Group theory

Question

The set of matrices $S = \left \{ \begin{bmatrix} x&-x \\ -x&x \end{bmatrix} | 0 \neq x \in R \right \}$ forms a group under matrix multiplication operation with identity element :

a) $\begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}$

b) $\begin{bmatrix} 1&-1 \\ -1&1 \end{bmatrix}$

c)$\begin{bmatrix} -1&1 \\ 1&-1 \end{bmatrix}$

d)$\begin{bmatrix} \frac{1}{2}&-\frac{1}{2} \\ -\frac{1}{2}&\frac{1}{2} \end{bmatrix}$

Comments

It seems bit strange that matrix of such type will have determinant 0 so how can we get inverse ?

$\left | S \right | = x * x - (-x)*(-x) = x^2 - x^2 = 0$

this way all the matrices of such sets will be singular.So no inverse possible. This set cannot form a group under matrix multiplication but while multiplying with option D matrix getting the same matrix so answer is D (identity)

So this set S can be a Monoid under matrix multiplication but it cant be a group under multiplication.

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Yup, but the answer given is option D.

Answer 1

Answer is D.

Comments

correct answer but wrong approach.

Answer 2

$\begin{bmatrix} x&-x \\ -x&x \end{bmatrix}$*$\begin{bmatrix} a&b \\ c&d \end{bmatrix}$ = $\begin{bmatrix} x&-x \\ -x&x \end{bmatrix}$

on solving we get, a - c = 1 and d – b = 1 so, any value of a,b in following matrix can be Identity matrix.

$\begin{bmatrix} a+1&b \\ a&b+1 \end{bmatrix}$