3 and 4 .Determinants and Matrices
medium

$k $ ની કેટલી કિંમતો માટે સમીકરણ સંહતી $\left( {k + 1} \right)x + 8y = 4k\;,\;kx + \left( {k + 3} \right)y $$= 3k - 1$ ને એક પણ ઉકેલ નથી.

A

અંનત

B

$1$

C

$2$

D

$3$

(JEE MAIN-2013) (IIT-2002)

Solution

${\rm{\Delta }} = \left| {\begin{array}{*{20}{c}}
{k + 1}&8\\
k&{k + 3}
\end{array}} \right|$$ = {k^2} + 4k + 3 – 8k$

$\qquad  = {k^2} – 4k + 3$

$\qquad  = (k – 3)(k – 1)$

${{\rm{\Delta }}_1} = \left| {\begin{array}{*{20}{c}}
{4k}&8\\
{3k – 1}&{k + 3}
\end{array}} \right|$$ = 4{k^2} + 12k – 24k + 8$

$\qquad  = 4{k^2} + 12k – 24k + 8$

$\qquad  = 4({k^2} – 3k + 2)$

$\qquad  = 4(k – 2)(k – 1)$

${{\rm{\Delta }}_2} = \left| {\begin{array}{*{20}{c}}
{k + 1}&{4k}\\
k&{3k – 1}
\end{array}} \right|$$ = 3{k^2} + 2k – 1 – 4{k^2}$

$\qquad  =  – {k^2} + 2k – 1$

$\qquad  =  – {(k – 1)^2}$

As given no solution 

${{\rm{\Delta }}_1}\;\;and\;\;{{\rm{\Delta }}_2} \ne 0$

but  ${\rm{\Delta }} = 0$

$k=3$

Standard 12
Mathematics

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