We know that {e1,e2} = {(1,0)T,(0,1)T} is the standard basis for R2. Further, T(e1) = A(1,0)T = (5,-3)T and T(e2) = A(0,1)T = (8,3)T. Hence, the standard matrix of T is M = [T(e1),T(e2)] =
5 |
8 |
-3 |
3 |
Further, since det(M)= 15+24 = 39 ? 0, hence M is invertible. Therefore, T-1 exists and T-1: R2? R2 is defined by T-1 (x1,x2)T = M-1 . (x1,x2)T.
Now, M-1 =
1/13 |
-8/39 |
1/13 |
5/39 |
so that T-1(x1,x2)T=M-1.(x1,x2)T=(x1/13-8x2/39, x1/13+5x2/39)T. Thus, y1= x1/13-8x2/39 and y2= x1/13+5x2/39.
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show steps! Let T:R2-R2 be multiplication by A. Determine whether T has an inverse; if so,...
Problem #4: Let A = زرا and let T: R2 R2 be multiplication by A. Determine whether T has an inverse; if so, find T-7[x y]T = [a b]7. If the inverse exists then enter the values of a and b into the answer box below, separated with commas. If I has no inverse then enter 0 for both values. Problem #4: Enter your answer as a symbolic function of x,y, as in these examples
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