Math 1b Practical — More on Determinan tsFebruary 17, 2010We pr ove, in these notes, the rules (i)–(iii) given for determinants on page 2 of thefirst set of Notes on Determinants. The current notes are provided for completeness of thematerial on determinants—but we won’t have time to do everything in class and you willnot be r esponsible for everything below. I will warn you about parts that you are to beresponsible for.Recall that the determinant of a 1 × 1 matrix is the entry of the matrix, and once wehave defined the determinant of (n − 1) × (n − 1) matrices, we define the determinant ofan n × n matrix A =bydet(A)=a11det(A11) − a12det(A12)+...+(−1)n−1a1ndet(A1n). (1)Here Aijdenotes the submatrix of A obtained by deleting row i and column j from A.It will be convenient to change notation somewhat. First, let us use a1,a2,...,anforthe top row of a square matrixA =a1a2··· anBand write BjforwhatwehavecalledA1j, the result of deleting row 1 and column j fromA. Now the definition of determinant for n ≥ 2readsdet(A)=a1det(B1) − a2det(B2)+...+(−1)n−1andet(Bn). (2)Prop osition 1. If Ais obtained from A by multiplying a row by a scalar t,thendet(A)=t det(A).Proof: By induction. This is clearly true for 1 × 1 matrices. Assume A is n × n and thatthe Proposition holds for (n − 1) × (n − 1) matrices.If row 1 is multiplied by t, then the top r ow of Ais (ta1,...,tan)anddet(A)=(ta1)det(B1) − (ta2)det(B2)+...+(−1)n−1(tan)det(Bn)=t det(A).Assume row j is multiplied by t to get A,wherej ≥ 2. If we us e Bjto denote thematrix obtained from Aby deleting row 1 and column j,thendet(Bj)=t det(Bj)foreach j =1, 2,...,n by the induction hypothesis, and sodet(A)=a1(t det(B1)) − a2(t det(B2)) + ...+(− 1)n−1an(t det(Bn)) = t det(A). Prop osition 2. If a r ow a of A is the s um s + t,andifS and T are obtained from A byreplacing the row containing a by s and t, respectively, then det(A)=det(S)+det(T ).For example,det⎛⎝23423 −74267 ♠∗⎞⎠=det⎛⎝11123 −74267 ♠∗⎞⎠+det⎛⎝12323 −74267 ♠∗⎞⎠.Proof: By induction. This is clearly true for 1 × 1 matrices. Assume A is n × n and thatthe Proposition holds for (n − 1) × (n − 1) matrices.Suppose row 1 of A is a = s + t, and that the i-th coordinates of a, s, t are denotedby ai,si,ti, respectively. Thendet(A)=a1det(B1) − a2det(B2)+...+(−1)n−1andet(Bn)=(s1+ t1)det(B1) − (s2+ t2)det(B2)+...+(−1)n−1(sn+ tn)det(Bn)=(s1det(B1) − ...+(−1)n−1sndet(Bn))+(t1det(B1) − ...+(−1)n−1tndet(Bn))=det(S)+det(T ).If row  of A is a = s + t, >1, then the result follows from the induction hypothesis, butwe om it writing down the details. The following formula (5) is need for our pro of of Proposition 3. It may be confusing,so to help understanding, we write the formula out completely when n =3andn =4atthe end of the notes.Suppose a1,a2,...,anis the top row and b1,b2,...,bnthe second row of a squarematrixA =a1a2··· anb1b2··· bnC.Let Cjkdenote the matrix obtained by deleting rows 1 and 2, and columns j and k,fromA. By the definition of determinant,det(A)=a1det(B1) − a2det(B2)+...+(−1)n−1andet(Bn)(3)anddet(Bj)=b1det(C1j) − b2det(C2j)+... where there is NO term involving bj. (4)The s ign of bidet(Cij) in this expression for det(Bj)is(−1)i−1if i<jand (−1)i−2wheni>j. If we substitute (4) into (3), we obtain the expressiondet(A)=(−1)j−i−1i<jdetaiajbibjdet(Cij). (5)Prop osition 3. If Ais obtained from A by interchanging two rows, then det(A)=− det(A).Proof: By induction, starting with n =2.If the interchanged rows are row 1 and row 2, then the result follows from (5), sincedetbibjaiaj= − detaiajbibj.Now assume that the proposition holds for (n − 1) × (n − 1) matrices and let A ben × n. We have already handled the case when the interchanged rows are row 1 and row2. If the interchanged rows ar e i and j both ≥ 2, then the result follows from (1) and theinduction hyp othesis. If the interchanged rows are row 1 and row j ≥ 3, then Acan beobtained from A by interchanging rows 1 and 2, then rows 2 and j,thenrows1and2.This is three interchanges and the sign changes after each one, because it is either rows 1and 2 that are interchanged, or rows i and j with both ≥ 2. So det(A)=− det(A). Prop osition 4. If A has a two equal rows, then det(A)=0.Proof: Suppose row i and row j of A are identical, i = j.LetAbe obtained from Aby interchanging rows i and j. By Proposition 3, det(A)=− det(A). But A= A,sodet(A)=det(A). The only wa y this can happen is if det(A)=0. Prop osition 5. If Ais obtained from A by replacing row i by the sum of row i and ttimes row j,wherei = j.Thendet(A)=det(A).Proof: Say row i and j of A contain the vectors a and b, respectively. So row i of Aisa + tb.LetB be the m atrix where row i of A is replaced by tb. By Prop os i tion 2,det(A)=det(A)+det(B).But det(B)ist times the determinant of a matrix with two equal rows (rows i and j), andso by Propos ition 4, det(B) = 0. The Proposition follows. ***The formula (5) says, in general,det(A)=deta1a2b1b2det(C12) − deta1a3b1b3det(C13)+deta2a3b2b3det(C13)+deta1a4b1b4det(C14) − deta2a4b2b4det(C24)+deta3a4b3b4det(C34)− deta1a5b1b5det(C15)+....In particular, for n =3andn =4:det⎛⎝a1a2a3b1b2b3c1c2c3⎞⎠=deta1a2b1b2c3− deta1a3b1b3c2+deta2a3b2b3c1.det⎛⎜⎝a1a2a3a4b1b2b3b4c1c2c3c4d1d2d3d4⎞⎟⎠=deta1a2b1b2detc3c4d3d4− deta1a3b1b3detc2c4d2d4+deta2a3b2b3detc1c4d1d4+deta1a4b1b4detc2c3d2d3−