Question

Let T : R3 —> R3 be defined by

T(xi, x2, x3) = (3x1 + x3, - 2x1 + x2,

- x1 + 2x2 + 4x3)

Show that T^(-1) exists. Give the expression for T^(-1)(x1 , x2, x3) for T above.

Answer 1

Let $T :\mathbb R^3\to\mathbb R^3$ be defined by $T(x_1, x_2, x_3) = (3x_1 + x_3, - 2x_1 + x_2, - x_1 + 2x_2 + 4x_3)$.

Let us fix arbitrary $(a,b,c)\in \mathbb R^3$ and consider the equation $T(x_1, x_2, x_3) =(a,b,c)$ which is equivalent to the system:

$\begin{cases}3x_1 + x_3=a\\ - 2x_1 + x_2=b\\ - x_1 + 2x_2 + 4x_3=c\end{cases}.$

Since the determinant $\left| \begin{array}{ccc}3 &0 & 1\\ -2 & 1 & 0\\ -1 & 2 & 4 \end{array} \right|=12-4+1=9\ne 0$ , the system has a unique solution for each $(a,b,c)\in \mathbb R^3$. Therefore, $T :\mathbb R^3\to\mathbb R^3$ is a bijection, and $T^{-1}$ exists. 

To give the expression for $T^{-1}$ let us solve the above system.

$\begin{cases} x_3=a-3x_1 \\ x_2=b+2x_1 \\ - x_1 + 2(b+2x_1) + 4(a-3x_1)=c\end{cases}$

$\begin{cases} x_3=a-3x_1 \\ x_2=b+2x_1 \\ - x_1 + 2b+4x_1 + 4a-12x_1=c\end{cases}$

$\begin{cases} x_3=a-3x_1 \\ x_2=b+2x_1 \\ x_1=\frac{1}{9}(4a+2b-c)\end{cases}$

$\begin{cases} x_3=a-\frac{3}{9}(4a+2b-c) \\ x_2=b+\frac{2}{9}(4a+2b-c) \\ x_1=\frac{1}{9}(4a+2b-c)\end{cases}$

$\begin{cases} x_3=a-\frac{4}{3}a-\frac{2}{3}b+\frac{1}{3}c \\ x_2=b+\frac{8}{9}a+\frac{4}{9}b-\frac{2}{9}c \\ x_1=\frac{1}{9}(4a+2b-c)\end{cases}$

$\begin{cases} x_3=-\frac{1}{3}a-\frac{2}{3}b+\frac{1}{3}c \\ x_2=\frac{8}{9}a+\frac{13}{9}b-\frac{2}{9}c \\ x_1=\frac{1}{9}(4a+2b-c)\end{cases}$

$\begin{cases} x_3=\frac{1}{3}(-a-2b+c) \\ x_2=\frac{1}{9}(8a+13b-2c) \\ x_1=\frac{1}{9}(4a+2b-c)\end{cases}$

Therefore, $T^{-1}(x_1,x_2,x_3)=\left(\frac{1}{3}( -x_1-2x_2+x_3) , \frac{1}{9}(8x_1+13x_2-2x_3),\frac{1}{9}(4x_1+2x_2-x_3)\right)$