Find $a\in\mathbb{R}$ that satisfies linear transformation, $T(1,−1,1) = (2,a,−1)$, $T(1,−1,2)=(a^2,−1,1)$, $T(1,-1,-1)=(5,-1,-7)$.

Question

Find all the possible $a\in\mathbb{R}$ such that there exists a linear transformation $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ that satisfies that $T(1,−1,1)=(2,a,−1)$, $T(1,−1,2)=(a^2,−1,1)$ and $T(1,−1,−2)=(5,−1,−7)$.

This is the first time that I've come across a problem like this one, could someone explain how to do it?

I was trying to find patterns between the given integers to find a rule that $T$ is defined by, but I didn't have any luck. The only other thing I can think of is maybe to set up a matrix? Not sure how it would look though.

Answer 1

By linearity of $T$:

$$T(0,0,1) = T(1,-1,2)-T(1,-1,1)=(a^2-2, -a-1, 2)$$

Now we can use this, along with $T(1,-1,-2)$, to see for which $a$ these values will be consistent.

Answer 2

I claim there is no such linear transformation.

I solved $x(1,-1,1)+y(1,-1,2)=(1,-1,-1)$ to get $x=3$ and $y=-2$.

Thus, if $T$ were a linear transformation, by linearity we would have

$T(0,0,0)=3T(1,-1,1)-2T(1,-1,2)-T(1,-1,-1)$

$=3(2,a,-1)-2(a^2,-1,1)-(5,-1,-7)$.

But we can see already that the third component is not $0$, so $T$ is not linear.

[If $T$ were linear, we would have $T(0,0,0)=(0,0,0)$.]