- Home
- Standard 12
- Mathematics
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
Similar Questions
medium