A. The linear transformation T1: ℝ^2 →ℝ^2 is given by: T1(x,...

a. The linear transformation $T_1: \mathbb{R}^2 \to \mathbb{R}^2$ is given by: $T_1(x, y) = (2x + 9y, 8x + 37y)$. Find $T_1^{-1}(x, y)$. $T_1^{-1}(x, y) = (\frac{37}{2}x + \frac{-9}{2}y, -4x + y)$ b. The linear transformation $T_2: \mathbb{R}^3 \to \mathbb{R}^3$ is given by: $T_2(x, y, z) = (x + 2z, 2x + y, 2y + z)$. Find $T_2^{-1}(x, y, z)$. $T_2^{-1}(x, y, z) = (\frac{1}{2}x + \frac{1}{2}y + \frac{-1}{2}z, \frac{-1}{2}x + \frac{1}{2}y + \frac{1}{2}z, \frac{1}{2}x + \frac{-1}{2}y + \frac{1}{2}z)$

Transcript

00:01 We have four parts in this problem.

00:04 The first part, a, the linear transformation t1, goes from r2 to r2 and is given by t1 at x, y equal the vector 2x plus 7y and 4x, sorry, 8x plus 29y.

00:20 So we want to find the inverse of this linear transformation and put the coefficients of x and y for each entry of that inverse of t1.

00:29 So let's say that t1 at any xy is just equal to some vector ab in r2 so the idea here is to find which is the xy which generate a vector ab any vector in r2 and that will be the definition of the inverse so what we want is then that the vector 2x plus 7y first component and 8x plus 29y second component be equal to a vector ab in r2 but then the equality of these two vectors means that each components are equal that is the corresponding components are equal so 2x plus 7y is equal to a and 8x plus 29y is b and from here we want to find x y so we got to solve this 2 by 2 linear system so to do that we are going to multiply the first equation by negative 4 so we get negative 8x minus 28y equal negative 4a.

01:57 Second equation stays the same, 8x plus 29y equal b.

02:09 And now we add together these two equations side by side, so we get 8x minus 8x cancel out, 29y minus 28y is y, y equal to b minus 4a.

02:28 So that's the value of y.

02:30 Now we can find x by using any of the two equations here.

02:34 We're going to use the first one.

02:36 So from 2x plus 7y equal a, and the fact that y y is b minus 4a, we get that 2 times x plus 7 times y, which is replaced by b minus 4a, is equal to a.

03:15 That is, we have replaced y, but the expression b minus 4a into this equation here.

03:23 And so we can do some algebra right here to x plus 7 times b minus 28a equal a.

03:34 So x is equal to negative 28a goes to the right as 28a plus a is 29a.

03:48 And then minus 7 times b and all that divided by 2 which is the coefficient of x here so that's the value of x so we have put indeed the pre -image xy in terms of the given vector ab and r2 so this is defining the inverse of x of t1 and that we write down not using variables a and b but x and y again that is a is replaced by x and b replaced by y to these formulas so t1 inverse at x y is just equal to the vector whose first component is this one here but writing x and y as i said before 29 x minus 7 y divided by 2 and the second component is this one here replacing a by x and b by y so we get negative 4x plus y i'm writing also in the other way around the terms in this order in order to have x first and y second.

05:09 So that's the formula.

05:10 We can even say this is 29 half x minus 7 half y, first component and second component, and put it down here, negative 4x plus y.

05:28 I write it this way in order to see clearly which are the coefficients of x and y in each of the components of the inverse so let's say the inverse of t is this one any of the two formulas let's say this one and since you get to answer this way putting the coefficients of x and y in each of the components so we know that the first component has coefficient 29 half for x that is here you get to write down and 29 half.

06:07 Here you have already written negative 7 half, which is the coefficient of y in the first component.

06:15 That's correct.

06:17 The coefficient of x now in the second component is negative 4.

06:23 So here you get to put negative 4...

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