3 3 Matrices ⇥ Much of this chapter is similar to the chapter on 2 2matrices.Themost substantial di↵erence between 2 2matricesand3⇥3matricesisthatit’s harder to write a 3 3matrixthanitistowritea2⇥ ⇥ 2matrix. ⇥ ⇥ 3 3matriceshave3rowsand3columns.Theyareasquareblockof9 numbers,⇥ such as 206 4 514 0 10− 3 4 1 − Two matrices are equal if the@ entry in any positionA of the one matrix equals the entry in the same position of the other matrix. Examples. 32 7 32 7 32 7 32 7 50 1 = 50 1 but 50 1 = 59 1 069− 41 069− 41 069− 41 6 060− 41 @ A @ A @ A @ A ************* Matrices as functions A3 3matrixdefinesafunctionwhosedomainisR3 and whose target is ⇥ R3. The function is defined as follows: abc u au + bv + cw def v = du + ev + fw 0ghi1 0w1 0gu + hv + iw1 Notice that the first,@ second,A @ or thirdA @ entry in the vectorA on the right side of the above equation can be found by multiplying the first, second, or third row of the matrix abc def 0ghi1 and the column @ A 278 u v 0w1 @ A Example. The matrix 531 224 0−701 1 − has 3 rows and 3 columns, so@ it is a functionA whose domain is R3, and whose target is R3. Because, 2 9 0 3 1 − is a vector in R3, @ A 531 2 224 9 0−701 1 0 3 1 − − is also a vector in R3. The@ vector it equalsA @ is A 531 2 5 2+39+ 1( 3) 224 9 = 2· 2+2· 9+ 4· (−3) 0−701 1 0 3 1 0−7 ·2+0· 9+(1)· −( 3)1 − − · · − · − @ A @ A @ 10 + 27 3 A = 4+18−12 0−14 + 0 +− 3 1 @ 34 A = 2 0171 @ A Identity matrix Notice that 100 x 1 x +0 y +0 z x 010 y = 0 · x +1· y +0· z = y 00011 0z1 00 · x +0· y +1· z1 0z1 · · · @ A @ A @ 279 A @ A Thus, 100 010 00011 is the identity function whose domain@ is AR3. We call this matrix the 3 3 identity matrix. ⇥ ************* Matrix multiplication You can “multiply” two 3 3matricestoobtainanother3 3matrix. ⇥ ⇥ Order the columns of a matrix from left to right, so that the 1st column is on the left, the 2nd column is directly to the right of the 1st,andthe3rd column is to the right of the 2nd. To multiply two matrices, call the columns of the matrix on the right “input columns”, and put each of the input columns into the matrix on the left (thinking of it as a function). The column that is assigned to the 1st input column by the matrix function will be the 1st column of the product you are trying to find. The column that is assigned to the 2nd input column by the matrix function will be the 2nd column of the product, and the column that is assigned to the 3rd input column by the matrix function will be the 3rd column of the product. Example. To find the product 273 301 158 210 00411 01241 separate the matrix on the@ right intoA its@ three “inputA columns”: 301 3 0 1 210 2 1 0 01241 7! 011 021 041 @ A 280@ A @ A @ A Then the product 273 301 158 210 00411 01241 @ A @ A will be a 3 3matrixwhosefirstcolumn(whenreadleft-to-right)equals ⇥ 273 3 23 158 2 = 21 00411 011 0 9 1 @ A @ A @ A whose second column equals 273 0 13 158 1 = 21 00411 021 0 6 1 @ A @ A @ A and whose third column is 273 1 14 158 0 = 33 00411 041 0 4 1 @ A @ A @ A To repeat the previous sentence, 273 301 23 13 14 158 210 = 21 21 33 00411 01241 0 9641 @ A @ A @ A Matrix multiplication is function composition If A and B are 3 3matrices,thentheresultofmultiplyingthematrices ⇥ AB would determine the same function AB : R3 R3 as the function that results from composition, namely A B. ! ◦ ************* 281 Inverse matrices 1 If B is a 3 3matrix,thenB− is the 3 3matrixwhere ⇥ ⇥ 100 1 BB− = 010 00011 and @ A 100 1 B− B = 010 00011 @ A Example. We can check that the two matrices 702 10 2 411 and 11− 1 03011 0−30 71 − are inverses of each other@ byA multiplying@ them in eitherA order and checking to see that their product is the identity matrix: 702 10 2 100 411 11− 1 = 010 03011 0−30 71 00011 − and @ A @ A @ A 10 2 702 100 11− 1 411 = 010 0−30 71 03011 00011 − @ A @ A @ A ************* 282 Exercises 1.) The matrix 11 1 A = 02 3 03 20− 1 − describes a function A : R3 R@3. A Find the vectors ! 11 1 0 11 1 2 02 3 1 and 02 3 −1 03 20− 1 041 03 20− 1 0 0 1 − − @ A @ A @ A @ A 2.) The matrix 32 2 B = 10− 10 04 571 − describes a function B : R3 R@3. A Find the vectors ! 32 2 1 32 2 3 10− 10 2 and 10− 10 0 04 571 0 1 1 04 571 021 − − − @ A @ A @ A @ A 3.) Find the product 010 2 15 202 03− 1 00301 000 4 1 − @ A @ A 4.) Find the product 2 15 010 03− 1 202 000 4 1 00301 − @ 283A @ A 5.) Find the product 3 17 5 100 17− 3 34 010 041 3 181 00011 @ A @ A For #6 and #7, determine whether the two matrices given are inverses of each other. 412 021 6.) 101 and 113 01 101 0 1021 − − @ A @ A 110 1 11 7.) −211 and 2 −11 0−0011 0001− 1 @ A @ A 284 I 2 J I r 1. Match the functions with their graphs. 8.) (x +1)2 +2 10.)(x 1)2 +2 I 2 −J r 2 2 I 9.)2 (x J+1) 211.)(x 1) 2 I − −a 1. − I r I 1. / A.) B.) a I 2 J r I I a / 1. I / i 7, C.) D.) a I / i 7, i I 7,4—’ \ —2 I i I 7,4—’ 4—’ \ —2 \ —2 290 27 285 I 285 4—’ \ —2 290 27 285 290 27 285 290 27 285.