1. Tardigrade
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  3. Mathematics
  4. If the matrix [1&3&λ+2 2&4&8 3&5&10] is singular, then λ =

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Q. If the matrix $\begin{bmatrix}1&3&\lambda+2\\ 2&4&8\\ 3&5&10\end{bmatrix} $ is singular, then $\lambda$ =

Determinants

Solution:

| A | = 0 as the matrix A is singular
$\therefore \ |A| = \begin{vmatrix}1&3&\lambda+2\\ 2&4&8\\ 3&5&10\end{vmatrix} = 0$
Apply $R_2 \to R_2 - 2 R_1$
and $R_3 \to R_3 - 3R_1$ and expand.
$ - 2(4 - 3\lambda ) + 4 (4 - 2 \lambda ) = 0$
$\Rightarrow \ 8 - 2 \lambda = 0 \ \Rightarrow \ \lambda = 4$
For $\lambda = 4$, the second and the third column are proportional.