# How to find the determinant of the matrix here ?

## Multiple options can be correct. Apr 15, 2018

(A) and (D)

#### Explanation:

$a \mathrm{dj} \left(P\right)$ $=$ $\left(\begin{matrix}1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3\end{matrix}\right)$

Edit:
|(color(red)1, color(blue)4, color(green)4), (color(green)2 ,color(red)1, color(blue)7), (color(blue)1, color(green)1, color(red)3)|

We multiply the same colors and add them and then we subtract when we multiply the same colors but from this scheme:

$| \left(\textcolor{b r o w n}{1} , \textcolor{\mathmr{and} a n \ge}{4} , 4\right) , \left(\textcolor{\mathmr{and} a n \ge}{2} , 1 , \textcolor{b r o w n}{7}\right) , \left(1 , \textcolor{b r o w n}{1} , \textcolor{\mathmr{and} a n \ge}{3}\right) |$

So we have:
$\textcolor{red}{3} + \textcolor{b l u e}{28} + \textcolor{g r e e n}{8} - 4 - \textcolor{\mathmr{and} a n \ge}{24} - \textcolor{b r o w n}{7} = 4$

Now, since this is the adjoint matrix, and the matrix is 3x3, there is a formula:

det(adj(A))=(det(A))^(n−1)

where n is the number of rows in the matrix.

So, we will have
$4 = \det {\left(A\right)}^{2}$
$\sqrt{4} = \sqrt{\det {\left(A\right)}^{2}}$
$2 = | \det \left(A\right) |$.........................{NOTE that $\sqrt{{x}^{2}} = | x |$}
$\therefore \det \left(A\right) = \pm 2$

So the answer is A and D.

I hope this helped you.