Question

If $\left|\begin{array}{ccc}-1& 7& 0\\ 2& 1& -3\\ 3& 4& 1\end{array}\right|=A$, then $\left|\begin{array}{ccc}13& -11& 5\\ -7& -1& 25\\ -21& -3& -15\end{array}\right|$ is

• 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 $A^2$

$A=\left|\begin{array}{ccc}-1& 7& 0\\ 2& 1& -3\\ 3& 4& 1\end{array}\right|=-1(1+12)-7(2+9)=-13-77=-90$

$B=\left|\begin{array}{ccc}13& -11& 5\\ -7& -1& 25\\ -21& -3& -15\end{array}\right|$

$=15\left|\begin{array}{ccc}13& -11& 1\\ -7& -1& 5\\ -7& -1& -1\end{array}\right|$

$=15\left|\begin{array}{ccc}13& -11& 1\\ -7& -1& 5\\ 0& 0& -6\end{array}\right|$

$R_3-R_2\to R_3$

$=15\left[\right(0-0)-6(-13-77\left)\right]=90\times 90=A^2$