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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants1 Mark
MCQ
The value of the determinant $\begin{vmatrix}\text{a}-\text{b}&\text{b}+\text{c}&\text{c}\\\text{b}-\text{c}&\text{c}+\text{b}&\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{c}\end{vmatrix}$ is:
  • A
    $a^3+b^3+c^3$
  • B
    $3bc$
  • ✓
    $a^3+b^3+c^3-3 a b c$
  • D
    None of these
✓

Answer

Correct option: C.
$a^3+b^3+c^3-3 a b c$
$\begin{vmatrix}\text{a}-\text{b}&\text{b}+\text{c}&\text{a}\\\text{b}-\text{c}&\text{c}+\text{b}&\text{b}\\\text{c}-\text{a}&\text{a}+\text{b}&\text{c}\end{vmatrix}$$=\begin{vmatrix}-\text{b}&\text{b}+\text{c}+\text{a}&\text{a}\\-\text{c}&\text{c}+\text{a}+\text{b}&\text{b}\\-\text{a}&\text{a}+\text{b}+\text{c}&\text{c}\end{vmatrix} [$Applying $C _1 \rightarrow C _1- C _3$ and $\left.C _2 \rightarrow C _2+ C _3\right]$
$=(-1)(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{b}&1&\text{a}\\\text{c}&1&\text{b}\\\text{a}&1&\text{c}\end{vmatrix} [$Taking $(-1)$ common from $C _1$ and $(a + b + c)$ common from $C _2]$
$=(-1)(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{b}&1&\text{a}\\\text{c}-\text{b}&0&\text{b}-\text{a}\\\text{a}-\text{b}&0&\text{c}-\text{a}\end{vmatrix} [$Applying $R_2 \rightarrow R_2-R_1$ and $\left.R_3 \rightarrow R_3-R_1\right]$
$=(-1)(a+b+c)[-(c-b)(c-a)+(b-a)(a-b)]$
$=(-1)(a+b+c)\left[-c^2+a c+b c-a b+b a-b^2-a^2+a b\right]$
$=(-1)(a+b+c)\left(-a^2-b^2-c^2+a b+b c+a c\right)$
$=(a+b+c)\left(a^2+b^2+c^2-a b-b c-a c\right)$
$=a^3+a b^2+a c^2-a^2 b-a b c-a^2 c+b a^2+b^3+b c^2-a b^2-b^2 c-a b c+c a^2+c b^2+c^3-a c b-b c^2-a c^2$
$=a^3+b^3+c^3-3 a b c$
Hence, the correct option is $(c)$

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