2.2 - The Inverse of a Matrix We've seen how to add matrices, multiply them by scalars, subtract them, and multiply one matrix by another. The question naturally arises: Can we divide one matrix by another? And the answer is: sort of. Consider the elementary algebra problem: An efficient way to solve for x is to multiply both sides by the reciprocal of, which is. Note that is the multiplicative inverse of, and we do something similar with matrices: if the matrix has an inverse, we can multiply by it to get the identity matrix, which is the linear algebra equivalent to 1. For example, the matrix has an inverse, which is: x = 12 x = 12 1x = 16 5 [ 6 5 ] 5 [ 6 5 ] 5 5 [ 6 5 ] [ ] = 1 0 6 5 [ 0 1 ] Theorem a b A = [ c d ] ad bc 0 Let. If, then is invertible and A A 1 1 = ad bc d b [ c a ] The quantity ad bc is called the determinant, and we write
det A = ad bc Note that this simple result only works for inverses of larger square matrices. 2 2 matrices, but we'll see other ways to find Theorem 5 A n n b R n Ax = b If is an invertible matrix, then for each in, the equation has the unique solution x = A 1 b. Theorem 6 If A is an invertible matrix, then A 1 is invertible and If and are invertible matrices, then so is, and the inverse of is the product of the inverses of and in the reverse order. That is If is an invertible matrix, then so is, and the inverse of is the transpose of A 1. That is The product of inverses in the reverse order. Multiply from the left by invertible matrices is invertible, and the inverse is the product of their on both sides. ( A 1 ) 1 = A A B n n AB AB A B (AB) 1 = B 1 A 1 A A T A T n n AB ( A T ) 1 = ( A 1 ) T (AB) 1 = B 1 A 1 (AB)(AB) 1 = (AB) B 1 A 1 I = A(B B 1 ) A 1 I = AIA 1 I = AA 1 I = I Elementary Matrices Elementary matrices are formed by doing elementary row operations on the identity matrix, and they have some useful properties.
First, we note that multiplying a matrix by the identity matrix (from the left or the right) produces no changes in the matrix (hence the name identity matrix): Now we create an elementary matrix by applying one of the elementary row operations to the identity matrix. If we scale the first row of the identity matrix by multiplying each element in the first row by 2, and then multiply it times another matrix, we get When we multiply by the elementary matrix from the left side, we see that the first row of the other matrix was also multiplied by 2. When we multiplied by the elementary matrix from the right side, we scaled the first column of the other matrix. What this means is that we can perform a scaling operation on a matrix, say in a row-reduction procedure, by multiplying the matrix from the left by a suitably scaled elementary matrix. In the same way, we can create an elementary matrix that does a row-swap operation, by swapping rows in an identity matrix: and we can create an elementary matrix that does a row-combine operation, by scaling and combining rows in an identity matrix, such as adding 2 times the first row to the second row:
The basic rule is: whatever elementary row operation we want to do on a matrix, we do, instead, to an identity matrix, creating an elementary matrix that we then multiply from the left times the matrix we want to modify. Hence we can do entire row-reduction procedure with a sequence of elementary matrix multiplications. E Each elementary matrix is invertible. The inverse of is the elementary matrix of the same type that transforms E I back into. For example, if we had an elementary matrix that scaled one row by a factor of 2, then the inverse matrix would be an elementary matrix that 1 scaled the same row by. 2 E Example of Row Reducing a Augmented Matrix with Elementary Matrix Multiplication First, we want to do a row-combine operation: add 1 2 row: times the first row to the second Now we want to add 2 times the first row to the third row:
Next, we want to add 7 times the second row to the third row: Now we want to scale the third row by 1 : 26 Next, we want to add 7 2 times the third row to the second row: Now we want to add -5 times the third row to the first row: Next, we want to scale the second row by 2:
Now we want to add - times the second row to the first row: 1 Finally, we want to scale the first row by : 2 Note that if we take all the elementary matrices we used and multiply them together from the left in the order they were applied, we get: which is the inverse of the coefficient matrix. Multiplying this inverse times our original augmented matrix gives us the solution at once:
Putting the augmented matrix in the form of a matrix equation, Ax = b, we get: Now we can multiply both sides by the inverse of the coefficient matrix to get
In other words, once we know the inverse A 1 of matrix A, then we can write: Ax = b Ax = A 1 b Ix = A 1 b x = A 1 b A 1 The importance of being able to find an inverse of a square matrix by multiplying by a suitable sequence of elementary matrices is to show that the inverse of only exists if is I row equivalent to, and the same sequence of elementary matrix multiplications turns the identity matrix into A 1. A 1 A = I I = A 1 A 1 A A A Theorem 7 n n A A I n A I n I n A 1 An matrix is invertible if and only if is row equivalent to, and in this case, any sequence of elementary row operations that reduces to, also transforms into. Algorithm for Finding A 1 [A I] A I n [A I] Row reduce the augmented matrix. If is row equivalent to, then is row equivalent to [I A 1 ]. Otherwise, A does not have an inverse. Practice 1) Use determinants to determine which of the following matrices are invertible. a) 9 [ 2 6 ]
b) c) 9 [ 0 5 ] 6 9 [ 6 ] A = 1 1 5 2 1 5 6 5 2) Find the inverse of the matrix, if it exists.