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3 and 4 .Determinants and Matrices
hard
रैंखिक समीकरण निकाय
$x + y + z = 2$
$2x + 3y + 2z = 5$
$2x + 3y + (a^2 -1)\,z = a + 1$
A
असंगत है जब $a =4$
B
का $| a |=\sqrt{3}$ के लिए मात्र एक हल है।
C
का $a =4$ के लिए अनन्त हल है।
D
असंगत है जब $| a |=\sqrt{3}$
(JEE MAIN-2019)
Solution
By applying Crammer's Rule
$D = \left| {\begin{array}{*{20}{c}}
1&1&1\\
2&3&2\\
2&3&{{a^2} – 1}
\end{array}} \right|$
$ = 3\left( {{a^2} – 1} \right) – 6 – 2\left( {{a^2} – 1} \right) + 4$
$ = {a^2} – 1 – 2 = {a^2} – 3$
If $\left| a \right| \ne \pm \sqrt 3 \Rightarrow $system has unique solution
If $\left| a \right| = \sqrt 3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \begin{array}{l}
x + y + z = 1\\
2x + 3y + 2z = 1\\
2x + 3y + 2z = \pm \sqrt 3 + 1
\end{array} \right\}$
Hence system is inconsistent for $\left| a \right| = \sqrt 3 $
Standard 12
Mathematics