3 and 4 .Determinants and Matrices
easy

If the matrix $\left[ {\begin{array}{*{20}{c}}0&1&{ - 2}\\{ - 1}&0&3\\\lambda &{ - 3}&0\end{array}} \right]$ is singular, then $\lambda $=

A

$-2$

B

$-1$

C

$1$

D

$2$

Solution

(d) The matrix $A = \left[ {\begin{array}{*{20}{c}}{\,\,0}&{\,\,1}&{ – 2}\\{ – 1}&{\,\,0}&{\,\,3}\\{\,\,\lambda }&{ – 3}&{\,\,0}\end{array}} \right]$ is singular

 $|A|$ = 0 ==> $0 – 1( – 3\lambda ) + ( – 2)(3) = 0$

$ \Rightarrow 3\lambda – 6 = 0 \Rightarrow \lambda = 2$.

Standard 12
Mathematics

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