Question

If $Z=\left|\begin{array}{ccc}2& 5-i& 7+i\\ 5+i& 2& 3-i\\ 7-i& 3+i& 7\end{array}\right|$ and arg (z) = $\theta$ then $\theta$ = ___

Answer 1

$Z=\left|\begin{array}{ccc}2& 5-i& 7+i\\ 5+i& 2& 3-i\\ 7-i& 3+i& 7\end{array}\right|$

Now we can solve this determinant either using Sarrus rule or using Cofactors.

Let’s do it using cofactors

$SoZ=\frac{2\left(\right(2\times 7)-(3+i\left)\right(3-i\left)\right)}{\downarrow }\text{Cofactor for}{a}_{11}=2+(5-i)\times (-1{)}^{1+2}\frac{\left(\right(5+i)\times 7-(3-i\left)\right(7-i\left)\right)}{\downarrow }\text{Cofactor for}{a}_{12}=5-i+(7+i\left)\right(-1{)}^{1+3}\frac{\left(\right(5+i\left)\right(3+i)-2(7-i\left)\right)}{\downarrow }\text{Cofactor for}{a}_{13}=7+i$

$+(5-i)x(-1)(35+7i-(21-3i-7i-1\left)\right)+(7+i)\left(1\right)(15+8i-1-14+2i)=-94$

Which is a negative real number and we know that for a number lying on negative real axis, the argument is $\pi$, which is correct answer.