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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
    $3bc$
  • ✓
    $a^3 + b^3 + c^3 - 3abc$
  • D
    None of these
✓

Answer

Correct option: C.
$a^3 + b^3 + c^3 - 3abc$
$\begin{vmatrix}\text{a}-\text{b}&\text{b}+\text{c}&\text{a}\\\text{b}-\text{c}&\text{c}+\text{b}&\text{b}\\\text{c}-\text{a}&\text{a}+\text{b}&\text{c}\end{vmatrix}$
$=\begin{vmatrix}-\text{b}&\text{b}+\text{c}+\text{a}&\text{a}\\-\text{c}&\text{c}+\text{a}+\text{b}&\text{b}\\-\text{a}&\text{a}+\text{b}+\text{c}&\text{c}\end{vmatrix}$ [Applying $C_1 \rightarrow C_1 - C_3$ and $C_2 \rightarrow C_2 + C_3]$
$=(-1)(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{b}&1&\text{a}\\\text{c}&1&\text{b}\\\text{a}&1&\text{c}\end{vmatrix} [$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 $R_3\rightarrow R_3 - R_1]$
$= (-1)(a + b + c)[-(c - b)(c - a) + (b - a)(a - b)]$
$= (-1)(a + b + c)[-c^2 + ac + bc - ab + ba - b^2 - a^2 + ab]$
$= (-1)(a + b + c)(-a^2 - b^2 - c^2 + ab + bc + ac)$
$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac)$
$= a^3 + ab^2 + ac^2 - a^2b - abc - a^2c + ba^2 + b^3 + bc^2 - ab^2-b^2c - abc + ca^2 + cb^2 + c^3 - acb - bc^2 - ac^2$
$= a^3 + b^3 + c^3- 3abc$
Hence, the correct option is $(c)$

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