# 3.1 - 3.1 Determinants For square matrix A associate a...

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Unformatted text preview: 3.1 Determinants For square matrix A, associate a number called the determinant of A, denoted detA. Deﬁnition: For n × n A, if n = 1, detA = a where A = [a]. a11 a12 · · · a1n if n > 1, A = ··· an1 an2 · · · ann detA = a11det A11 − a12 detA12 + · · · + (−1)n+1 a1n detA1n where Aij is the submatrix obtained from A by deleting the ith row and j th column. ∴ for n = 2 and A = a11 a12 a21 a22 detA = a11det[a22 ] − a12det[a21 ] = a11a22 − a12 a21. EX: A = 2 −1 1 3 =⇒ detA = 2 · 3 − (−1 · 1) = 7. n=3: a11 a12 a13 A = a21 a22 a23 a31 a32 a33 a22 a23 a32 a33 −a12det a21 a23 a31 a33 +a13det a21 a22 a31 a32 detA = a11det 1 = a11(a22 a33 − a32 a23) − a12 (a21a33 − a31 a23) +a13(a21 a32 − a31a22) = a11a22 a33 + a12a23a31 + a13 a21a32 −a13a22 a31 − a12a21a33 − a11a23a32 ∴ Trick for 3 × 3 matrices: Add products in “downward” direction, subtract them in “upward” direction: 3 2 1 0 0 1 2 3 4 5 EX: det 111 0 1 2 −1 3 1 − 1 · det 02 −1 1 + 1 · det 01 −1 3 = 1 · det 12 31 = 1(1 − 6) − 1(2) + 1(1) = −5 − 2 + 1 = −6 or trick: detA = (1 · 1 · 1) + (1 · 2 · −1) + (0 · 3 · 1) −(−1 · 1 · 1) − (0 · 1 · 1) − (3 · 2 · 1) = −6. 2 “downward” “upward” products Question: What about for A = a11 a12 a13 0 a11 a12 0 a22 Answer: From the above formulas, for these, detA is the product of the diagonal entries. Check this, especially for 3 × 3 case! or A = 0 a22 a23 ? 0 a33 Deﬁnition: The (i, j )-cofactor of A is the number + − + − + − + − + ··· − + − ··· ··· ··· ··· Cij = (−1)i+j detAij , where the (n − 1) × (n − 1) matrix Aij is A with row i and column j removed. detA = a11C11 + a12C12 + · · · + a1n C1n = n k=1 a1k C1k . Note: In the deﬁnition of determinant, we expanded in terms of cofactors along the ﬁrst row. It turns out that no matter which row or column we use for the expansion, we get the same number at the end. That is, 3 Theorem For n × n A, detA can be computed along any row or column: detA = ai1Ci1 + ai2 Ci2 + · · · + ainCin = a1j C1j + a2j C2j + · · · + anj Cnj EX: 1 −2 5 0 0 3 Notation: For A = 2 −6 −7 5 0 4 1 −2 write det A = 0 5 0 0 52 30 44 , so 2 0 , 5 4 2 −6 −7 5 4 1 −2 0 5 0 0 52 30 44 1 −2 2 = 0 · C21 + 0 · C22 + 3 · C23 + 0 · C24 2 −6 −7 5 = 3(−1)2+3 2 −6 5 04 ˆ ˆ ˆ = −3(5 · C31 + 0 · C32 + 4 · C33 ) = −3 5(−1)3+1 −2 2 −6 5 + 4(−1)3+3 1 −2 2 −6 5 = −3(5(−10 + 12) + 4(−6 + 4)) = −3(5(2) + 4(−2)) = −6 EX: 214 043 002 7 5 1 = 2 · C11 + 0 · C21 + 0 · C31 + 0 · C41 0 0 0 −1 43 = 2(−1)1+1 0 2 2 5 ˆ ˆ ˆ 1 = 2(4 · C11 + 0 · C21 + 0 · C31 ) 1 0 0 −1 = 2 4(−1)1+1 0 −1 = 2 · 4 · (2 · (−1)) = −16 Theorem For a (lower or upper) triangular matrix A detA = a11 a22· · ·ann 5 EX: Find determinants of A = 10 k1 ab cd a ab cd b and EA = = ka + c kb + d for this elementary matrix E : detA = a b ab cd = ad − bc ka + c kb + d = a(kb + d) − b(ka + c) = ad − bc = detA Question: Does this work for other elementary matrices? We’ll see. Question: What is the geometric interpretation of detA? Consider 2 × 2 matrix A = . Draw pictures for cd some simple cases to see that it is the area of the parallelogram formed by the column vectors of A. ab 6 ...
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