Determinants. Dr. Doreen De Leon Math 152, Fall 2015

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1 Determinants Dr. Doreen De Leon Math 52, Fall 205 Determinant of a Matrix Elementary Matrices We will first discuss matrices that can be used to produce an elementary row operation on a given matrix A. Definition. An elementary matrix, typically denoted E, is a matrix that is obtained by performing a single elementary row operation on the identity matrix. As such, there are three types of elementary matrices.. The first type is formed by exchanging row i of I with row j of I, which the author denotes as E i,j. 2. The second type is formed by multiplying row i of I by a nonzero scalar α, which the author denotes E i (α). 3. The third type is formed by adding a nonzero multiple α of row i to row j, which the author denotes E i,j (α). Example: For each of the following, determine if it is an elementary matrix (a) 0 0 (b) 0 0 (c) (d) Solution: 0 0 (e) (a) Yes. Row is formed by adding row 3 of the identity matrix to row.

2 (b) No. Rows and 3 of I are exchanged, and then the resulting rows and 2 are exchanged. (c) Yes. Row of the identity matrix is multiplied by 3. (d) No. Row is multiplied by 3 and then twice row 3 of the identity matrix is added to row. (e) Yes. Row 2 is formed by adding -2 times row of the identity matrix to row 2. We have the following theorem. textbook. Note that this is a summary of Theorem EMDRO in the Theorem. If an elementary row operation is performed on an m n matrix A, the resulting matrix can be written as EA, where the m m elementary matrix E is created by performing the same row operation on I m. Elementary matrices have another useful property. Theorem 2. If E is an elementary matrix, then E is nonsingular. Proof. The idea is that we can row reduce E to the identity matrix by reversing the row operation that formed E. If E = E i,j, then exchanging rows i and j again will give the identity matrix. If E = E i (α), multiply row i of E by to obtain the identity matrix. Finally, α if E = E i,j (α), perform the row operation that multiplies row i by α and adds it to row j. Therefore, each elementary matrix is row equivalent to the identity matrix and is thus nonsingular. In fact, we have the following useful theorem. Theorem 3. Suppose that A is a nonsingular matrix. Then there exist elementary matrices E, E 2,..., E t so that A = E t E t E 2 E. Proof. Since A is nonsingular, it is row equivalent to the identity matrix. Therefore, there is a sequence of t row operations that converts I to A. For each of these row operations, form the associated elementary matrix and denote these matrices by E, E 2,..., E t. Applying the first row operation to I gives the matrix E I. The second row operation gives E 2 (E I) = E 2 E I. and so on. The result of the full sequence of t operations will yield A, so we have A = E t E t E 2 E I = E t E t E 2 E. 2

3 Definition of the Determinant We need a few definitions first. Definition. Suppose that A is an m n matrix. Then the submatrix A ij (denoted A(i j) in the textbook) is the (m ) (n ) matrix obtain from A by removing row i and column j. Examples: Given 3 2 A = 0 4 5, [ ] 0 5 A 2 = 2 0 [ ] 3 A 23 =. 2 4 Exercise: Given find A 2 and A 34. Solution: A = , A 2 = 2, A 34 = Definition. Suppose A is an n n matrix. Then its determinant, det(a) = A, is an element of C defined recursively by. If n =, then det(a) = a. 2. If n 2, then det(a) = a det(a ) a 2 det(a 2 ) + + ( ) +n a n det(a n ) n = ( ) +j a j det(a j ). i= 2 5 Example: Find det(a) for A =

4 Solution: Using the formula given in the definition, we have = ( )+ () ( )+2 (2) ( )+3 (5) = 3[( ) + ( 3) + ( ) +2 (7) 0 ] 2[( ) + (0) + ( ) +2 (7) ] + +5[( ) + (0) 0 + ( ) +2 ( 3) ] = 3( 3 0) 2( 7) + 5(3) = 26. Note that this definition also leads to the standard formula for the determinant of a 2 2 matrix. [ ] a b Theorem 4. Let A =. Then det(a) = ad bc. c d Definition. Given an n n matrix A, the (i, j) cofactor of A, denoted C ij, is given by C ij = ( ) i+j det(a ij ). Then, using this, we can define det(a) = a C + a 2 C a n C n. () Equation () is a cofactor expansion across the first row. Computing Determinants There are a number of ways to compute the determinant. Theorem 5. The determinant of an n n matrix A can be computed by a cofactor expansion across any row i, det(a) = ( ) i+ a i det(a i ) + ( ) i+2 a i2 det(a i2 ) + + ( ) i+n a in det(a in ), called a cofactor expansion along row i, or down any column j, det(a) = ( ) +j a j det(a j ) + ( ) i+2 a 2j det(a 2j ) + + ( ) n+j a nj det(a nj ), called a cofactor expansion along column j. Example: Compute det(a), where A = ,

5 using a cofactor expansion. Solution: We choose to do a cofactor expansion along row 2. det(a) = ( ) 2+ a 2 det(a 2 ) + ( ) 2+2 a 22 det(a 22 ) + ( ) 2+3 a 23 det(a 23 ) + ( ) 2+4 a 24 det(a 24 ) 2 2 = 0 det(a 2 ) + 0 det(a 22 ) + ( ) det(a 24) 2 2 = We will use a cofactor expansion along row 3 to compute this determinant. ( det(a) = 3 ( ) ( ) ) 2 6 = 3[5( 0 ( 2)) + 4( 6 ( 4))] = 6. Another property of the determinant follows. Theorem 6. Suppose that A is a square matrix. Then det(a t ) = det(a). Example: Let Then, 3 4 A = det(a) = 0 + ( ) det(a 22 ) + 0 = = 2((2) 4(0)) = 4. And, since 0 0 A t = 2 2, det(a t ) = ( ) = (2(2) (0)) = 4 = det(a). Consider the following example. 5

6 Example: Let Find det(a). Solution: A = = ( ) = ( ) = ()( ) = ()(6)( ) = ()(6)(2)(7(5) 2(0)) = ()(6)(2)(7)(5) = 420. Notice that this is equivalent to multiplying the numbers on the diagonal of A. We can generalize this to the theorem following this requisite definition. Definition.. An upper triangular matrix is a square matrix whose entries below the main diagonal are zero. 2. A lower triangular matrix is a square matrix whose entries above the main diagonal are zero. 3. A diagonal matrix is a square matrix whose entries above and below the main diagonal are zero. Theorem 7. If A is a triangular matrix, then det(a) is the product of the entries on the main diagonal. 6

7 Example: Evaluate Solution: = ()(6)(2)(7)(5) = Properties of Determinants of Matrices Theorem 8. Suppose that A is a square matrix with a row with every entry a zero or a column with every entry a zero. Then det(a) = 0. Proof. Suppose that A is an n n matrix and that every entry in row i is 0. Then, we can find det(a) by doing a cofactor expansion along row i, giving det(a) = = = n ( ) i+j a ij det(a ij ) j= n ( ) i+j 0 det(a ij ) j= n 0 = 0. j= The proof for the case of a column consisting entirely of zeros is similar. Theorem 9 (Row Operations and the Determinant). Let A be a square matrix. (a) If a multiple of one row of A is added to another row to produce a matrix B, then det B = det(a). (b) If two rows of A are interchanged to produce B, then det B = det(a). (c) If one row of A is multiplied by a nonzero scalar α to produce B, then det B = α det(a). 7

8 2 4 Example: Let A = We form matrix B by r 2 2r 0 4 = B Then, det(a) = 0 C C C 33 = 5( ) = 5( 4) = 20. det(b) = ( 4)(5) = 20. So, we see that det(b) = 20 = det(a). 2 2 Example: Let A = We form matrix B as follows: r 2r = B Then, det(a) = (3)(6) = 9 2 and det(b) = (3)(6) = 8. So, we see that det(b) = 2 det(a). 2 3 Example: Let A = 0. We form matrix B by r 2 r Then, So, we see that det(b) = det(a). det(a) = ()(5) = 5. det(b) = 0 C C C 23 = 5( ) = 5. Note that as a consequence of the above properties, we can show the following. 8

9 Theorem 0. Suppose that A is a square matrix with two equal rows or two equal columns. Then det(a) = 0. Proof. Suppose A is an n n matrix such that rows i and j are equal. Let B be the matrix from by subtracting row i from row j. Then det(a) = det(b) = 0. Example: Compute det(a) by row reducing to echelon form, where A = Solution: = r 2 r 2 +2r r 3 r 3 3r r 4 r 4 r = r 3 27 r 3 = 27()()()(4) = = r 3 r 3 +4r 2 r 4 r 4 +3r 2 = r 4 r 4 8r 3 Determinants, Row Operations, and Elementary Matrices First, we will prove a few theorems about elementary matrices, which we will use to prove an important theorem in a bit. Theorem. For every n, det(i n ) =. Proof. We can see that I n is a triangular matrix. Therefore, det(i n ) is equal to the product of the entries on the diagonal, or det(i n ) = (n times) =. 9

10 Theorem 2 (Determinants of Elementary Matrices). For the three possible versions of an elementary matrix, we have the determinants () det(e i,j ) =, (2) det(e i (α)) = α, (3) det(e i,j (α)) =. This theorem is proved by using the fact that each elementary matrix is obtained by performing a single elementary row operation on the identity matrix, and then applying the theorem on elementary row operations and the determinant Theorem 3. If E is an elementary matrix, then det(ea) = det(e) det(a). The proof of this theorem uses the theorem on determinants of elementary matrices and the fact that if we let B = EA, then B is the matrix formed by performing the row operation from E on A. A Quick Note on Column Operations (not in text) We can perform column operations on a matrix in the same way we perform row operations. Column operations have the same effect on determinants as row operations. NOTE: Do NOT perform column operations when solving systems of equations. Determinants, Nonsingular Matrices, Matrix Multiplication Theorem 4. Let A be a square matrix. Then A is singular if and only if det(a) = 0. This is proved in the text, so I will prove the following equivalent theorem. Theorem 5. A square matrix A is nonsingular if and only if det(a) 0. Proof. A can be reduced to reduced row echelon form U with a finite number of row operations, so U = E k E k E A, where E i represents an elementary matrix. 0

11 Then, det(u) = det(e k E k E A) = det(e k ) det(e k ) det(e ) det(a). Since det(e i ) 0 for all i, det(a) = 0 if and only if det(u) = 0. If A is nonsingular, then U = I, so det(u) = = det(a) 0. If det(a) = 0, then det(u) = 0 = U contains a row consisting entirely of zeros (since det U = u u 22 u nn ). Therefore, A is singular. Theorem 6 (Nonsingular Matrix Equivalences, Round 5). Suppose that A is a square matrix. Then the following are equivalent. () A is nonsingular. (2) A row reduces to the identity matrix. (3) The null space of A contains only the zero vector (i.e., N (A) = {0}). (4) The linear system LS(A, b) has a unique solution for every b. (5) The columns of A are linearly independent. (6) A is invertible. (7) The column space of A is C n (Col(A) = C n ). (8) The determinant of A is nonzero, i.e., det(a) 0. Finally, we have the following property of determinants. Theorem 7. If A and B are n n matrices, then det(ab) = det(a) det(b). Proof. If A or B is singular, then so is AB. So, det(ab) = 0 = det(a) det(b). If A is nonsingular, then Therefore, we have, A = E E 2 E t. det(ab) = det(e E 2 E t B) = det(e ) det(e 2 ) det(e t ) det(b) = det(e E 2 E t ) det(b) = det(a) det(b).

12 Corollary. Let A be an n n matrix. Then det(a k ) = [det(a)] k for k a nonnegative integer Example: Let A = Find det(a 3 ) Solution: Since det(a 3 ) = [det(a)] 3, det(a) = 2(9)(2) = 36. det(a 3 ) = [36] 3 = Example: Show that if A is invertible, then Solution: Since A is invertible, det(a ) = det(a). AA = I = det(aa ) = det(i) = (det(a))(det(a) ) = = det(a ) = det(a). 3 Cramer s Rule We first need the following notation. For an n n matrix A and any b R n, let A i (b) be the matrix obtained from A by replacing column i of A by the vector b; so, A i (b) = [ a a i b a i+ a n ]. Theorem 8 (Cramer s Rule). Let A be an invertible n n matrix. For any b R n, the unique solution x of Ax = b has entries given by x i = det(a i(b)), for i =, 2,..., n. det(a) Proof. Let A = [ a a 2 a n ] and I = [ e e 2 e n ]. If Ax = b, then AI i (x) = A [ e e i x e i+ e n ] = [ Ae Ae i Ax Ae i+ Ae n ] = [ Ae Ae i b Ae i+ Ae n ] = [ a a i b a i+ a n ] = Ai (b). 2

13 Then, (det(a))(det(i i (x)) = det(a i (b), and x i det(a) = det(a i (b)) = x i = det(a i(b)). det(a) Note that det(a) 0 since A is invertible. Example: Use Cramer s rule to solve 2x + x 2 = 7 3x + x 3 = 8 x 2 + 2x 3 = 3. Solution: For this problem, 2 0 A = 3 0 and det(a) = Applying Cramer s rule, we have x = 4 = 2 4 = x 2 = 4 = 4 4 = x 3 = 4 = 4 4 =. 3 So, x =. 3

14 Use Cramer s Rule for Engineering Applications Systems of first order differential equations solved using Laplace transforms can lead to systems of equations like 6sx + 4x 2 = 5 9x + 2sx 2 = 2. We need to know (a) for what values of s the is solution unique, and (b) what the solution is for these values of s. First, we know from Cramer s rule that if the coefficient matrix A is invertible, the solution is unique. Since [ ] 6s 4 A =, 9 2s we have 6s 4 9 2s = 2s2 36 = 2(s 2 3) = 2(s + 3)(s 3). Therefore, the system has a unique solution if s ± 3. For such an s, we have s x = A = 0s + 8 2(s 2 3). 6s x 2 = A 2s 45 = 2(s 2 3). A Formula for A Cramer s rule leads to a general formula for the inverse of an n n matrix as follows. The j th column of A is a vector x that satisfies Ax = e j. The i th entry of x is the (i, j) th entry of A. 4

15 By Cramer s rule, the We can show that (i, j) th entry of A = x i = det(a i(e j )). det(a) det(a i (e j )) = ( ) i+j det(a ji ) = C ji, where C is the cofactor matrix (the matrix of cofactors of A), and So, A ji = (n ) (n ) matrix formed by deleting row j and column i of A. A = det(a) C C 2 C n C 2 C 22 C n2... = det(a) Ct. C n C 2n C nn The matrix C t is the adjugate of A, adj(a). This leads us to the following theorem. Theorem 9. Let A be an invertible n n matrix. Then A = det(a) adj(a) Example: Find A if A = Solution: First, note that det(a) = 3()(2) = 6. Then, C = ( ) = 2 C 2 = ( ) = 2 C 3 = ( ) = 3 ( 2) = C 2 = ( ) = 0 C 22 = ( ) = 6 C 23 = ( ) = 9 C 3 = ( ) = 0 C 32 ( ) = 0 C 33 = ( ) = 3 5

16 So, A = =