Q68 / 155 · WBJEE 2018 · Mathematics
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 \begin{array}{l} \hline \end{array} $\left(I_{3}\right.$ denotes the det of the identity matrix of order 3 )
- Correct answerA$A^{2}$
- B$A^{2}-A+1_{3}$
- C$A^{2}-3 A+l_{3}$
- D$3 A^{2}+5 A-4 l_{3}$
Solution
We have.
$$
\begin{aligned}
A=&\left|\begin{array}{ccc}
-1 & 7 & 0 \\
2 & 1 & -3 \\
3 & 4 & 1
\end{array}\right| \\
&=-1(1+12)-7(2+9) \\
&=-13-77=-90
\end{aligned}
$$
And let $\quad B=\left|\begin{array}{ccc}13 & -11 & 5 \\ -7 & -1 & 25 \\ -21 & -3 & -15\end{array}\right|$
$$
=3 \times 5\left|\begin{array}{ccc}
13 & -11 & 1 \\
-7 & -1 & 5 \\
-7 & -1 & -1
\end{array}\right|
$$
$$
\begin{array}{l}
=15\left|\begin{array}{ccc}
13 & -11 & 1 \\
-7 & -1 & 5 \\
0 & 0 & -6
\end{array}\right|\left(R_{3} \rightarrow R_{3}-R_{2}\right) \\
=15\{0-0-6(-13-77)\} \\
=15\{(-6)(-90)\}=90 \times 90
\end{array}
$$
From Eqs. (i) and (ii),
$$
B=A^{2}
$$
Hence, the correct answer is option: 'A'.