Let T : R 3 → R 3 T :\mathbb R^3\to\mathbb R^3 T : R 3 → R 3 T ( x 1 , x 2 , x 3 ) = ( 3 x 1 + x 3 , − 2 x 1 + x 2 , − x 1 + 2 x 2 + 4 x 3 ) T(x_1, x_2, x_3) = (3x_1 + x_3, - 2x_1 + x_2, - x_1 + 2x_2 + 4x_3) T ( x 1 , x 2 , x 3 ) = ( 3 x 1 + x 3 , − 2 x 1 + x 2 , − x 1 + 2 x 2 + 4 x 3 )
Let us fix arbitrary ( a , b , c ) ∈ R 3 (a,b,c)\in \mathbb R^3 ( a , b , c ) ∈ R 3 T ( x 1 , x 2 , x 3 ) = ( a , b , c ) T(x_1, x_2, x_3) =(a,b,c) T ( x 1 , x 2 , x 3 ) = ( a , b , c )
{ 3 x 1 + x 3 = a − 2 x 1 + x 2 = b − x 1 + 2 x 2 + 4 x 3 = c . \begin{cases}3x_1 + x_3=a\\ - 2x_1 + x_2=b\\ - x_1 + 2x_2 + 4x_3=c\end{cases}. ⎩ ⎨ ⎧ 3 x 1 + x 3 = a − 2 x 1 + x 2 = b − x 1 + 2 x 2 + 4 x 3 = c .
Since the determinant ∣ 3 0 1 − 2 1 0 − 1 2 4 ∣ = 12 − 4 + 1 = 9 ≠ 0 \left| \begin{array}{ccc}3 &0 & 1\\ -2 & 1 & 0\\ -1 & 2 & 4 \end{array} \right|=12-4+1=9\ne 0 ∣ ∣ 3 − 2 − 1 0 1 2 1 0 4 ∣ ∣ = 12 − 4 + 1 = 9 = 0 ( a , b , c ) ∈ R 3 (a,b,c)\in \mathbb R^3 ( a , b , c ) ∈ R 3 T : R 3 → R 3 T :\mathbb R^3\to\mathbb R^3 T : R 3 → R 3 T − 1 T^{-1} T − 1
To give the expression for T − 1 T^{-1} T − 1
{ x 3 = a − 3 x 1 x 2 = b + 2 x 1 − x 1 + 2 ( b + 2 x 1 ) + 4 ( a − 3 x 1 ) = c \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} ⎩ ⎨ ⎧ x 3 = a − 3 x 1 x 2 = b + 2 x 1 − x 1 + 2 ( b + 2 x 1 ) + 4 ( a − 3 x 1 ) = c
{ x 3 = a − 3 x 1 x 2 = b + 2 x 1 − x 1 + 2 b + 4 x 1 + 4 a − 12 x 1 = c \begin{cases} x_3=a-3x_1 \\ x_2=b+2x_1 \\ - x_1 + 2b+4x_1 + 4a-12x_1=c\end{cases} ⎩ ⎨ ⎧ x 3 = a − 3 x 1 x 2 = b + 2 x 1 − x 1 + 2 b + 4 x 1 + 4 a − 12 x 1 = c
{ x 3 = a − 3 x 1 x 2 = b + 2 x 1 x 1 = 1 9 ( 4 a + 2 b − c ) \begin{cases} x_3=a-3x_1 \\ x_2=b+2x_1 \\ x_1=\frac{1}{9}(4a+2b-c)\end{cases} ⎩ ⎨ ⎧ x 3 = a − 3 x 1 x 2 = b + 2 x 1 x 1 = 9 1 ( 4 a + 2 b − c )
{ x 3 = a − 3 9 ( 4 a + 2 b − c ) x 2 = b + 2 9 ( 4 a + 2 b − c ) x 1 = 1 9 ( 4 a + 2 b − c ) \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} ⎩ ⎨ ⎧ x 3 = a − 9 3 ( 4 a + 2 b − c ) x 2 = b + 9 2 ( 4 a + 2 b − c ) x 1 = 9 1 ( 4 a + 2 b − c )
{ x 3 = a − 4 3 a − 2 3 b + 1 3 c x 2 = b + 8 9 a + 4 9 b − 2 9 c x 1 = 1 9 ( 4 a + 2 b − c ) \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} ⎩ ⎨ ⎧ x 3 = a − 3 4 a − 3 2 b + 3 1 c x 2 = b + 9 8 a + 9 4 b − 9 2 c x 1 = 9 1 ( 4 a + 2 b − c )
{ x 3 = − 1 3 a − 2 3 b + 1 3 c x 2 = 8 9 a + 13 9 b − 2 9 c x 1 = 1 9 ( 4 a + 2 b − c ) \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} ⎩ ⎨ ⎧ x 3 = − 3 1 a − 3 2 b + 3 1 c x 2 = 9 8 a + 9 13 b − 9 2 c x 1 = 9 1 ( 4 a + 2 b − c )
{ x 3 = 1 3 ( − a − 2 b + c ) x 2 = 1 9 ( 8 a + 13 b − 2 c ) x 1 = 1 9 ( 4 a + 2 b − c ) \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} ⎩ ⎨ ⎧ x 3 = 3 1 ( − a − 2 b + c ) x 2 = 9 1 ( 8 a + 13 b − 2 c ) x 1 = 9 1 ( 4 a + 2 b − c )
Therefore, T − 1 ( x 1 , x 2 , x 3 ) = ( 1 3 ( − x 1 − 2 x 2 + x 3 ) , 1 9 ( 8 x 1 + 13 x 2 − 2 x 3 ) , 1 9 ( 4 x 1 + 2 x 2 − x 3 ) ) 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) T − 1 ( x 1 , x 2 , x 3 ) = ( 3 1 ( − x 1 − 2 x 2 + x 3 ) , 9 1 ( 8 x 1 + 13 x 2 − 2 x 3 ) , 9 1 ( 4 x 1 + 2 x 2 − x 3 ) )
