Answer to Question #146827 in Linear Algebra for Sourav Mondal

Question #146827

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.


1
Expert's answer
2020-12-01T01:59:22-0500

Let T:R3R3T :\mathbb R^3\to\mathbb R^3 be defined by T(x1,x2,x3)=(3x1+x3,2x1+x2,x1+2x2+4x3)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)R3(a,b,c)\in \mathbb R^3 and consider the equation T(x1,x2,x3)=(a,b,c)T(x_1, x_2, x_3) =(a,b,c) which is equivalent to the system:

{3x1+x3=a2x1+x2=bx1+2x2+4x3=c.\begin{cases}3x_1 + x_3=a\\ - 2x_1 + x_2=b\\ - x_1 + 2x_2 + 4x_3=c\end{cases}.


Since the determinant 301210124=124+1=90\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)R3(a,b,c)\in \mathbb R^3. Therefore, T:R3R3T :\mathbb R^3\to\mathbb R^3 is a bijection, and T1T^{-1} exists. 


To give the expression for T1T^{-1} let us solve the above system.


{x3=a3x1x2=b+2x1x1+2(b+2x1)+4(a3x1)=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}


{x3=a3x1x2=b+2x1x1+2b+4x1+4a12x1=c\begin{cases} x_3=a-3x_1 \\ x_2=b+2x_1 \\ - x_1 + 2b+4x_1 + 4a-12x_1=c\end{cases}


{x3=a3x1x2=b+2x1x1=19(4a+2bc)\begin{cases} x_3=a-3x_1 \\ x_2=b+2x_1 \\ x_1=\frac{1}{9}(4a+2b-c)\end{cases}


{x3=a39(4a+2bc)x2=b+29(4a+2bc)x1=19(4a+2bc)\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}


{x3=a43a23b+13cx2=b+89a+49b29cx1=19(4a+2bc)\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}


{x3=13a23b+13cx2=89a+139b29cx1=19(4a+2bc)\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}


{x3=13(a2b+c)x2=19(8a+13b2c)x1=19(4a+2bc)\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, T1(x1,x2,x3)=(13(x12x2+x3),19(8x1+13x22x3),19(4x1+2x2x3))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)



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