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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 $a, b, c \in R$ be all non-zero and satisfy $a^{3}+b^{3}+c^{3}=2 .$ If the matrix $A=\left(\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right)$ satisfies $\mathrm{A}^{\mathrm{T}} \mathrm{A}=\mathrm{I},$ then a value of $abc$ can be
  • A
    $\frac{2}{3}$
  • B
    $-\frac{1}{3}$
  • C
    $3$
  • ✓
    $\frac{1}{3}$
✓

Answer

Correct option: D.
$\frac{1}{3}$
d
$A^{T} A=I$

$\Rightarrow a^{2}+b^{2}+c^{2}=1$

and $a b+b c+c a=0$

Now, $(a+b+c)^{2}=1$

$\Rightarrow a+b+c=\pm 1$

So, $a^{3}+b^{3}+c^{3}-3 a b c$

$=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$

$=\pm 1(1-0)=\pm 1$

$\Rightarrow 3 a b c=2 \pm 1=3,1$

$\Rightarrow \quad a b c=1, \frac{1}{3}$

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