Unit 19 Properties of Determinants

Transcription

1 Unit 9 Properties of Determinants Theorem 9.. Suppose A and B are identical n n matrices with the exception that one row (or column) of B is obtained by multiplying the corresponding row (or column) of A by some nonzero constant c. Then det B = c(det A) This theorem may be used in a variety of situations to make a seemingly difficult determinant (due to large numbers or fractions involved) into a much easier one, as the following examples demonstrate. Example. Find det A where A = Solution: The numbers in the first column have a common factor of. The arithmetic will be easier if we factor it out. That is, we have the matrix = so the matrix can be obtained from the matrix by multiplying the first column by. Thus, Theorem 9. tells us that det A = det = det

2 and so, expanding along row, we have 5 det A = (4 ( ) + det ) = 84 5 = 40 5 Notice: What we saw here is that we can use Theorem 9. by factoring a constant, c, out of every entry in some particular row or column of the matrix and then multiplying the determinant of the resulting matrix by the constant. For instance, for a 3 3 matrix, we can factor out a common factor in row and find the determinant as det c a c a c a 3 a a a 3 a 3 a 3 a 33 = c det a a a 3 a a a 3 a 3 a 3 a 33 Example. Find det A where A = Solution: We can see that det A is most easily found by expanding along column. However, we can simplify the arithmetic by factoring out common factors in some rows and columns. To begin with, we see that column 3 contains only even numbers (and is the largest integer factor shared by these numbers) so we can factor out a from this column: det A = det = det Now, we can see fairly easily that row contains only multiples of 3, so we can factor 3 out of row to get: 3 4 det A = (3 ) det Also, we can factor a 3 out of column and after that, a 4 out of row 3.

3 This gives: det A = 3 6 det Finally, we expand along column. det A = 6 = det ( ( ) + det ) = 3 ( ) (4 3) = 64 = 64 Example 3. Find det B, where B = Solution: We know we will want to expand along column 3, but let s see if we can get rid of the fractions inside the matrix. Notice that the (, )-entry is = 3, so there is a common factor of in row. Likewise, we can factor 6 6 out of row and out of row 3. We get: 3 4 (3) ( ) (6) 6 det B = det = det ( ) () (0) () ( ) (0) = det 0 0 Now, expanding along column 3, we get: det B = = ( 6 ( ) +3 det = = ) Notice: In general, if c is a constant then det(ca) c det A. Since ca is formed by multiplying every entry and hence every row of A by c, we need to take a factor of c out of each row. 3

4 This means that for any n n matrix A: det(ca) = c n det A For instance, if A is a 3 3 matrix, then det(a) = 3 det A = 8 deta. Theorem 9.. More Facts About Determinants Suppose A and B are square n n matrices. Then:. If A has one row (column) that is a scalar multiple of another row (column) of A then det(a) = 0.. If B is identical to A except that two rows (or columns) from A are interchanged, then det B = det A. (i.e. when rows or columns of a matrix are interchanged, the sign of the determinant is reversed). 3. If B is obtained from A by adding a scalar multiple of one row (column) of A to another row (column) of A, then det B = det A. (i.e., adding a scalar multiple of one row (or column) to another row (or colunm) does not change the value of the determinant.) 4. det(ab) = det A det B Notice: Two of these facts, along with Theorem 9., tell us how the determinant is affected when elementary row operations (or the corresponding column operations) are performed on a matrix. They are meant to be used to take a matrix whose determinant at first glance seems difficult, and transform it, through elementary row or column operations, to a matrix whose determinant is easy to calculate. For instance, we may obtain a row or column with many zeroes, or transform the entire matrix into one which is upper or lower triangular. Example 4. Find the determinant of A = 3 4 Solution: A is a 4 4 matrix and contains no zeroes. To calculate det A directly, expanding along any row or column would involve determinants of four 3 3 matrices, and evaluating each of these 3 3 determinants would involve determinants of three matrices, so in total, we would need to identify and find determinants of twelve matrices. We can significantly reduce 4

5 the work required to calculate det A by using elementary row (or column) operations to get rows or columns with lots of zeroes. We will work towards making an upper triangular matrix, but we don t have to go all the way there. To begin with, we zero-out column by replacing R by R 3R, replacing R3 with R3 R and replacing R4 with R4 R. Since each of these operations involves adding a scalar multiple of Row to another row, these 3 operations all leave the determinant unchanged, so we see that: det A = det 3 4 = det The arithmetic will be slightly easier if we interchange rows and 4. Of course, interchanging rows changes the sign of the determinant, so we get: det A = det Now, we get zeroes in the (3, ) and (4, ) entries using R3 R3 3 R and R4 R4 R. Again, these are operations which don t change the value of the determinant, so: det A = det We can get rid of the fractions by factoring a out of row 3. We get: det A = det We could continue until we were working with an upper triangular matrix. However, the determinant can easily be calculated at the current stage, by expanding along column and then expanding the only 3 3 matrix we need to deal with along column as well. 5

6 det A = det = ( ) + det = 3 (( ) () + det ) = ()(3)( 3) ( )(7) = ( ) = Example 5. Find det A, where A = Solution: This matrix looks very unfriendly. That is, it would appear that calculating the determinant requires calculating determinants of at least three 3 3 matrices, all involving not-exactly-trivial arithmetic. However, if we first do some factoring from each row, to get some friendlier numbers, we can then easily transform the matrix to an upper triangular matrix, using elementary row operations. The determinant will then be easy to find. We can factor 5 out of Row, 9 out of Row, 5 of row 4. We see that det A is given by: out of Row 3 and out det = (5 9 5 ) det

7 We now perform elementary row operations to get an upper triangular matrix. First, replace row with R ( R), and replace row 3 with (R3 R). We know that these operations have no effect on the determinant. That is, the determinant of the transformed matrix will be the same as the determinant of the matrix before the transformations. So: det A = ( 90) det Next, we can interchange Rows and 4 (so that when we subtract a multiple of one from the other we won t need to introduce fractions into the matrix). Interchanging rows changes the sign of the determinant, so to counteract this change, we need to change the sign of the scalar we re multiplying the determinant by. That is, the determinant of the matrix before interchanging the rows is equal to ( ) times the determinant of the matrix which has the rows interchanged. We have: det A = (90) det Now, we can replace row 4 by R4 ( R) to complete the process of transforming the matrix to an upper triangular matrix. Once again, this kind of elementary row operation has no effect on the determinant. That is, the determinant of the new matrix is equal to the determinant of the old matrix. So we get: det A = (90) det = (90) (4 7) = 50 This procedure was much easier than calculating det A by expanding along some row or column of A would have been. 7

8 Notice: We could have omitted the interchange of Rows and 4 above, and instead simply subtracted R from row 4. This would have given the final calculation of the determinant as ( 90) ( 4 ( )) 7 = 50. We have seen that:. When we interchange two rows of a matrix, the sign of the determinant changes,. when we add a scalar multiple of row to another row of the matrix, the determinant stays the same, and 3. when we multiply a row by a non-zero scalar, the determinant is multiplied by the same scalar. Thus the only effects on the determinant when we row reduce a matrix are sign changes and multiplication by non-zero scalars. That is, we see that: Lemma 9.3. Let A be any square matrix and let B be the RREF of A. Then det A = c det B for some non-zero scalar c. Notice: The scalar c in this lemma is the cumulative effect of all of the ero s performed in transforming the matrix to RREF, i.e., of some sign changes and some multiplications by non-zero scalars. Thus c is the product of some ( ) s and some non-zeroes, and hence c 0. This is important because it gives us the following theorem. Theorem 9.4. A square matrix A is invertible if and only if det A 0. Proof: From our list of equivalent statements in Unit 7, we know that a matrix A is invertible if and only if the row reduced form of A is the identity matrix I. Let B be the row reduced form of A. Therefore det A = c detb, where c is some nonzero scalar (by Lemma 9.3). 8

9 Suppose A is invertible. Then B = I, and of course, det I =, so det A = c detb = c() = c 0 We see that if A is invertible then det A 0. Suppose now that A is not invertible. Then r(a) < n, where n is the order of A, (because r(a) n and we know that A is invertible and r(a) = n are equivalent statements for any square matrix A, i.e., A is invertible if and only if r(a) = n). Of course, we have r(a) = r(b), so r(b) < n. But a square matrix which is in RREF and has less than full rank has, by definition, fewer than n non-zero rows, so B contains at least one row containing only zeroes. And we know that a matrix containing a zero row has determinant 0, so we have det B = 0, and we see that: det(a) = c det(b) = c(0) = 0 Therefore if A is not invertible, then det A = 0. So if det A 0, it cannot be true that A is non-invertible, i.e., it must be true that A is invertible. Thus we have shown that if A is invertible, then det A 0, and if det A 0, then A is invertible, i.e., we have shown that A is invertible if and only if det A 0. We can now add det A 0 to our list of equivalent statements from Unit 7. The new version of Theorem 7. is thus: Theorem 9.5. If A is a square matrix of order n, then the following statements are equivalent to one another (i.e. if any one statement is true then all are true).. A is invertible (i.e., nonsingular).. r(a) = n (i.e., A has full rank). 3. The RREF of A is I n (i.e., A is row-equivalent to the identity matrix). 4. A x = b has a unique solution for any b. 5. A x = 0 has only the trivial solution. 6. det A 0. 9

10 Cramer s Rule/The Classical Adjoint We now introduce new techniques for (i) solving the SLE A x = b, and (ii) finding A, when A is a square matrix with det A 0. Both of these new techniques involve using determinants. Definition 9.6. Given a linear system of equations A x = b, where A is a square matrix, we use A(i) to denote the matrix obtained by replacing the i th column of A with b. Example 6. Find A(), A() and A(3), where A is the coefficient matrix of the linear system: x +x x 3 = 6 x x +x 3 = x x 3 = 0 Solution: Writing this system in matrix equation form, A x = b, we have x 6 x = 0 x 3 0 We see that: A = 0 and b = To get A(), we replace the first column of A by the column vector b. 6 We get A() = Similarly, we have A() = 0 6 and A(3) =

11 Definition 9.7. A square linear system is a linear system in which the coefficient matrix is a square matrix, i.e., a SLE in which the number of equations is the same as the number of variables. Theorem 9.8. Cramer s Rule Suppose A x = b is a square linear system in the variables x, x, x 3,..., x n with the property that det A 0. Then the (unique) solution to the system is given by: x = det A() det A, x = det A() det A,..., x det A(n) n = det A Notice: We can only use Cramer s rule when det(a) 0. However, this means that Cramer s Rule can be used to solve any square linear system which has a unique solution, since we know (from Theorem 9.5) that A x = b has a unique solution if and only if det A 0. Example 7. Solve the following system using Cramer s rule: x +x x 3 = 6 x x +x 3 = x x 3 = 0 Solution: This is the same system as in the previous example, so we have the matrices: 6 A =, A() =, A() = 6 0, A(3) = We need to find the determinant of each of these matrices.

12 For A, we expand along column to get det A = ()( ) det + ( )( ) det = ( 3) ( ) = Notice that det A 0, so Cramer s Rule may be used to solve this system. We can find det A() most easily by expanding along row 3. 6 det A() = ( )( ) 3+3 det = ( 6 ) = 6 For A(), we again expand along column : det A() = (6)( ) det + ()( ) det = ( 6)( 3) + ()( ) = 6 For A(3), we exapand along row 3 and get det A(3) = ()( ) 3+ det = ( + 6) = Thus, by Cramer s rule we get: x = and x = and x 3 = det A() det A = 6 4 = 4 det A() det A = 6 4 = 4 det A(3) det A = 8 4 = We see that the unique solution to the given system is (x, x, x 3 ) = (4, 4, ).

13 Example 8. Use Cramer s Rule, if possible, to solve the system: x +x +x 3 = 4 x x +x 3 = 5 3x +x +4x 3 = 3 Solution: The coefficient matrix is A = the first row, we see that det A = det det det 3. By expanding along = 6 ( ) + 4 = 0 Since det A = 0, Cramer s Rule cannot be used to solve this system. The system does not have a unique solution. Gauss-Jordan elimination would be used to solve the system. Recall that given a square matrix A, we defined the (i, j)-minor, M ij, of A to be the determinant of A ij, where A ij was the submatrix of A obtained by deleting the i th row and j th column of A. We then defined the (i, j)-cofactor, C ij, of A to be ( ) i+j M ij. So C ij = ( ) i+j M ij = ( ) i+j det A ij Definition 9.9. Given a square matrix A, we define the cofactor matrix of A, denoted cof A to be cof A = c ij where c ij = C ij i.e. The (i, j)-entry of cof A is the (i, j)-cofactor of A. Definition 9.0. The classical adjoint, or simply the adjoint, of A is denoted AdjA and is the transpose of the cofactor matrix. That is: AdjA = cof A T 3

14 Example 9. Find cof A and AdjA, for A = Solution: We have: = = det det det cof A = det C C C 3 C C C 3 C 3 C 3 C det 0 0 det det det det 0 Therefore we have: AdjA = cof A T = T = Theorem 9.. The cofactor matrix of A T is equal to the transpose of the cofactor matrix of A. That is, AdjA = cof A T = cof (A T ) Notice: This means that to find AdjA, we can first take the transpose of A and then find the cofactors, so that we don t need to remember to take the transpose at the end. 4

15 Example 0. Find AdjA for A = 3 4 Solution: We have A = We get AdjA = Example. For A = 3 4 det4 det det3 det 3 4 3, so A T = 4 =. 4 3, find A(AdjA) and (AdjA)A. Also, find det A. Solution: From the previous example, we have AdjA = 4 0 So A(AdjA) = = and (AdjA)A = = Also, det A = det = 4 6 = Notice what we have here. We see that A (AdjA) = (AdjA) A. Moreover, the matrix product is just a scalar multiple of I, and that scalar is det A. This is, of course, not a coincidence. For any square matrix A, it is always true that A (AdjA) = (AdjA) A = (det A)I. So if det A 0, then we can divide through by det A to get: That is, we see that: A det A (AdjA) = But then, this means that det A A(AdjA) = det A (AdjA)A = I det A (AdjA) A = I (AdjA) is the inverse of A! We have: det A 5

16 Theorem 9.. The Adjoint Form of the Inverse Suppose A is a square matrix with det A 0. Then A is invertible, and: A = det A AdjA This gives us a new way to calculate A. It also gives a way to calculate individual entries of A, without having to find the whole matrix. We simply need to find the corresponding entry of AdjA, and multiply it by det A. Example. Consider A = (a)find the (,)-entry of A 3 0, with cof A = (b) Find A Solution: (a) We know that A = AdjA, so the (,)-entry of det A A is given by times the (,)-entry of AdjA, which is times the (,)-cofactor of det A det A A. Now, expanding the determinant of A along row, we have: det A = (0) + ()( 4) + ( )( )( ) = 6 and from the given matrix of cofactors we see that the (,)-cofactor is, so the (,)-entry of A is 6 () = 3. (b) Since det A = 6 and A = AdjA = cof det A det A AT, we have: A = T = =