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 $$\det \left( A \right) = - 1$$
Statement - 2 :
If $$A \ne I$$ and $$A \ne - I,$$ then $${\text{tr}}\left( A \right) \ne 0.$$
A.
Statement -1 is false, Statement-2 is true
B.
Statement -1 is true, Statement-2 is true; Statement -2 is
a correct explanation for Statement-1
C.
Statement -1 is true, Statement-2 is true; Statement -2
is not a correct explanation for Statement-1
D.
Statement -1 is true, Statement-2 is false
Answer :
Statement -1 is true, Statement-2 is false
Solution :
\[{\rm{Let }}\,\,A = \left[ \begin{array}{l}
a\,\,\,\,\,b\\
c\,\,\,\,\,d
\end{array} \right]{\rm{then }}\,\,{A^2} = I\]
$$\eqalign{
& \Rightarrow \,\,{a^2} + bc = 1\,\,\,\,\,\,\,\,\,ab + bd = 0 \cr
& \,\,\,\,\,\,ac + cd = 0\,\,\,\,\,\,\,\,\,bc + {d^2} = 1 \cr} $$
From these four relations,
$$\eqalign{
& \,\,\,\,\,\,\,\,\,{a^2} + bc = bc + {d^2} \Rightarrow \,{a^2} = {d^2} \cr
& {\text{and }}b\left( {a + d} \right) = 0 = c\left( {a + d} \right) \Rightarrow \,\,a = - d \cr} $$
We can take $$a = 1,b = 0,c = 0,d = - 1\,\,{\text{as one}}$$
possible set of values, then \[A = \left[ \begin{array}{l}
1\,\,\,\,\,\,\,\,0\\
0\,\,\,\, - 1
\end{array} \right]\]
Clearly $$A \ne I$$ and $$A \ne - I$$ and det $$A = - 1$$
∴ Statement 1 is true.
Also if $$A \ne I$$ then tr$$(A) = 0$$
∴ Statement 2 is false.