Question

First row of the matrix $A$ is $\left[\begin{array}{ccc}1& 3& 2\end{array}\right]$. If $adj\left(A\right)$ =$$\left[\begin{array}{ccc}-2& 4& a\\ -1& 2& 1\\ 3a& -5& -2\end{array}\right]$$ then a possible value of $det\left(A\right)$ is

• A. $1$
• B. $2$
• C. $-1$
• D. $-2$

Answer 1

The correct option is A $1$

$|A|=a_{11}{A}_{11}+a_{12}{A}_{21}+a_{13}{A}_{31}$

$|A|=1(-2)+3(-1)+2\left(3\alpha \right)$

$|A|=-5+6\alpha$

$adjA=\left[\begin{array}{ccc}-2& 4& \alpha \\ -1& 2& 2\\ 3\alpha & -5& -2\end{array}\right]$

$|adjA|=-2(-4+5)-4(2-3\alpha )+\alpha (5-6\alpha )$

$=-2-8+12\alpha +5\alpha -6{\alpha }^2$

$|A{|}^2=-6{\alpha }^2+17\alpha -10$

$(6\alpha -5{)}^2=-6{\alpha }^2+17\alpha -10$

$36{\alpha }^2+25-60\alpha =-6{\alpha }^3+17\alpha -10$

$42{\alpha }^2-77\alpha +35=0$

$6{\alpha }^2-11\alpha +5=0$

$(6\alpha -5)(\alpha -1)=0$

$\alpha =\frac{5}{6}$ or $\alpha =1$

$|A|=-5+6\alpha$

When, $\alpha =\frac{5}{6},|A|=0$

When, $\alpha =1,|A|=1$