3 and 4 .Determinants and Matrices
easy

आव्यूह $\left[ {\begin{array}{*{20}{c}}2&\lambda &{ - 4}\\{ - 1}&3&4\\1&{ - 2}&{ - 3}\end{array}} \right]$ व्युत्क्रमणीय होगा, यदि

A

$\lambda \ne - 2$

B

$\lambda \ne 2$

C

$\lambda \ne 3$

D

$\lambda \ne - 3$

Solution

(a) दिया गया आव्यूह $A = \left[ {\begin{array}{*{20}{c}}2&\lambda &{ – 4}\\{ – 1}&3&4\\1&{ – 2}&{ – 3}\end{array}} \right]$ एक व्युत्क्रमणीय आव्यूह है, यदि  $|A|\, \ne 0$         

$|A|\,\, = \left| {\,\begin{array}{*{20}{c}}2&\lambda &{ – 4}\\{ – 1}&3&4\\1&{ – 2}&{ – 3}\end{array}\,} \right|\,$=$\left| {\,\begin{array}{*{20}{c}}1&{\lambda  + 3}&0\\{ – 1}&3&4\\1&{ – 2}&{ – 3}\end{array}\,} \right|\,$,$\{ {R_1} \to {R_2} + {R_1}\} $        

  = $\left| {\,\begin{array}{*{20}{c}}1&{\lambda  + 3}&0\\0&1&1\\0&{ – \lambda  – 5}&{ – 3}\end{array}\,} \right|$,        $\left\{ \begin{array}{l}{R_2} \to {R_2} + {R_3}\\{R_3} \to {R_3} – {R_1}\end{array} \right\}$         

= $1\,( – 3 + \lambda  + 5) \ne 0$                    

$ \Rightarrow \lambda  + 2 \ne 0$ $ \Rightarrow \lambda \,\, \ne  – 2.$

Standard 12
Mathematics

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