of I is

( )T ei = 0 ⋅⋅⋅ 0 1 0 ⋅⋅⋅ 0

Thus, if AB = I, b_i, the i^th column of B must satisfy the equation Ab_i = e_i. The augmented matrix for finding b_i is

. Thus, by doing row operations till A becomes I, you end up with where b_i is the solution to Ab_i = e_i. Now the same sequence of row operations works regardless of the right side of the agumented matrix and so you can save trouble by simply doing the following.

row operations (A|I) → (I|B )

and the i^th column of B is b_i, the solution to Ab_i = e_i. Thus AB = I.

This is the reason for the following simple procedure for finding the inverse of a matrix. This procedure is called the Gauss Jordan procedure. It produces the inverse if the matrix has one. Actually, it produces the right inverse.

As described above, the following is a description of what you have just done.

where those R_i sympolize row operations. It follows that you could undo what you did by doing the inverse of these row operations in the opposite order. Thus Here R⁻¹ is the row operation which undoes the row operation R. Therefore, if you form and do the inverse of the row operations which produced I from A in the reverse order, you would obtain . By the same reasoning above, it follows that A is a right inverse of B and so BA = I also. It follows from Proposition 2.1.23 that B = A⁻¹. Thus the procedure produces the inverse whenever it works.

If it is possible to do row operations and end up with A

I, then the above argument shows that A has an inverse. Conversely, if A has an inverse, can it be found by the above procedure? In this case there exists a unique solution x to the equation Ax = y. In fact it is just x = Ix = A⁻¹y. Thus in terms of augmented matrices, you would expect to obtain

( ) (A |y ) → I|A−1y

That is, you would expect to be able to do row operations to A and end up with I.

The details will be explained fully when a more careful discussion is given which is based on more fundamental considerations. For now, it suffices to observe that whenever the above procedure works, it finds the inverse.

Form the augmented matrix

( ) 1 0 1 1 0 0 |( 1 − 1 1 0 1 0 |) . 1 1 − 1 0 0 1

Now do row operations until the n×n matrix on the left becomes the identity matrix. This yields after some computations,

( ) 1 0 0 0 12 12 |( 0 1 0 1 − 1 0 |) 0 0 1 1 − 1 − 1 2 2

and so the inverse of A is the matrix on the right,

( 1 1 ) | 0 2 2 | ( 1 − 1 0 ) . 1 − 12 − 12

Checking the answer is easy. Just multiply the matrices and see if it works.

( ) ( 1 1 ) ( ) | 1 0 1 | | 0 2 2 | | 1 0 0 | ( 1 − 1 1 ) ( 1 − 1 0 ) = ( 0 1 0 ) . 1 1 − 1 1 − 12 − 12 0 0 1

Always check your answer because if you are like some of us, you will usually have made a mistake.

Set up the augmented matrix

( ) 1 2 2 1 0 0 |( 1 0 2 0 1 0 |) 3 1 − 1 0 0 1

Next take

times the first row and add to the second followed by times the first row added to the last. This yields

( ) | 1 2 2 1 0 0 | ( 0 − 2 0 − 1 1 0 ) . 0 − 5 − 7 − 3 0 1

Then take 5 times the second row and add to −2 times the last row.

( ) 1 2 2 1 0 0 |( 0 − 10 0 − 5 5 0 |) 0 0 14 1 5 − 2

Next take the last row and add to

times the top row. This yields

( ) − 7 − 14 0 − 6 5 − 2 |( 0 − 10 0 − 5 5 0 |) . 0 0 14 1 5 − 2

Now take

times the second row and add to the top.

( ) | − 7 0 0 1 − 2 − 2| ( 0 − 10 0 − 5 5 0 ) . 0 0 14 1 5 − 2

Finally divide the top row by −7, the second row by -10 and the bottom row by 14 which yields

( ) 1 0 0 − 17 27 27 |( 0 1 0 12 − 12 0 |) . 0 0 1 -1 5- − 1 14 14 7

Therefore, the inverse is

( ) − 17 27 27 |( 1 − 1 0 |) 21- 52 − 1 14 14 7

Write the augmented matrix

( ) | 1 2 2 1 0 0 | ( 1 0 2 0 1 0 ) 2 2 4 0 0 1

and proceed to do row operations attempting to obtain

. Take times the top row and add to the second. Then take times the top row and add to the bottom.

( ) 1 2 2 1 0 0 |( 0 − 2 0 − 1 1 0 |) 0 − 2 0 − 2 0 1

Next add

times the second row to the bottom row.

( ) | 1 2 2 1 0 0 | ( 0 − 2 0 − 1 1 0 ) 0 0 0 − 1 − 1 1

At this point, you can see there will be no inverse because you have obtained a row of zeros in the left half of the augmented matrix

. Thus there will be no way to obtain I on the left. In other words, the three systems of equations you must solve to find the inverse have no solution. In particular, there is no solution for the first column of A⁻¹ which must solve

( ) ( ) x 1 A |( y |) = |( 0 |) z 0

because a sequence of row operations leads to the impossible equation, 0x + 0y + 0z = −1.