I recently encountered a challenging determinant problem in an Olympiad Exam (BdJSO) and have been struggling to find an efficient solution. The matrix in question is:
$A=\begin{vmatrix} 3 & 0 & 0 & -2 & 4 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & -1 & 0 & 5 & -3 \\ -4 & 0 & 1 & 0 & 6 \\ 0 & -1 & 0 & 3 & 2 \\ \end{vmatrix}$; Find the value of $A$
I have attempted to compute the determinant using traditional methods, but the complexity of the matrix has made the process quite intricate. According to a simulator, the result is $-114$, but I am eager to understand the steps involved in reaching this solution or any shortcut or determinant property that can be applied.
Could someone kindly provide insights into a more efficient or shortcut approach to compute the determinant of this specific $5$$\times$$5$ matrix? Any explanations, hints, or alternative methods would be greatly appreciated.
Thank you in advance for your assistance!
I believe by "traditional methods", you mean expanding along the first row.
The main idea is that we can expand a determinant along any row/column we want and arrive at the correct answer.
For the ease of computation, we choose to expand along rows/columns which have abundant zeroes.
$A=\begin{vmatrix} 3 & 0 & \color{red}{0} & -2 & 4 \\ 0 & 2 & \color{red} 0 & 0 & 0 \\ 0 & -1 & \color{red} 0 & 5 & -3 \\ -4 & 0 & \color{red} 1 & 0 & 6 \\ 0 & -1 & \color{red} 0 & 3 & 2 \\ \end{vmatrix}$
Expanding along the third column... You ignore the zeroes and get to the only non-zero entry i.e., $a_{43}=1$ and extract its minor by "removing" the fourth row and the third column.
$A=1\times (-1)^{4+3}\cdot \begin{vmatrix} 3 & 0 & -2 & 4 \\ \color{red}0 & \color{red}2 & \color{red}0 & \color{red}0 \\ 0 & -1 & 5 & -3 \\ 0 & -1 & 3 & 2 \\ \end{vmatrix}$
I'll add explanation about the factor $1\times (-1)^{4+3}$ if you want.
Now continue this process choosing the second row.
$A=1\times (-1)^{4+3}\times 2\times (-1)^{2+2}\cdot \begin{vmatrix} 3 & -2 & 4 \\ 0 & 5 & -3 \\ 0 & 3 & 2 \\ \end{vmatrix}$
Repeat this exercise, expanding along the first column.
$A=1\times (-1)^{4+3}\times 2\times (-1)^{2+2}\times 3\times (-1)^{1+1}\cdot \begin{vmatrix} 5 & -3 \\ 3 & 2 \\ \end{vmatrix}$