Question

matrix , if A =
1² 2² 3²
2² 32 4²
3² 4² 5²
then |Adj A|​

Answer 1

We know that,

$\longrightarrow A^{-1}=\dfrac{1}{|A|}\,adj(A)$

$\longrightarrow adj(A)=|A|A^{-1}$

Then,

$\longrightarrow |adj(A)|=\Big||A|A^{-1}\Big|$

If $A$ is a square matrix of order $n,$

$\longrightarrow |adj(A)|=|A|^n\left|A^{-1}\right|$

$\longrightarrow |adj(A)|=|A|^{n-1}$

Here $n=3.$ So,

$\longrightarrow |adj(A)|=|A|^2$

Let us find $|A|.$

$\longrightarrow |A|=\left|\begin{array}{ccc}1^2&2^2&3^2\\2^2&3^2&4^2\\3^2&4^2&5^2\end{array}\right|$

Performing operations $R_2\to R_2-R_1$ and $R_3\to R_3-R_2,$

$\longrightarrow |A|=\left|\begin{array}{ccc}1^2&2^2&3^2\\2^2-1^2&3^2-2^2&4^2-3^2\\3^2-2^2&4^2-3^2&5^2-4^2\end{array}\right|$

$\longrightarrow |A|=\left|\begin{array}{ccc}1&4&9\\3&5&7\\5&7&9\end{array}\right|$

Performing operation $R_3\to R_3-R_1,$

$\longrightarrow |A|=\left|\begin{array}{ccc}1&4&9\\3&5&7\\4&3&0\end{array}\right|$

Expanding along $C_3,$

$\longrightarrow |A|=9(3\cdot 3-5\cdot 4)-7(1\cdot 3-4\cdot 4)$

$\longrightarrow |A|=9(9-20)-7(3-16)$

$\longrightarrow |A|=-9\cdot11+7\cdot13$

$\longrightarrow |A|=-8$

Therefore,

$\longrightarrow |adj(A)|=(-8)^2$

$\longrightarrow\underline{\underline{|adj(A)|=64}}$