Matrix algebra for beginners, Part I matrices, determinants
matrices with capital letters, like A, B, etc, although we will sometimes use lower case letters for one dimensional matrices (ie: 1 ×m or n ×1 matrices) One dimensional matrices are often called vectors, as in row vector for a n ×1 matrix or column vector for a 1 ×m matrix but we are going
Matrices ch 3 311006
MATRICES 59 ANote In this chapter 1 We shall follow the notation, namely A = [aij] m × n to indicate that A is a matrix of order m × n 2 We shall consider only those matrices whose elements are real numbers or
CHAPTER 8: MATRICES and DETERMINANTS
SECTION 8 1: MATRICES and SYSTEMS OF EQUATIONS PART A: MATRICES A matrix is basically an organized box (or “array”) of numbers (or other expressions) In this chapter, we will typically assume that our matrices contain only numbers Example Here is a matrix of size 2 3 (“2 by 3”), because it has 2 rows and 3 columns: 10 2 015
LVandenberghe ECE133A(Fall2019) 3Matrices
Addition:sumoftwom n matricesA andB (realorcomplex) A+ B = 2 6 6 6 6 6 6 4 A11 + B11 A12 + B12 A1n + B1n A21 + B21 A22 + B22 A2n + B2n::: :: :: Am1 + Bm1 Am2 + Bm2 Amn + Bmn 3 7 7 7 7 7 7 5 Matrices 3 11
matrix identities
0 9 block matrices for conformably partitioned block matrices, addition and multiplication is performed by adding and multiplying blocks in exactly the same way as scalar elements of regular matrices however, determinants and inverses of block matrices are very tricky; for 2 blocks by 2 blocks the results are: 11 11 A A 12 A 21 A 22 22= jA jjF
EXERCISES OF MATRICES OPERATIONS Question 1
EXERCISES OF MATRICES OPERATIONS 3 (24) If A,Bare both n×nmatrices and ABis singular, then Ais singular or Bis singular (25) Ais diagonalizable, then Ais non-singular (26) Ais symmetric, then Ais non-singular (27) If all the eigenvalues of Aare 1, then Ais similar to the identity matrix (28) If all the eigenvalues of Aare 1, then Ais non
LINEAR RECURRENCES AND MATRICES - Purdue University
LINEAR RECURRENCES AND MATRICES 3 This is a quadratic equation, so it can be solved using highschool algebra Either we get two distinct solutions 1; 2 or just one solution Let us assume the rst case Then THEOREM 3 2 If (3) has two distinct solutions 1; 2, then all the solutions to (2) are given by a n = b 1 n 1 + b 2 n 2 for constants b 1;b
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