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3 and 4 . determinant and metrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths3 and 4 . determinant and metrices1 Mark
MCQ
An ordered pair $(\alpha , \beta )$ for which the system of linear equations

$\left( {1 + \alpha } \right)x + \beta y + z = 2$ ; $\alpha x + \left( {1 + \beta } \right)y + z = 3$ ; $\alpha x  + \beta y + 2z = 2$ has a unique solution, is

  • ✓
    $(2, 4)$
  • B
    $(-3, 1)$
  • C
    $(-4, 2)$
  • D
    $(1, -3)$
✓

Answer

Correct option: A.
$(2, 4)$
a
$\left( {1 + \alpha } \right)x + \beta y + z = 0$

$\alpha x + \left( {1 + \beta } \right)y + z = 0$

$\alpha x + \beta y + 2z = 0$

$D = \left| {\begin{array}{*{20}{c}}
{1 + \alpha }&\beta &1\\
\alpha &{1 + \beta }&1\\
\alpha &\beta &2
\end{array}} \right|$

${C_1} \to {C_1} + {C_2} + {C_3}$

$D = \left( {\alpha  + \beta  + 2} \right)\left| {\begin{array}{*{20}{c}}
1&\beta &1\\
1&{1 + \beta }&1\\
1&\beta &2
\end{array}} \right|$

${R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1}$

$D = \left( {\alpha  + \beta  + 2} \right)\left| {\begin{array}{*{20}{c}}
1&\beta &1\\
0&1&0\\
0&0&1
\end{array}} \right| = \alpha  + \beta  + 2$

For unique solution $\alpha  + \beta  + 2 \ne 0 \Rightarrow \alpha  + \beta  \ne  - 2$

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