Question-30
Compute \(\text{det}(A)\) where \(A = \begin{bmatrix}1 & 2010 & 2020 \times 2030\\1 & 2020 & 2030 \times 2010\\1 & 2030 & 2010 \times 2020\end{bmatrix}\)
\(2000\)
\(\text{det}(A)\) is of the form: \[ \begin{vmatrix} 1 & a & bc\\ 1 & b & ca\\ 1 & c & ab \end{vmatrix} \] We shall now compute this determinant:
Step-1: \(R_2 \rightarrow R_2 - R_1\)
\[ \begin{vmatrix} 1 & a & bc\\ 1 & b & ca\\ 1 & c & ab \end{vmatrix} = \begin{vmatrix} 1 & a & bc\\ 0 & b - a & c(a - b)\\ 1 & c & ab \end{vmatrix} \]
Step-2: \(R_3 \rightarrow R_3 - R_1\)
\[ \begin{vmatrix} 1 & a & bc\\ 1 & b & ca\\ 1 & c & ab \end{vmatrix} = \begin{vmatrix} 1 & a & bc\\ 0 & b - a & c(a - b)\\ 0 & c - a & b(a - c) \end{vmatrix} \]
Step-3: \(R_2 \rightarrow \cfrac{1}{b - a} R_2\) and \(R_{3} \rightarrow \cfrac{1}{c - a} R_{3}\)
We can divide by \(b - a\) and \(c - a\) as \(a, b, c\) are distinct.
\[ \begin{vmatrix} 1 & a & bc\\ 1 & b & ca\\ 1 & c & ab \end{vmatrix} = (b - a)(c - a)\begin{vmatrix} 1 & a & bc\\ 0 & 1 & -c\\ 0 & 1 & -b \end{vmatrix} \]
We can now expand along the first column. After moving some minus signs around, we get:
\[ \begin{vmatrix} 1 & a & bc\\ 1 & b & ca\\ 1 & c & ab \end{vmatrix} = (a - b)(b - c)(c - a) \]
We can now plug in \(a = 2010, b = 2020, c = 2030\) to get the desired determinant. This turns out to be \(\boxed{2000}\).