DETERMINANTS AND EIGENVALUES 1. Introduction Gauss ...

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80 II. DETERMINANTS AND EIGENVALUESadjacent rows. All told, to switch rows i and j in this way requires 2k + 1 switchesof adjacent rows. The net effect is to multiply the determinant by (−1) 2k+1 = −1as required.There is one important consequence of rule (iii) which we shall use later in theproof of rule (i).Rule (iiie): If an n × n matrix has two equal rows, then det A =0.This is not too hard to see. Interchanging two rows changes the sign of det A,but if the rows are equal, it doesn’t change anything. However, the only numberwith the property that it isn’t changed by changing its sign is the number 0. Hence,det A =0.We next verify rule (ii). Suppose A ′ is obtained from A by multiplying the ithrow by c. Consider the recursive definition(1) det A ′ = a ′ 11D 1 (A ′ )+···+(−1) i+1 a ′ i1D i (A ′ )+···+(−1) n+1 a n1 D n (A).For any j ≠ i, D j (A ′ )=cD j (A) since one of the rows appearing in that determinantis multiplied by c. Also, a ′ j1 = a j1 for j ≠ i. On the other hand, D i (A ′ )=D i (A)since the ith row is deleted in calculating these quantities, and, except for the ithrow, A ′ and A agree. In addition, a ′ i1 = ca i1 so we pick up the extra factor ofc in any case. It follows that every term on the right of (1) has a factor c, sodet A ′ = c det A.Finally, we attack the proof of rule (i). It turns out to be necessary to verify thefollowing stronger rule.Rule (ia): Suppose A, A ′ , and A ′′ are three n × n matrices which agree except inthe ith row. Suppose moreover that the ith row of A is the sum of the ith row of A ′and the ith row of A ′′ . Then det A = det A ′ + det A ′′ .Let’s first see why rule (ia) implies rule (i). We can add c times the jth row ofA to its i row as follows. Let B ′ = A, let B ′′ be the matrix obtained from A byreplacing its ith row by c times its jth row, and let B be the matrix obtained formA by adding c times its jth row to its ith row. Then according to rule (ia), we havedet B = det B ′ + det B ′′ = det A + det B ′′ .On the other hand, by rule (ii), det B ′′ = c det A ′′ where A ′′ has both ith and jthrows equal to the jth row of A. Hence, by rule (iiie), det A ′′ = 0, and det B = det A.Finally, we establish rule (1a). Assume it is known to be true for (n−1)×(n−1)determinants. We have(2) det A = a 11 D 1 (A) −···+(−1) i+1 a i1 D i (A)+···+(−1) n+1 a n1 D n (A).For j ≠ i, the the sum rule (ia) may be applied to the determinants D i (A) becausethe appropriate submatrix has one row which breaks up as a sum as needed. Hence,D j (A) =D j (A ′ )+D j (A ′′ ).
2. DEFINITION OF THE DETERMINANT 81Also, for j ≠ i, we have a j1 = a ′ j1 = a′′ j1 since all the matrices agree in any rowexcept the ith row. Hence, for j ≠ i,a i1 D i (A) =a i1 D i (A ′ )+a i1 D i (A ′′ )=a ′ i1D i (A ′ )+a ′′i1D i (A ′′ ).On the other hand, D i (A) =D i (A ′ )=D i (A ′′ ) because in each case the ith rowwas deleted. But a i1 = a ′ i1 + a′′ i1 ,soa i1 D i (A)=a ′ i1D i (A)+a ′′i1D i (A) =a ′ i1D i (A ′ )+a ′′i1D i (A ′′ ).It follows that every term in (2) breaks up into a sum as required, and det A =det A ′ + det A ′′ .Exercises for Section 2.1. Find the determinants of each of the following matrices. Use whatever methodseems most ⎡ convenient, but seriously consider the use of elementary row operations.(a) ⎣ 1 1 2⎤1 3 5⎦.6 4 1⎡⎤1 2 3 4⎢ 2 1 4 3⎥(b) ⎣⎦.1 4 2 34 3 2 1⎡⎤0 0 0 0 31 0 0 0 2(c) ⎢ 0 1 0 0 1⎥⎣⎦ .0 0 1 0 4⎡0 0 0 1 2(d) ⎣ 0 x y⎤−x 0 z⎦.−y −z 02. Verify the following rules for 2 × 2 determinants.(i) If A ′ is obtained from A by adding a multiple of the first row to the second,then det A ′ = det A.(ii) If A ′ is obtained from A by multiplying its first row by c, then det A ′ =c det A.(iii) If A ′ is obtained from A by interchanging its two rows, then det A ′ = − det A.Rules (i) and (ii) for the first row, together with rule (iii) allow us to derive rules(i) and (ii) for the second row. Explain.3. Derive the following generalization of rule (i) for 2 × 2 determinants.[ ] [ ] [ ]adet′ + a ′′ b ′ + b ′′ a′b= det′ a′′b+ det′′.c dc d c dWhat is the corresponding rule for the second row? Why do you get it for free ifyou use the results of the previous problem?
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