System of equations λ x + y + z = 0, - x + λ y + z = 0, - x

English

Gujarati

Hindi

3 and 4 .Determinants and Matrices

medium

The system of equations $\lambda x + y + z = 0,$ $ - x + \lambda y + z = 0,$ $ - x - y + \lambda z = 0$, will have a non zero solution if real values of $\lambda $ are given by

$0$

$1$

$3$

$\sqrt 3 $

(IIT-1984)

Solution

(a) Accordingly, $\left| {\,\begin{array}{*{20}{c}}\lambda &1&1\\{ – 1}&\lambda &1\\{ – 1}&{ – 1}&\lambda \end{array}\,} \right| = 0 \Rightarrow {\lambda ^3} + 3\lambda = 0$

Therefore $\lambda = 0$, since $\lambda = i\sqrt 3 $ does not exist.

Standard 12

Mathematics

If $\omega$ is one of the imaginary cube roots of unity, then the value of the determinant $\left| {\begin{array}{*{20}{c}}1&{{\omega ^3}}&{{\omega ^2}}\\ {{\omega ^3}}&1&\omega \\{{\omega ^2}}&\omega &1\end{array}} \right|$ $=$

medium

If $a,b,c$ are respectively the ${p^{th}},{q^{th}}{r^{th}}$terms of an $A.P.,$ the $\left| {\,\begin{array}{*{20}{c}}a&p&1\\b&q&1\\c&r&1\end{array}\,} \right| = $

medium

$\left| {\,\begin{array}{*{20}{c}}{19}&{17}&{15}\\9&8&7\\1&1&1\end{array}\,} \right| = $

easy

If ${2^{{a_1}}},{2^{{a_2}}},{2^{{a_3}}},{……2^{{a_n}}}$ are in $G.P.$ then $\left| {\begin{array}{*{20}{c}}
{{a_1}}&{{a_2}}&{{a_3}} \\
{{a_{n + 1}}}&{{a_{n + 2}}}&{{a_{n + 3}}} \\
{{a_{2n + 1}}}&{{a_{2n + 2}}}&{{a_{2n + 3}}}
\end{array}} \right|$ is equal to

normal

$\left| {\,\begin{array}{*{20}{c}}{1/a}&1&{bc}\\{1/b}&1&{ca}\\{1/c}&1&{ab}\end{array}\,} \right| = $

easy

The system of equations $\lambda x + y + z = 0,$ $ - x + \lambda y + z = 0,$ $ - x - y + \lambda z = 0$, will have a non zero solution if real values of $\lambda $ are given by

Solution

Similar Questions

If $\omega$ is one of the imaginary cube roots of unity, then the value of the determinant $\left| {\begin{array}{*{20}{c}}1&{{\omega ^3}}&{{\omega ^2}}\\ {{\omega ^3}}&1&\omega \\{{\omega ^2}}&\omega &1\end{array}} \right|$ $=$

If $a,b,c$ are respectively the ${p^{th}},{q^{th}}{r^{th}}$terms of an $A.P.,$ the $\left| {\,\begin{array}{*{20}{c}}a&p&1\\b&q&1\\c&r&1\end{array}\,} \right| = $

$\left| {\,\begin{array}{*{20}{c}}{19}&{17}&{15}\\9&8&7\\1&1&1\end{array}\,} \right| = $

If ${2^{{a_1}}},{2^{{a_2}}},{2^{{a_3}}},{……2^{{a_n}}}$ are in $G.P.$ then $\left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}} \\ {{a_{n + 1}}}&{{a_{n + 2}}}&{{a_{n + 3}}} \\ {{a_{2n + 1}}}&{{a_{2n + 2}}}&{{a_{2n + 3}}} \end{array}} \right|$ is equal to

$\left| {\,\begin{array}{*{20}{c}}{1/a}&1&{bc}\\{1/b}&1&{ca}\\{1/c}&1&{ab}\end{array}\,} \right| = $

Start a Free Trial Now

If ${2^{{a_1}}},{2^{{a_2}}},{2^{{a_3}}},{……2^{{a_n}}}$ are in $G.P.$ then $\left| {\begin{array}{*{20}{c}}
{{a_1}}&{{a_2}}&{{a_3}} \\
{{a_{n + 1}}}&{{a_{n + 2}}}&{{a_{n + 3}}} \\
{{a_{2n + 1}}}&{{a_{2n + 2}}}&{{a_{2n + 3}}}
\end{array}} \right|$ is equal to