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USC If A and B are both m × n matrices then the sum of A and B, denoted A + B, is a matrix obtained by adding corresponding corresponding elements of elements of AA and and B. B. add these add these add these add these add these add these 111 222 222 AA A 11 22 22 AA000 111 333 0 1 3 0 1 3 3 0 4 3 3 0 0 4 4 B 3 0 4 3 0 4 BB 44 BB 222311 01 B 2 2 1 1 44 44 2 1 4 2 2 222 A B 2 A B AA AABB BB 2 22 2 6 2222 666 0 1 0 USC A B B A A ( B C ) ( A B) C USC If A is an m × n matrix and s is a scalar, then we let kA denote the matrix obtained by multiplying every element of A by k. This procedure is called scalar multiplication. 1 2 2 31 3 2 32 3 6 6 A 3A 0 1 3 3 0 3 1 3 3 0 3 9 PROPERTIES OF SCALAR MULTIPLICATION k hA kh A k h A kA hA k A B kA kB USC The m × n zero matrix, denoted 0, is the m × n matrix whose elements are all zeros. 0 0 0 0 0 0 0 1×3 2×2 A0 A A ( A) 0 0 A 0 USC The multiplication of matrices is easier shown than put into words. You multiply the rows of the first matrix with the columns of the second adding products Find AB 3 2 1 A 0 4 1 322 2211 1 3 5 3 2 4 B 1 3 3 1 First we multiply across the first row and down the first column adding products. We put the answer in the first row, first column of the answer. USC Find AB 3 2 1 A 0 4 1 2 4 B 1 3 3 1 55 77 AB AB 0 0 0 0 2 4 4 2 4 4 4 4 3 3 1 1 1 1 3 11 1 3 3 4 2 2 3 3 1 1 7 1 11 Notice the sizes of A and B and the size of the product AB. Now we multiply across the first second rowrow andand down down the the second first We multiplied across first row and down first column column column second and we’lland putwe’ll the answer put the in answer the second first in the row,row, second second firstrow, so we put the answer in the first row, first column. USC column.column. second To multiply matrices A and B look at their dimensions mn n p MUST BE SAME SIZE OF PRODUCT If the number of columns of A does not equal the number of rows of B then the product AB is undefined. USC Now let’s look at the product BA. 2 4 3 2 1 A B 1 3 0 4 1 3 1 3132121213123223114134034140104410 9 6 414 2312 across acrossthird second second third row row row row across first second row row as as aswe we godown down we as we go go go down down first third third first second second third column: column: column: column: column: second column: first column: column: 32 23 12 12 12 12 2 6666 12 222 BA BA BA BA BA 333 14 14 14 4 4 4 99 10 4 10 Commuter's Beware! Completely different than AB! AB BA USC PROPERTIES OF MATRIX MULTIPLICATION ABC ABC AB C AB AC A B C AC BC AB BA Is it possible for AB = BA ?,yes it is possible. USC What is AI? 2 1 2 0 1 5 A 2 2 3 1 0 0 I 3 20 1 12 0 What is IA? 0 1 5 A 0 0 1 2 2 3 1 2 2 A 0 1 5 2 2 3 Multiplying a matrix by the identity gives the matrix back again. 1 0 0 I 3 0 1 0 0 0 1 an n n matrix with ones on the main diagonal and zeros elsewhere USC Can we find a matrix to multiply the first matrix by to get the identity? 1 3 1 1 2 1 ? 3 4 22 0 2 1 1 3 1 1 2 3 4 2 0 2 2 0 1 0 1 Let A be an n n matrix. If there exists a matrix B such that AB = BA = I then we call this matrix the inverse of A and denote it A-1. USC If A has an inverse we say that A is nonsingular. If A-1 does not exist we say A is singular. To matrix A, A, a a To find find the the inverse inverse of of a a matrix matrix we we put put the the matrix line identity matrix. matrix. We We then then perform perform row row line and and then then the the identity operations toturn turnititinto intothe theidentity. identity. We We operations on on matrix matrix A A to carry right hand hand side side carry the the row row operations operations across across and and the the right will turn into into the inverse. will turn the inverse. 3 1 3 1 0 1 2 7 0 1 1 3 1 0 2r1+r2 0 1 2 1 A 2 7 r2 r 1 r2 0 1 3 1 0 1 2 1 3 1 0 7 0 1 2 USC1 3 1 A 2 7 3 7 A 2 1 1 Check this answer by multiplying. We should get the identity matrix if we’ve found the inverse. 1 0 AA 0 1 1 USC We can use A-1 to solve a system of equations x 3y 1 2x 5 y 3 To see how, we can re-write a system of equations as matrices. Ax b coefficient matrix variable matrix 1 3 2 5 x y constant matrix 1 3 USC Ax b 1 left multiply both sides by the inverse of A 1 A Ax A b This is just the identity 1 Ix A b but the identity times a matrix just gives us back the matrix so we This then gives us a formula have: for finding the variable matrix: Multiply A inverse by the constants. 1 x A b USC x 3y 1 2x 5 y 3 1 3 1 0 2 5 0 1 1 3 1 0 -r2 0 1 2 1 1 3 A 2 5 -2r1+r2 find the inverse 1 0 1 3 0 1 2 1 0 5 3 0 1 2 1 r1-3r2 1 5 3 1 4 1 A b 2 1 3 1 x y This is the answer to the system USC Your calculator can compute inverses and determinants of matrices. To find out how, refer to the manual or click here to check out the website. USC Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. www.slcc.edu Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar www.ststephens.wa.edu.au USC