Question

Two matrices A and B are given below :

$A=\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]$ $B=\left[\begin{array}{cc}{p}^2+q^2& pr+qs\\ pr+qs& r^2+s^2\end{array}\right]$

If the rank of matrix A is N, then the rank of matrix B is

• A. $\frac{N}{2}$
• B. $N-1$
• C. $N$
• D. $2N$

Answer 1

The correct option is C $N$

Method 1: $\left|A\right|=\left|\begin{array}{cc}p& q\\ r& s\end{array}\right|=ps-qr$

$\left|B\right|=$ $\left|\begin{array}{cc}{p}^2+q^2& pr+qs\\ pr+qs& r^2+s^2\end{array}\right|$

= $(ps-qr{)}^2$ = ${\left|A\right|}^2$

$\therefore$ So A & B becomes singular or non-singular simultaneously.
So $\rho \left(A\right)=\rho \left(B\right)=N$

Method 2:

$A=\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]$ and $B=\left[\begin{array}{cc}{p}^2+q^2& rp+sq\\ pr+qs& r^2+s^2\end{array}\right]$

Now let us calculate

$A{A}^T=\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]\left[\begin{array}{cc}p& r\\ q& s\end{array}\right]$

= $\left[\begin{array}{cc}{p}^2+q^2& rp+sq\\ pr+qs& r^2+s^2\end{array}\right]$

= B

we know that $\rho \left(A\right)=\rho \left(A{A}^T\right)=N$

$\Rightarrow \rho \left(A\right)=\rho \left(B\right)=N$