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3 and 4 .Determinants and Matrices
hard
જો રેખીય સમીકરણો $x + y+ z = 5$ ; $x + 2y + 3z = 9$ ; $x + 3y + \alpha z = \beta $ એ અનંત ઉકેલ ધરાવે છે તો $\beta - \alpha $ મેળવો.
A
$21$
B
$8$
C
$18$
D
$5$
(JEE MAIN-2019)
Solution
$x + y – z = 5$
$x + 2y + 3z = 9,$
$x + 3y + \alpha z = \beta $
$D = \left| {\begin{array}{*{20}{c}}
1&1&1\\
1&2&3\\
1&3&0
\end{array}} \right| = 0 \Rightarrow \left( {2\alpha – 9} \right) + \left( {3 – \alpha } \right) + \left( {3 – 2} \right) = 0 \Rightarrow \alpha = 5$
Now, ${D_3} = \left| {\begin{array}{*{20}{c}}
1&1&5\\
1&2&9\\
1&3&\beta
\end{array}} \right| = 0 \Rightarrow 2\beta – 27 + 9\beta – 5\left( {3 – 2} \right) = 0 \Rightarrow \beta = 13$
$ \Rightarrow $ at $\alpha = 5,b = 13$ above $3$ planes from common line
Standard 12
Mathematics
