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

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.