Then three elimination matrices and elimination matrix eij row

The basic elimination step subtracts a multiple ℓij of equation j from equation i.

We always speak about subtractions as elimination proceeds. If the first pivot is a11 = 3

Row 3, column 1, multiplier 2 E31 =−2 0 1

0 1 0
1 0 0

that produces zero in row 3, column 2 of U. E32 gives the last step of 3 by 3 elimination.

Example 1 E21 and then E31 subtract multiples of row 1 from rows 2 and 3 of A :

0
0

1−3

6
1

4 column 1 zeros in (2) 1 0 0 1 0 0 3 1 0 3 1 0 two new

0
2

6
1

At the end, the pivots 3, 2, 1 are on the main diagonal of U : zeros below that diagonal.

The inverse of each matrix Eij adds back ℓij(row j) to row i. This leads to

that A−1times A equals I. Whatever A does, A−1undoes. Their product is the identity

matrix—which does nothing to a vector, so A−1Ax = x. But A−1might not exist.

Matrices are more complicated and interesting. The matrix A−1is called “A inverse”.

DEFINITION The matrix A is invertible if there exists a matrix A−1that “inverts” A :

calculate A−1. In most problems we never compute it ! Here are seven “notes” about A−1.

This shows that a left inverse B (multiplying A from the left) and a right inverse C

A has dependent columns. It cannot have an inverse. No matrix can bring 0 back to x.

If A is invertible, then Ax = 0 only has the zero solution x = A−10 = 0.

ad − bc 1 �−c d −b a�. (6)

This number ad−bc is the determinant of A. A matrix is invertible if its determinant is not zero (Chapter 5). The test for n pivots is usually decided before the determinant appears.

0

0
×

×

×   then A−1= 
1/d1

0
×

×

Elimination turns the second row of this matrix A into a zero row. No pivot.

Example 3 Three of these matrices are invertible, and three are singular. Find the inverse

7� � 6
6

0� � 6
6

1
1

1
0

36� 0 −6 6 � S−1=−1 0 −1 1 0

1
1 0 0

For two nonzero numbers a and b, the sum a + b might or might not be invertible. The

numbers a = 3 and b = −3 have inverses1 3and − 1 3. Their sum a + b = 0 has no inverse.

It is also common sense: If you put on socks and then shoes, the first to be taken off

are the . The same reverse order applies to three or more matrices :

F =0

0 −4
1 0

1 0 0 1 0 0

Now multiply F by the matrix E in Example 4 to find FE. Also multiply E−1times F−1

0 4
1 0

1 . (9) 1 0 0 1 0 0

row 3. After that, E−1adds 5 times row 1 to row 2. There is no 20, because row 3 doesn’t

change again. In this order E−1F−1, row 3 feels no effect from row 1.

2.2. Elimination Matrices and Inverse Matrices 53

L is the Inverse of E

inverse steps E−1 ij. Remember that those steps come in the opposite order.

With n = 3, the complication for E = E32E31E21 is in the bottom left corner:

0 1=(ℓ32ℓ21 − ℓ31) −ℓ21

−ℓ32
1

0 1ℓ31 0 1

0 1 0

All the multipliers ℓij appear in their correct positions in L. The next section will show

that this remains true for all matrix sizes. Then EA = U becomes A = LU.

Problems 1–11 are about elimination matrices.

1 Write down the 3 by 3 matrices that produce these elimination steps :

5 Suppose a33 = 7 and the third pivot is 5. If you change a33 to 11, the third pivot is

. If you change a33 to , there is no third pivot.

(c) If the reverse step is applied first (and then E) show that EE−1= I.

matrix M = P23E21 does both steps at once?

(b) P23 exchanges rows 2 and 3 and then E31 subtracts row 1 from row 3. What

11 Create a matrix that has a11 = a22 = a33 = 1 but elimination produces two negative

pivots without row exchanges. (The first pivot is 1.)

2.2. Elimination Matrices and Inverse Matrices 55

17 If A has column 1 + column 2 = column 3, show that A is not invertible:

(a) Find a nonzero solution x to Ax = 0. The matrix is 3 by 3.

(b) Find singular matrices A and B such that A + B is invertible.

20 If the product C = AB is invertible (A and B are square), then A itself is invertible.

23 Prove that a matrix with a column of zeros cannot have an inverse.

24

2 to row 3, add row 1 to row 3, then add row 1 to row 2.

26 If B is the inverse of A2, show that AB is the inverse of A.

(c) If A is invertible then A−1and A2are invertible.

1 and solve Ax =1 .

 1 1
1
36 Suppose the matrices P and Q have the same rows as I but in any order. They are “permutation matrices”. Show that P − Q is singular by solving (P − Q) x = 0.