Question

For each transformation T and basis B and C, find the corresponding matrix representation M of T from basis B to basis C. 1) Let T6 = la + 2b + 4c 3a +86 + 16c la + 3b + 6c be a linear transformation. -2a +(-7) + (-14)c] с 1 Let B= 2 > -1 4 0 2 Let C = [11] [32] [] [1] The matrix M for transformation T from basis B to C would be: 2) Let T(a + bx + cx? + dr3) = la +(-1) b+3c+(-1)d - 1a + 3b + (-5)c + Od - 1a + 2b + (-4)c + Od la +(-1) b + 3c +(-2)] Let B= (1 + 1x + 1x2 + (-1) 2*), (2 + 3x + 4x2 + (-3) 23), (1 + 2x + 4x2 + (-3)x),(-1+0x + 3x2 + (-1)x3) Let C= [% 1] [E4 1] [31] [1, 1] The matrix M for transformation T from basis B to C would be:

Answer 1

B to C, we arrange the coefficients of the image of B under T as columns, after arranging linear combination of vectors of c. them as a T (a b c la +26+40 3 a +8t + loc da+36+60 -29+67)6+(140) B = (21)3 (-40)" ( 42) (-2), ( -5 -3) (-9 -6.) ; ( 3 ) First, note that for ( a be ( a ) e M2x2 :) ift5.) tis -1 -52-34+39 (a la 2) +2 2 3 - 26 - 34-60+2's ---2-20s -p+39 + 55-23 -> ၆၉ il b C --52-9 +35 2 - 39 -6 or +23 - p +39 +5r-2g þ & -28-3 al solving these tequations * - 59 t 6-5.6 td 3 3 q = -26 + 76-200 + 4 de 8= sarub + 4cd = a + 2d CA Sa CS Scanned with CamScanner

Again let - است ناحیا ده ر ) ، . :-) - ای پی - Cit في --3. - ق.) T (at txt cx=dx) = a +36-sc. سے ہوا۔ 24 - ام۔ 3 - رنو 4 ---3 ا ا 3- ا ا - ( ) ( 50- = 3- 3 S T1+x+x²x 19 - 2 ولا هكدا ) - ر 43 - ہوتا ۔ 1/3+2+92- 3) = -13 17 ۱۶ -3 دل )) - ا ا - 2619 - في 6- کا ۲(27+7 - 3x )= ر كز - و ہلا 4 تا 36 - 9 - 2 - = 11- ۲( ( +0+ 33) - (-۱۲ C so - 19 - (26 و 2 سے .ا - 14 = م - 19 -9 2. -) 56 36 -۱۷ -6 . ۔ As CS Scanned with CamScanner

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