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STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 3 and 4 . determinant and metrices1 Mark
MCQ
Let $A$ be a $3 \times 3$ matrix of non-negative real elements such that $A\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=3\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$. Then the maximum value of $\operatorname{det}(\mathrm{A})$ is........................
  • A
    $49$
  • B
    $54$
  • ✓
    $27$
  • D
    $50$
✓

Answer

Correct option: C.
$27$
c
$\begin{aligned} \text { Let } A & =\left[\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right] \\ A\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right] & =3\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]\end{aligned}$

$ \Rightarrow a_1+a_2+a_3=3 $

$ \Rightarrow b_1+b_2+b_3=3 $

$ \Rightarrow c_1+c_2+c_3=3 $

Now,

$ |A|=\left(a_1 b_2 c_3+a_2 b_3 c_1+a_3 b_1 c_2\right) $

$ -\left(a_3 b_2 c_1+a_2 b_1 c_3+a_1 b_3 c_2\right)$

$\therefore$ From above in formation, clearly $|\mathrm{A}|_{\max }=27$, when $a_1=3, b_2=3, c_3=3$

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