Evaluate $\left|\begin{array}{ccc}102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6\end{array}\right|$

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Note that $\left|\begin{array}{ccc}102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6\end{array}\right|=\left|\begin{array}{ccc}6(17) & 6(3) & 6(6) \\ 1 & 3 & 4 \\ 17 & 3 & 6\end{array}\right|=6\left|\begin{array}{ccc}17 & 3 & 6 \\ 1 & 3 & 4 \\ 17 & 3 & 6\end{array}\right|=0$

Similar Questions

Which of the following values of $\alpha$ satisfy the equation

$\left|\begin{array}{lll}(1+\alpha)^2 & (1+2 \alpha)^2 & (1+3 \alpha)^2 \\ (2+\alpha)^2 & (2+2 \alpha)^2 & (2+3 \alpha)^2 \\ (3+\alpha)^2 & (3+2 \alpha)^2 & (3+3 \alpha)^2\end{array}\right|=-648 \alpha$ ?

$(A)$ $-4$ $(B)$ $9$ $(C)$ $-9$ $(D)$ $4$

  • [IIT 2015]

If ${D_r} = \left| {\begin{array}{*{20}{c}}{{2^{r - 1}}}&{{{2.3}^{r - 1}}}&{{{4.5}^{r - 1}}}\\x&y&z\\{{2^n} - 1}&{{3^n} - 1}&{{5^n} - 1}\end{array}} \right|$, then the value of $\sum\limits_{r = 1}^n {{D_r} = } $

Using properties of determinants, prove that:

$\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|=(1+p x y z)(x-y)(y-z)(z-x),$ where $p$ is any scalar.

The total number of distinct $x \in \mathbb{R}$ for which $\left|\begin{array}{ccc}x & x^2 & 1+x^3 \\ 2 x & 4 x^2 & 1+8 x^3 \\ 3 x & 9 x^2 & 1+27 x^3\end{array}\right|=10$ is

  • [IIT 2016]

Let $P=\left[a_{\|}\right]$be $a \times 3$ matrix and let $Q=\left[b_1\right]$, where $b_1=2^{1+j} a_{\|}$for $1 \leq i, j \leq 3$. If the determinant of $P$ is $2$ , then the determinant of the matrix $Q$ is

  • [IIT 2012]