5 indicate whether each of the following functions is linear...

5. Indicate whether each of the following functions is linear and if it is, find the standard matrix that represents the linear transformation. (a) T : R^2 => R^2 given by T(x, y) = (1 + x, y) (b) T : R^2 => R^2 given by T(x, y)=(x-y, 0) (c) T : R^2 => R^2 given by T(x, y) = (0, xy) (d) T : R^3 => R^2 given by T(x, y, z) = (0, 0) (e) T : R^3 => R^2 given by T(x, y, z)=(x, y) (f) T : R^3 => R^4 given by T(x, y, z)=(z^2, x, y)

Transcript

00:01 Hello students, we know that a transformation t from r2 to r2 is linear if t of u plus v is equal to t of u plus t of v for some vectors u comma v belong to r2 and t of c of u is equal to c into t of u.

00:16 Also if t of 0 .0 is equal to audit pair 0 .0, we say that t is linear.

00:24 Thus if t of 0 .0 is not equal to 0 0, then we can see that t is linear.

00:26 Thus if t of 0 .0 is not equal to 0 0, then we can see that t.

00:30 Is not linear.

00:31 We use these facts for the proofs of above.

00:35 So, a t from r2 to r2 by t of x comma y is equal to 1 plus x comma y is not linear because t of 0 .0 is equal to 1 .0 which is not equal to 0 comma 0.

01:05 Now for b we have t from r2 to r2 given by t of x comma y is equal to audit pair x minus y comma 0.

01:21 So let u is equal to audit pair x1.

01:25 Y1 and b is equal to audit pair x2 2, y2.

01:32 Let these two be the elements of r2.

01:35 So then we have u plus v is equal to audit x1 plus x2 comma y1 plus y2.

01:46 Now we have to find t of u plus v so t of u plus v will be equal to t of x1 plus x2 comma y1 plus y 2 which will be equal to audit pair x1 plus x2 minus y1 plus y2 comma 0 which can be written as x1 minus y1 plus x2 which can be again written as x1 minus y1 plus x2 minus y2 comma 0 which can be again written as x1 minus y1 0 plus x2 minus y2 comma 0 which is equal to t of u plus t of v also t of c u is equal to t of c x1 comma c y 1 which is equal to c x1 minus c y 1 0 which is equal to c into x1 minus y1 comma 0 which can be written as c of t u so therefore we can say that t is linear to find the matrix we have to find t of 1 .0 and t of 0 .1 .1 and arrange them into columns so t of 1 .0 is equal to 1 .0 and t of 0 .0 is equal to 1 .0 and t of 0.

03:27 Is equal to minus 1 comma 0.

03:30 So the matrix a will be equal to 1 0 minus 1 0.

03:39 Now for c, t from r2 to r2 is given by t of x comma y is equal to 0...

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