Question

(4
1. If A = 0
Lo
0
4
0
0
0 then find the value of adj Al.
4]​

Answer 1

Answer:

so, here's your answer....

I solved it :-)

please, mark my answer brainliest & rate 5 stars too

Answer 2

Given matrix is,

$\mathrm{A=\left[\begin{array}{ccc}4&0&0\\0&4&0\\0&0&4\end{array}\right]}$

Taking its cofactor matrix,

$\left|\begin{array}{cc}4&0\\0&4\end{array}\right|=16\quad;\quad-\left|\begin{array}{cc}0&0\\0&4\end{array}\right|=0\quad;\quad\left|\begin{array}{cc}0&4\\0&0\end{array}\right|=0$

$-\left|\begin{array}{cc}0&0\\0&4\end{array}\right|=0\quad;\quad\left|\begin{array}{cc}4&0\\0&4\end{array}\right|=16\quad;\quad-\left|\begin{array}{cc}4&0\\0&0\end{array}\right|=0$

$\left|\begin{array}{cc}0&0\\4&0\end{array}\right|=0\quad;\quad-\left|\begin{array}{cc}4&0\\0&0\end{array}\right|=0\quad;\quad\left|\begin{array}{cc}4&0\\0&4\end{array}\right|=16$

So the cofactor matrix will be,

$\left[\begin{array}{ccc}16&0&0\\0&16&0\\0&0&16\end{array}\right]$

We know the transpose of this one is the adjugate of A. So,

$\mathrm{adj(A)=\left[\begin{array}{ccc}16&0&0\\0&16&0\\0&0&16\end{array}\right]}$

Well, the same.

Now we have to find the determinant of this one, which is asked!

So,

$\mathrm{|adj(A)|=16\left|\begin{array}{cc}16&0\\0&16\end{array}\right|-0\left|\begin{array}{cc}0&0\\0&16\end{array}\right|+0\left|\begin{array}{cc}0&16\\0&0\end{array}\right|=16^3=}\ \mathbf{4096}$

I guess this would be the answer, but I got a short method now from this!

$\boxed{\begin{minipage}{11.4cm}If a matrix \mathrm{A=nI} where \mathrm{n}\in\mathbb{R} and \mathrm{I} is the identity matrix,\\\\then \mathbf{|A|=n^3\quad\&\quad adj(A)=n^2I} \\\\So \mathbf{|adj(A)|=(n^2)^3=n^6} \end{minipage}}$

We were actually given $\mathrm{A=4I}.$

Hence we could say that $\mathrm{|adj(A)|=4^6=}\ \mathbf{4096}.$