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શ્રેણિક $\left[ {\begin{array}{*{20}{c}}2&\lambda &{ - 4}\\{ - 1}&3&4\\1&{ - 2}&{ - 3}\end{array}} \right]$ એ સામાન્ય શ્રેણિક હોય હોય તો . .
$\lambda \ne - 2$
$\lambda \ne 2$
$\lambda \ne 3$
$\lambda \ne - 3$
Solution
(a) The given matrix $A = \left[ {\begin{array}{*{20}{c}}2&\lambda &{ – 4}\\{ – 1}&3&4\\1&{ – 2}&{ – 3}\end{array}} \right]$ is non singular, if $|A|\, \ne 0$
$|A|\,\, = \left| {\,\begin{array}{*{20}{c}}2&\lambda &{ – 4}\\{ – 1}&3&4\\1&{ – 2}&{ – 3}\end{array}\,} \right|\,$=$\left| {\,\begin{array}{*{20}{c}}1&{\lambda + 3}&0\\{ – 1}&3&4\\1&{ – 2}&{ – 3}\end{array}\,} \right|\,$,$[{R_1} \to {R_2} + {R_1}]$
= $\left| {\,\begin{array}{*{20}{c}}1&{\lambda + 3}&0\\0&1&1\\0&{ – \lambda – 5}&{ – 3}\end{array}\,} \right|$$\left[ {\begin{array}{*{20}{c}}{{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.$
