Add a multiple of a row to another row, replacing the row which was not multiplied.

Transcription

1 Determinants Properties involving elementary row operations There are a few sections on properties. Rirst, we ll simply state collections of properties, provide some examples, and talk about why they are useful. As you look at the examples, start thinking about why these properties would have to be true. At the end, the proofs are attched - please read through carefully! Recall the three elementary row operations used in elimination: Multiply a row by a constant. Swap rows. Add a multiple of a row to another row, replacing the row which was not multiplied. We know that performing these operations on a matrix does not affect the solution to the system it represents. The question is, do they affect the determinant, and how? Theorem: If B is obtained by multiplying a row (or column) of A by a constant k, then B = k A If B is obtained by exchanging two rows (or columns) of A, (corollary) If A has two identical rows (or columns) B = A A =0 If B is obtained by adding a multiple of a row (or column) of A to another row (or column) of A (replacing the row which has not been multiplied), the value of the determinant remains unchanged: B = A Note that at this point, we ve described the effect that the elementary row operations have on the determinant: Multiply a row by a constant, and it ll multiply the determinant by a constant (E2). Swap two rows, and it ll reverse the sign (E1). Add a multiple of a row to another row, replacing the unmultiplied row, and the determinant will be unchanged(e3).

2 Example: Let Use cofactor expansion to compute A : A = Now, without doing any more computations (other than maybe a little multiplication in your head), what are the determinants of B = C = D = (Figure out how the original A was transformed through a row or column operation (or operations) into each of these matrices. Then apply the appropriate rule for what happens to the determinant after that row operation.)

3 Example: Let A = a b c d e f g h i and suppose that A = 5. What are the values of the determinants of B = a +2b b c d +2e e f C = 3d f e 3a c b g +2h h i 3g i h (Again, figure out how the original was transformed. Might take more than one row (or column) operation.) The row operation properties essentially tell us it s OK to do Gaussian elimination to a matrix carefully in order to make it easier to evalulate the determinant. The idea is that we can use elimination to produce zeros, and matrices with lots of zeros are easy to find determinants of. The key is carefully - we re always safe with the add a multiple of a row (no change to the determinant), but if we swap rows or multiply by a constant, we need to keep up with the effect that had on the determinant. We ll see a little of this in the next section.

4 The proofs: If B is obtained by multiplying a row (or columns) of A by a constant k, then B = k A Proof: Let A be an n n matrix. Suppose we multiply the ith row of A by a constant k, resulting in the matrix B. Consider the cofactor expansion of B along that ith row: B = b i1 C i1 + b i2 C i b in C in The coefficients b i1,b i2,...b in are the entries in the ith row of B... which was obtained multiplying the ith row of A by the scalar k. So b i1 = ka i1 b i2 = ka i2... b in = ka in The cofactors of B (the C i1 etc) are formed from elements NOT in the ith row of B. So these are identical to the same cofactors of the matrix A, since none of these elements are changed. So B = ka i1 C i1 + ka i2 C i ka in C in = k(a i1 C i1 + a i2 C i a in C in ) But (a i1 C i1 + a i2 C i a in C in ) is just the cofactor expansion of A on the ith row. Therefore B = k A If B is obtained by exchanging two rows (or columns) of A, B = A Proof: This one is oddly tricky to prove, and beyond our scope to write the formal notation (I ll at least explain it). You can t do it by looking at a cofactor expansion. You have to look at the permutation definition: Let A be an n by n matrix. Then A = ±a 1j1 a 2j2...a njn where the sum is over all permutations j 1 j 2...j n of the set {1, 2,..., n}. Even permutations are signed (+); odd permutations are signed ( ). When you swap rows, you are permuting the indices once. This means it takes one more flip to restore the permutations in the formula to the original state (or maybe one less). For the matrix A, a certain number of permutations will be odd, and the others even. For the matrix B, allofa s even permutations are now odd, and all of A s odd permutations are now even. That means that while none of the numbers in the terms of the sum have changed, all of the signs have switched. So B = A.

5 (corollary) If A has two identical rows (or columns) A =0 I squeezed this corollary in here because (1) it follows immediately from the previous result, and (2) allows us to prove the next result! Proof: Suppose A is a matrix with two identical rows. Construct a matrix B by swapping the two identical rows of A. From the previous property, we know that B = A. However, since we have swapped identical rows, B is identical to A, so B = A. So A = A 2 A = 0 A = 0 If B is obtained by adding a multiple of a row (or column) of A to another row (or column) of A (replacing the row which has not been multiplied), the value of the determinant remains unchanged: B = A Suppose B is obtained by performing the operation krj + Ri Ri on the matrix A (note we re taking a multiple of the jth row, adding it to the ith row, and replacing the ith row). Consider the cofactor expansion of B along the ith row: B = b i1 C i1 + b i2 C i b in C in Consider how all the coefficients were obtained from krj + Ri Ri: b i1 = ka j1 + a i1 b i2 = ka j2 + a i2. =. b in = ka jn + a in B = b i1 C i1 + b i2 C i b in C in = (ka j1 + a i1 )C i1 +(ka j2 + a i2 )C i (ka jn + a in )C in = ka j1 C i1 + a i1 C i1 + ka j2 C 12 + a i2 C i ka jn C in + a in C in = [ka j1 C i1 + ka j2 C i ka jn C in ]+[a i1 C i1 + a i2 C i a in C in ] = k[a j1 C i1 + a j2 C i a jn C in ]+[a i1 C i1 + a i2 C i a in C in ] The quantity [a i1 C i1 + a i2 C i a in C in ] is just the cofactor expansion of A on the ith row, so we have B = k[a j1 C i1 + a j2 C i a jn C in ]+ A (now, the sneaky bit...) Consider the matrix obtained by taking the ith row of A and replacing it with a copy of the j row of A. You know two things about it: first, the cofactor expansion on the i th row now looks like this: a j1 C i1 + a j2 C i a jn C in (all the a iq elements have been replaced by a jq elements). Second, its determinant must be zero, because it has two identical rows! So a j1 C i1 + a j2 C i a jn C in =0

6 and therefore, B = k[a j1 C i1 + a j2 C i a jn C in ]+ A = k[0] + A = A