Search any question & find its solution
Question:
Answered & Verified by Expert
Let $A$ be a non-singular square matrix of order $3 \times 3$. Then $|\operatorname{Adj} A|$ is equal to :
(a) $|\mathbf{A}|$
(b) $|\mathbf{A}|^2$
(c) $|\mathbf{A}|^3$
(d) $\mathbf{3}|\mathbf{A}|$
(a) $|\mathbf{A}|$
(b) $|\mathbf{A}|^2$
(c) $|\mathbf{A}|^3$
(d) $\mathbf{3}|\mathbf{A}|$
Solution:
1273 Upvotes
Verified Answer
$\begin{aligned}
&\text { Let } \mathrm{A}=\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right] \\
&\text { Let Adj. } \mathrm{A}=\left[\begin{array}{lll}
\mathrm{A}_{11} & \mathrm{~A}_{21} & \mathrm{~A}_{31} \\
\mathrm{~A}_{12} & \mathrm{~A}_{22} & \mathrm{~A}_{32} \\
\mathrm{~A}_{13} & \mathrm{~A}_{23} & \mathrm{~A}_{33}
\end{array}\right] \\
&\text { A. Adj. } \mathrm{A}=\left[\begin{array}{ccc}
a_{11} & a_{12} & a_{12} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]\left[\begin{array}{lll}
A_{11} & A_{21} & A_{31} \\
A_{12} & A_{22} & A_{32} \\
A_{13} & A_{23} & A_{33}
\end{array}\right] \\
&=\left[\begin{array}{ccc}
|\mathrm{A}| & 0 & 0 \\
0 & |\mathrm{~A}| & 0 \\
0 & 0 & |\mathrm{~A}|
\end{array}\right]=[|\mathrm{A}|]^3
\end{aligned}$
Dividing by $|\mathrm{A}|, \mid$ Adj. $\left.\mathrm{A}|=| \mathrm{A}\right|^2$
Hence, Part (b) is the correct answer.
&\text { Let } \mathrm{A}=\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right] \\
&\text { Let Adj. } \mathrm{A}=\left[\begin{array}{lll}
\mathrm{A}_{11} & \mathrm{~A}_{21} & \mathrm{~A}_{31} \\
\mathrm{~A}_{12} & \mathrm{~A}_{22} & \mathrm{~A}_{32} \\
\mathrm{~A}_{13} & \mathrm{~A}_{23} & \mathrm{~A}_{33}
\end{array}\right] \\
&\text { A. Adj. } \mathrm{A}=\left[\begin{array}{ccc}
a_{11} & a_{12} & a_{12} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]\left[\begin{array}{lll}
A_{11} & A_{21} & A_{31} \\
A_{12} & A_{22} & A_{32} \\
A_{13} & A_{23} & A_{33}
\end{array}\right] \\
&=\left[\begin{array}{ccc}
|\mathrm{A}| & 0 & 0 \\
0 & |\mathrm{~A}| & 0 \\
0 & 0 & |\mathrm{~A}|
\end{array}\right]=[|\mathrm{A}|]^3
\end{aligned}$
Dividing by $|\mathrm{A}|, \mid$ Adj. $\left.\mathrm{A}|=| \mathrm{A}\right|^2$
Hence, Part (b) is the correct answer.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.