Question

If A and B are two matrices of order '3' such that $3A+4B{B}^T=I$ and $B^{-1}=A^T$, then identify which of the following statements is/are correct?

• A. $T_r(A^{-1}-4B^3-BA+I)=9$
• B. $T_r(A^{-1}-4B^3+BA)=12$
• C. $|A^2-3A^3|=64$
• D. $|A^{-1}-4B^3-BA|=8$

Answer 1

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

The correct options are
A $T_r(A^{-1}-4B^3-BA+I)=9$
B $T_r(A^{-1}-4B^3+BA)=12$
C $|A^2-3A^3|=64$
D $|A^{-1}-4B^3-BA|=8$

$3A+4B{B}^T=I$
$\Rightarrow 3A=I-4B{B}^T$
$\Rightarrow 3A^T=I^T-4(B{B}^T{)}^T$
$\Rightarrow 3A^T=I-4B{B}^T=3A$
So A is symmetric.
So $B^{-1}=A^T=A$
So $B^{-1}=A\Rightarrow B=A^{-1}$
$\begin{array}{cc}3A+4B.B=I& 3A+4(A^{-1}{)}^2=I\\ 3B^{-1}+4B^2=I& 3A^2+4A^{-1}=A\\ 3I+4B^3=B=A^{-1}& 3A^3+4I=A^2\\ A^{-1}-4B^3=3I& A^2-3A^3=4I\end{array}$
So $Tr(A^{-1}-4B^3-BA+I)=Tr(3I-I+I)=9$
$T_r(A^{-1}-4B^3+BA)=T_r(3I+I)=12$
$|A^2-3A^3|=|4I|=64$
$|A^{-1}-4B^3-BA|=|3I-I|=|2I|=8$