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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants1 Mark
MCQ
Choose the correct answer If $a, b, c,$ are in $A.P,$ then the determinant$:\ \begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+2\text{a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$ is
  • ✓
    $0$
  • B
    $1$
  • C
    $x$
  • D
    $2x$
✓

Answer

Correct option: A.
$0$
$\triangle=\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+\text{2a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$
$\triangle=\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+\text{2a}\\\text{x}+3&\text{x}+4&\text{x}+(\text{a}+\text{c})\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix} (2b = a + c$ as $a,b$ and $c$ are in $A.P.)$
Applying $R_1 \rightarrow R_1-R_2$ and $R_3 \rightarrow R_3-R_2,$ we have::
$\triangle=\begin{vmatrix}-1&-1&\text{a}-\text{c}\\\text{x}+3&\text{x}+4&\text{x}+(\text{a}+\text{c})\\1&1&\text{c}-\text{a}\end{vmatrix}$
Applying $R _1 \rightarrow R _1+ R _3,$ we have:
$\triangle=\begin{vmatrix}0&0&0\\\text{x}+3&\text{x}+4&\text{x}+(\text{a}+\text{c})\\1&1&\text{c}-\text{a}\end{vmatrix}$

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