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Home  >>  CBSE XII  >>  Math  >>  Determinants
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If $A=\begin{vmatrix}2 & \lambda & 3\\0 & 2 & 5\\1 & 1 & 3\end{vmatrix},$ then$\;A^{-1}\;$exists if $\lambda = ?$

\[\begin{array}{1 1}(A)\quad \lambda=2 & (B)\quad \lambda\neq 2\\(C)\quad \lambda\neq -2 & (D)\quad None\;of\;these\end{array}\]

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  • If $A=\begin{vmatrix}a_{11} &a_{21} & a_{31}\\a_{12} &a_{22} &a_{32}\\a_{13} &a_{23} &a_{33}\end{vmatrix}$
  • $|A|=a_{11}(a_{22}\times a_{33}-a_{32}\times a_{23})-a_{21}(a_{12}\times a_{33}-a_{32}\times a_{13})+a_{31}(a_{12}\times a_{23}-a_{22}\times a_{13})$
  • If the value of a determinant is 0,then its said to be a singular matrix.
Let $A=\begin{vmatrix}2 & \lambda & 3\\0 & 2 & 5\\1 & 1 & 3\end{vmatrix}$
If $A^{-1}$ exists then the value of the determinant $\neq$ 0.
(i.e)$|A|\neq 0$
Expanding along $R_1$
$|A|=2(2\times 3-5\times 1)-\lambda (0\times 3-5\times 1)+3(0\times 1-2\times 1)$
If |A| is singular then |A|=0.
Therefore $\lambda=\frac{4}{5}$
Hence if A should be non-singular
$\lambda\neq \frac{4}{5}$
Hence D is the correct answer.
answered Mar 26, 2013 by sreemathi.v

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