Answer to Question #285333 in Linear Algebra for Sabelo Xulu

Question

13. Suppose T, S : R^2$\to$ R^2 are linear defined by T(u, v) =(3u + v, u + 2v) and S(x, y) =(2x - y, x + y). Also the matrices of T and S with respect to the standard bases of R^2 and R^2 are given as

M(T) =$\begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}$ and M(S) =$\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}$Then M(TS) =

(i)$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

(ii)$\begin{bmatrix} 3 & 1 \\ 1 & 5 \end{bmatrix}$

(iii)$\begin{bmatrix} 3 & 1 \\ 4 & 1 \end{bmatrix}$

(iv) None

14. Suppose T : R^2$\to$ R^2 is linear defined by T(x, y) = (y, x). Then the eigenvalues of T is...

(i) 1 and - 1

(ii) 0 and 2

(iii) Does not exist

(iv) None

Answer 1

13)

$M(TS)=\begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 2 & -1\\ 1 & 1 \end{bmatrix}=\begin{bmatrix} 7 & -2 \\ 4 & 1 \end{bmatrix}$

Correct option: None

14)

$T(x,y)=(y,x)$

$\implies\>T(x,y)=\lambda(x,y)$

$\therefore\>\lambda(x,y)=(y,x)$

$\lambda\>x=y........(i)$

$\lambda\>y=x........(ii)$

Substituting $(ii)$ in $(i)$

$\lambda(\lambda\>y)=y$

$\lambda^2=1$

$\lambda=^+_-1$

$\therefore\lambda=1\>or\>-1$