Linear Algebra 3. Determinants CSIE NCU 1

3. Determinants

3.1 Introduction of determinants …………….. 2 3.2 Properties of determinants ………………. 9 3.3 Cramer’s rule, volume, and linear transformations …………………… 18

Linear Algebra 3. Determinants CSIE NCU 2  Determinants are tools for analytic geometry and other parts of mathematics. For example, (1) provide an invertibility criterion for a square matrix. (2) give formulas for A-1 and A-1b. (3) derive the geometric interpretation of determinants.

3.1 Introduction of determinants  Notation

Assume that A is a square matrix. Let Aij denote the submatrix formed by deleting the i th row and j th column of A. Linear Algebra 3. Determinants CSIE NCU 3  Definition

For n ≥ 2, the determinant of an nxn matrix A = [aij] is the sum of n terms of the form ±aij detAij, with plus and minus signs alternative. That is

 Ex.1.

 Another notation det A = | A |.

Linear Algebra 3. Determinants CSIE NCU 4  For n ≤ 3, the determinant can be computed by

 Note The formula can only be used for the cases of n ≤ 3. Why the formula can not be used for the cases of n > 3 ? Linear Algebra 3. Determinants CSIE NCU 5 Example,

It only contains partial terms; not complete.  Definition

Given A = [aij], the (i,j)-cofactor of A is the number cij given i+j by cij = (-1) det Aij .

Linear Algebra 3. Determinants CSIE NCU 6  According to the definition of cofactor

det A = a11c11 + a12c12 +… + a1nc1n The formula is called the cofactor expansion along the first row.  Theorem 1

The determinant of Anxn may be computed by a cofactor expansion along any row or down any column. The expansion across the i th row is given

det A = ai1ci1 + ai2ci2 + … + aincin . The expansion across the j th column is given

det A = a1j c1j + a2j c2j + … + anj cnj .

 Theorem 1 is helpful for computing the determinant of a matrix that contains many zeros. The expansion is done across the row or column with most zeros. Linear Algebra 3. Determinants CSIE NCU 7  Ex.3.

Linear Algebra 3. Determinants CSIE NCU 8  Theorem 2 If A is a triangular matrix, then det A is the product of the entries on the main diagonal of A.

 Exercise of Section 3.1. Linear Algebra 3. Determinants CSIE NCU 9 3.2 Properties of determinants  Theorem 3 (row operations) Let A be a square matrix. (a) replacement A multiplication of one row of A is added to another to produce a matrix B, then det B = det A.

Er  det Er = 1. (b) interchange Two rows of A are interchanged to produce B, then det B = - det A.

Ei  det Ei = -1. (c) scaling One row of A is multipled by k to produce B, then

det B = k det A. Es  det Es = k.

Linear Algebra 3. Determinants CSIE NCU 10  Ex.2. Linear Algebra 3. Determinants CSIE NCU 11  Proof of Theorem 3 By induction. Assume A is an nxn matrix and E is an nxn elementary matrix. To prove that n = 2, it is true.

Assume that n = m, it is true. To prove that n = m+1, it is true. EA = B. The action of E on A involves either two rows or only one row. So we may expand det EA along the non-involving row, say, row i. Let Aij be the matrix obtained by deleting row i and column j from A. Then the rows of Bij are obtained from the rows of Aij by the same type of elementary row operation as E performs on A and α α then Bij = EAij  det Bij = det Aij , where = 1, -1 or k. Linear Algebra 3. Determinants CSIE NCU 12

α EA= B  Bij = EAij  det Bij = det Aij (m+1)×(m+1) m×m

A B = EA ← i th row

↑ j th column Aij Bij Linear Algebra 3. Determinants CSIE NCU 13

 Theorem 4 A square matrix A is invertible if and only if det A ≠ 0.

 Corollary

det Anxn = 0 if and only if the rows or columns of A are linearly dependent.

Linear Algebra 3. Determinants CSIE NCU 14  Ex.3.

 Ex.4. Linear Algebra 3. Determinants CSIE NCU 15 Column operations  Theorem 5 T Anxn is a square matrix. det A = det A. proof. By induction. If n = 2, trivial. Let n = m be true, to show n = m+1 is true. 1+j The cofactor of a1j in A (c1j = (-1) det A1j) T j+1 T = The cofactor of a’j1 in A (c’j1 = (-1) det A 1j )

LT a1j

A AT a’j1 LR RT

since a1j = a’j1 and c1j = c’j1 for j =1, 2, …, n.

Linear Algebra 3. Determinants CSIE NCU 16 Matrix products  Theorem 6 If A and B are nxn matrices, then det AB = (detA) (detB). proof. If A is not invertible, then neither is AB. (If AB is invertible, we take C = B (AB) -1; then AC = AB (AB) -1 = I  A is invertible.) In this case, (detA) (detB) = 0 = det A B. If A is invertible, A is a product of elementary matrices,

A = EpEp-1…E1.

|AB| = |EpEp-1…E1B| = |Ep||Ep-1…E1B| = …

= |Ep||Ep-1|…|E1||B| = |EpEp-1…E1||B| = |A||B|. Linear Algebra 3. Determinants CSIE NCU 17  Note that det (A + B) ≠ det A + det B .

For example, .

A linearity property of the determinant function

 Assume that A = [a1 a2 … an]. det A = f (a1, …, an), f is a linear function. Suppose that the j th column of A is allowed to vary, and other column are held fixed. We write

A = [a1 a2 … aj-1 xaj+1 … an ]. Define a transformation T from R n to R by ← T(x) = det [a1 a2 … aj-1 xaj+1 … an ]. (new definition) Then T(kx) = kT(x) for all scalars k and all x in R n. T(u + v) = T(u) + T(v) for all u, v in R n.  Exercises of Section 3.2.

Linear Algebra 3. Determinants CSIE NCU 18 3.3 Cramer’s rule, volume, and linear transformations  Notation n For any nxn matrix A and b in R . Let Ai(b) be the matrix obtained from A by replacing column i by the vector b,

Ai (b) = [a1 a2 … b … an ].

 Theorem 7 (Cramer’s rule) n Anxn is invertible, for any b in R , then the unique solution x of Ax = b has entries Proof.

Denote A = [a1 a2 … an ] and I = [e1 e2 …en ] AIi (x) = A [e1 e2 .. x .. en ] = [Ae1 Ae2 .. Ax .. Aen ] = [a1 a2 .. b .. an ] = Ai (b) By Theorem 6 ( det AB = det A det B )

(det A) (det Ii (x)) = det Ai (b)  (det A) xi = det Ai (b)  Linear Algebra 3. Determinants CSIE NCU 19  Ex.2.

3sx1 –2x2 = 4

-6x1 + sx2 = 1 has a unique solution. to find s and the solution. Answer. det A ≠ 0  exist a unique solution.

Linear Algebra 3. Determinants CSIE NCU 20 A formula for A -1 (the 3rd method) x1j x2j  Theorem 8 : x = Anxn is invertible, then j xij

xnj

Page 16 in Ch.2 -1 A A = I = [e1 e2 .. en] -1 A = [x1 x2 .. xn] Proof. A xj = ej -1 By Cramer’s rule, the j th column of A is a vector xj that -1 satisfies Axj = ej and the (i, j) entry of A is

row column Linear Algebra 3. Determinants CSIE NCU 21

the j th row

 Examples. omitted.

Linear Algebra 3. Determinants CSIE NCU 22 Determinants as area or volume

 2-dimensional cases (e, f) (g, h) (e-a, f-b) (g-a, h-b)

(a, b)(c, d) (0, 0) (c-a, d-b)

 3-dimensional cases

c b a Linear Algebra 3. Determinants CSIE NCU 23 Linear transformation  Theorem 10 (a) Let T: R 2 → R 2 be the linear transformation determined by a 2×2 matrix A. If S is a parallelogram in R 2, then {area of T(S)} = |det A| {area of S}. (b) If T: R 3 → R 3 determined by a 3×3 matrix A and S is a parallelepiped in R 3, then {volume of T(S)} = |det A| {volume of S}. b2 Proof. S

(a) A =[a1 a2] b1 S = { s1b1 + s2b2 : 0 ≤ s1 ≤ 1, 0 ≤ s2 ≤ 1 }

area of S = | det [b1 b2] |

T(S) = T(s1b1 + s2b2) = s1T(b1) + s2T(b2) = s1Ab1 +s2Ab2 is the parallelogram determined by the columns of matrix [Ab1 Ab2]. {area of T(s)} = | det A| { area of S }.

Linear Algebra 3. Determinants CSIE NCU 24

 The conclusion of Theorem 10 hold whenever S is a region in R 2 with finite area or a region in R 3 with finite volume.

 Ex. 5. Find the area of the region E bounded by the ellipse x2 b Answer. a x1 E is an image of the unit disk D under the linear transformation T determined by

{ area of ellipse } = { are of T(D)} = |det A| { area of D } = ab π 1 2 = ab π.  Exercise of Section 3.3.