Let A be [a,2 2] matrix with real entries.

Question

Let $A$ be $$a\,2 \times 2$$ matrix with real entries. Let $I$ be the $$2 \times 2$$ identity matrix. Denote by tr$(A)$, the sum of diagonal entries of $a$. Assume that ${a^2} = I.$

Statement-1 : If $$A \ne I$$ and $$A \ne  - I$$, then det$(A)=-1$

Statement- 2 : If $$A \ne I$$ and $$A \ne  - I$$, then tr $(A)$ $$ \ne 0$$.

Accepted Answer

Let $$A = \left[ {\begin{pmatrix} a & b  \
   c & d  \

\end{pmatrix} } \right]$$ $$\,\,\,$$ then ${A^2} = 1$

$$ \Rightarrow {a^2} + bc = 1\,\,\,\,ab + bd = 0$$

$$ac + cd = 0\,\,\,\,bc + {d^2} = 1$$

From these four relations,

$${a^2} + bc = bc + {d^2} \Rightarrow {a^2} = {d^2}$$

and $$\,\,b\left( {a + d} \right) = 0 = c\left( {a + d} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a =  - d$$

We can take $a = 1,b = 0,c = 0,d =  - 1$

as one possible set of values, then

$$A = \left[ {\begin{pmatrix} 1 & 0  \
   0 & { - 1}  \

\end{pmatrix} } \right]$$

Clearly $$A \ne I\,\,\,$$ and $$\,\,\,\,A \ne  - I\,\,$$ and $$\,\,\,A =  - 1$$

$$\therefore$$ $$\,\,\,\,\,$$ Statement $1$ is true.

Also if $$A \ne I\,\,\,\,\,tr\left( A \right) = 0$$

$$\therefore$$ $$\,\,\,\,\,$$ Statement $2$ is false.

Solutions · 1