Question

The value of the determinant $\Delta =\left| \begin{matrix} 0 & i-100 & i-500 \\\ 100-i & 0 & 1000-i \\\ 500-i & i-1000 & 0 \\\ \end{matrix} \right|$ is equal to
(a) 100
(b) 500
(c) 1000
(d) 0

Answer 1

Assume (i - 100) = a, (i – 500) = b and (i - 1000) = c. Substitute these values in the given expression of the determinant. Expand it along the first row. Simplify the expanded form by performing simple arithmetic operations and get the answer.

Complete step-by-step solution:
Here, we have been provided with an expression of determinant given as $\Delta =\left| \begin{matrix} 0 & i-100 & i-500 \\\ 100-i & 0 & 1000-i \\\ 500-i & i-1000 & 0 \\\ \end{matrix} \right|$ and we have been asked to simplify it and find its value.
Now, we can clearly see that the given terms in the determinant are very large numbers, so we need to make the terms small. The provided determinant is
$\Delta =\left| \begin{matrix} 0 & i-100 & i-500 \\\ 100-i & 0 & 1000-i \\\ 500-i & i-1000 & 0 \\\ \end{matrix} \right|$
The above expression can be written as
$\Rightarrow \Delta =\left| \begin{matrix} 0 & i-100 & i-500 \\\ -\left( i-100 \right) & 0 & -\left( i-1000 \right) \\\ -\left( i-500 \right) & i-1000 & 0 \\\ \end{matrix} \right|$
Here, we are assuming (i - 100) = a, (i – 500) = b and (i - 1000) = c. So, the given expression becomes,
$\Rightarrow \Delta =\left| \begin{matrix} 0 & a & b \\\ -a & 0 & -c \\\ -b & c & 0 \\\ \end{matrix} \right|$
Taking (-1) common from the first column, we get
$\Rightarrow \Delta =\left( -1 \right)\times \left| \begin{matrix} 0 & a & b \\\ a & 0 & -c \\\ b & c & 0 \\\ \end{matrix} \right|$
Expanding the determinant along the first row, we get
\begin{aligned} & \Rightarrow \Delta =\left( -1 \right)\times \left[ 0\times \left\\{ \left( 0\times 0 \right)-\left( c\times -c \right) \right\\}-a\times \left\\{ \left( a\times 0 \right)-\left( b\times -c \right) \right\\}+b\times \left\\{ \left( a\times c \right)-\left( b\times 0 \right) \right\\} \right] \\\ & \Rightarrow \Delta =\left( -1 \right)\times \left[ 0-a\times \left\\{ bc \right\\}+b\times \left\\{ ac \right\\} \right] \\\ & \Rightarrow \Delta =\left( -1 \right)\times \left[ -abc+abc \right] \\\ \end{aligned}
Cancelling the like terms inside the square bracket, we get
$\begin{aligned} & \Rightarrow \Delta =\left( -1 \right)\times \left[ 0 \right] \\\ & \Rightarrow \Delta =0 \\\ \end{aligned}$
Therefore, the value of the determinant $\left( \Delta \right)=0$ .
Hence, option (d) is the correct answer.

Note: One may note that we have assumed the given terms as a, b and c because the terms were large and if we would have been expanded the determinant without the substitution, then the chances of making calculation mistakes would have been more. So, to make our calculation easy we have expanded the determinant along the first row, you may expand it along any row or column, the answer will be the same. The only important thing to note is that you have to take care of the signs.