# Lecture 15 Recursive and Iterative Formula for Determinants Shang-Hua Teng.

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Lecture 15 Recursive and Iterative Formula for Determinants Shang-Hua Teng

Pseudo-Hypercube or Pseudo-Box n-Pseudo-Hypercube For any n affinely independent vectors

Determinant of Square Matrix How to compute determinant or the volume of pseudo-cube?

Properties of Determinant 1.det I = 1 2.The determinant changes sign when sign when two rows are changed (sign reversal) 1.Determinant of permutation matrices are 1 or -1 3.The determinant is a linear function of each row separately 1.det [a 1, …,ta i,…, a n ] = t det [a 1, …,a i,…, a n ] 2.det [a 1, …, a i + b i,…, a n ] = det [a 1, …,a i,…, a n ] + det [a 1, …, b i,…, a n ]

Properties of Determinant and Algorithm for Computing it [4] If two rows of A are equal, then det A = 0 –Proof: det […, a i,…, a j …] = - det […, a j,…, a i …] –If a i = a j then –det […, a i,…, a j …] = -det […, a i,…, a j …]

Properties of Determinant and Algorithm for Computing it [5] Subtracting a multiple of one row from another row leaves det A unchanged –det […, a i,…, a j - t a i …] = det […, a i,…, a j …] + det […, a i,…, - t a i …] One can compute determinant by elimination –PA = LU then det A = det U

Properties of Determinant and Algorithm for Computing it [6] A matrix with a row of zeros has det A = 0 [7] If A is triangular, then –det [A] = a 11 a 22 … a nn The determinant can be computed in O(n 3 ) time

Determinant and Inverse [8] If A is singular then det A = 0. If A is invertible, then det A is not 0

Determinant and Matrix Product [9] det AB = det A det B (|AB| = |A| |B|) –Proof: consider D(A) = |AB| / |B| 1.(Determinant of I) A = I, then D(A) = 1. 2.(Sign Reversal): When two rows of A are exchanged, so are the same two rows of AB. Therefore |AB| only changes sign, so is D(A) 3.(Linearity) when row 1 of A is multiplied by t, so is row 1 of AB. This multiplies |AB| by t and multiplies the ratio by t – as desired. –Therefore D(A) = |A|, so |AB| = |A| |B|

Determinant and Matrix Product [9’] | A -1 | = 1/|A| –Proof: 1 = det A A -1 = det A det A -1

Determinant and Matrix Product [10] | A -T | = |A| –Proof: PA = LU, so |P| |A| = |L| |U| = |U| A T P T = U T L T, so| A T | | P T | = | U T | |L T | = | U| (why?) P T = P -1. So P T P = I. So | P T | | P | = 1. So | P T | = | P | Therefore, | A T | = |A|

Determinants and Permutations Apply property [3]: the linearity rule on a 3 by 3 matrix

Permutations of n Items There are n! of them Basis for Permutation: pair-wise swaps Odd / Even permutation (1 2 3) ( 2 3 1) (3 1 2) (1 3 2) (2 3 1) (3 2 1)

Big Formula det A = sum over all n! column permutations

Cofactors and Recursive Formula Expanding the linearity decomposition by only one level. Matrix Minors –M ij Cofactors –C ij = (-1) i+j | M ij |

Cofactor Formula of Determinant i j

Signs

Determinants and Linear System Cramer’s Rule

Cramer’s Rule If det A is not zero, then Ax = b has the unique solution

Cramer’s Rule for Inverse Proof:

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