Question

$\text{If}\left|\begin{array}{ccc}-1& 7& 0\\ 2& 1& -3\\ 3& 4& 1\end{array}\right|=A,\text{then}\left|\begin{array}{ccc}13& -11& 5\\ -7& -1& 25\\ -21& -3& -15\end{array}\right|\text{is}$
$(I_3$ denotes the det of the identity matrix of order $3)$

• A. $A^2$
• B. $A^2-A+I_3$
• C. $A^2-3A+I_3$
• D. $3A^2+5A-4I_3$

Answer 1

The correct option is A $A^2$

Finding the determminant by expanding along $1^{st}$ row, we get
$A=\left|\begin{array}{ccc}-1& 7& 0\\ 2& 1& -3\\ 3& 4& 1\end{array}\right|$
$=-1(1+12)-7(2+9)=-13-77=-90$

$\text{Let}B=\left|\begin{array}{ccc}13& -11& 5\\ -7& -1& 25\\ -21& -3& -15\end{array}\right|$
Now, taking $3$ common form $C_3$ and $5$ from $R_3$, we get
$B=15\left|\begin{array}{ccc}13& -11& 1\\ -7& -1& 5\\ -7& -1& -1\end{array}\right|$
Now, expanding along row $R_3$, we get
$B=15(0-0-6(-13-77\left)\right)=-90\times -90$
Now, we know that $A=-90\Rightarrow B=A\times A=A^2$
Hence, Option A. is correct