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3 and 4 .Determinants and Matrices
normal
Let $A = \left[ {\begin{array}{*{20}{c}}
1&2&3\\
2&2&{ - 1}\\
3&0&k
\end{array}} \right]$ and $f(x) = {x^3} - 2{x^2} - \alpha x + \beta = 0$ . If $A$ satisfies $f(x)=0$ ,then
A
$k = 1, \alpha = 14$
B
$\alpha = 13,\beta = 22$
C
$k = - 1,\beta = 22$
D
$\alpha = - 14,\beta = - 22$
Solution
Sum of roots $=$ sum of diagonal elements
$\Rightarrow 2=1+2+k \Rightarrow k=-1$
and product of roots $=$ value of the determinant.
$\Rightarrow-\beta=\left|\begin{array}{ccc}{1} & {2} & {3} \\ {2} & {2} & {-1} \\ {3} & {0} & {\mathrm{k}}\end{array}\right|=-2 \mathrm{k}-24$
$\Rightarrow \beta=22$
Standard 12
Mathematics
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