Question

If $A$ and $B$ are two matrices of the same order, then

• A. $A^2-B^2=(A+B)(A-B)$
• B. $A^2=I\iff (A+I)(A-I)=0$
• C. $(A^{\prime }{)}^{\prime }=A$
• D. $A+A^{\prime }$ is symmetric

Answer 1

Answer: (B), (C), (D)

The correct options are
B $A^2=I\iff (A+I)(A-I)=0$
C $(A^{\prime }{)}^{\prime }=A$
D $A+A^{\prime }$ is symmetric

A and B$\to$ Two matrices of same order

Option A: $A^2-B^2=(A+B)(A-B)$

RHS$=A^2-AB+BA+B^2$

RHS$=$LHS if and only if $AB=BA$

$\to$ Option A is not true for all cases

Option B: $A^2=I\iff (A+I)(A-I)=0$

RHS$=A^2-A+A-I=0$

$\Rightarrow A^2=I$

$\Rightarrow$ Option B is correct

Option C: ${\left(A^T\right)}^T=A$

Transpose means we reverse the columns and rows

Let C be columns and R be rows

$A^T$ means C becomes rows

${\left(A^{{}^{\prime }}\right)}^{{}^{\prime }}$ mean columns and rows are again interchanged $C\to$columns

$R\to$rows

$\therefore {\left(A^{{}^{\prime }}\right)}^{{}^{\prime }}=A$

Option C is correct

Option D Let $A+A^T=P{P}^T={(A+A^{{}^{\prime }})}^{{}^{\prime }}=A^{{}^{\prime }}+A$

$\therefore P=P^T\therefore A+A^T={(A+A^T)}^T\Rightarrow$symmetric

Option D is correct