Can I get some help in computing the determinant of this 5x5 matrix? I am confused with regards to the triangular form and how I go about deriving the determinant from that form in terms of a, b and c.

$$\left[\begin{array}{cc} 1&2&0&0&0\\ 0&1&2&a&0\\ 0&0&3&-1&0\\ 0&0&0&b&1\\ 5&0&0&0&c \end{array} \right] $$

  • $\begingroup$ Expand in terms of "minors" is quite quick because there are so many zeros $\endgroup$ – Mark Bennet Sep 10 '18 at 10:08

Alternatively, Laplace expansion along the highlighted rows/columns: $$\begin{align}\begin{vmatrix} \color{red}{1}&\color{red}{2}&\color{red}{0}&\color{red}{0}&\color{red}{0}\\ 0&1&2&a&0\\ 0&0&3&-1&0\\ 0&0&0&b&1\\ 5&0&0&0&c \end{vmatrix} &= \\ =1\cdot (-1)^{1+1}\cdot 3bc+2\cdot (-1)^{1+2} \begin{vmatrix} \color{red}{0}&2&a&0\\ \color{red}{0}&3&-1&0\\ \color{red}{0}&0&b&1\\ \color{red}{5}&0&0&c \end{vmatrix}&=\\ =3bc-2\cdot 5\cdot (-1)^{4+1} \begin{vmatrix} 2&a&\color{red}{0}\\ 3&-1&\color{red}{0}\\ 0&b&\color{red}{1} \end{vmatrix}&=\\ =3bc+10\cdot 1\cdot (-1)^{3+3} \begin{vmatrix}2&a\\ 3&-1\end{vmatrix}&=\\ =3bc +10(-2-3a)&=\\ =3bc-20-30a&. \end{align}$$


Hint: Do the following steps where in, $R$ is a row of the matrix. At the end, we will have a upper triangle matrix:

  1. $-5R_1+R_5\to R_5~~$
  2. $10R_2+R_5\to R_5~~$
  3. $-\frac{20}{3}R_3+R_5\to R_5~~$
  4. $\frac{-10a+20/3}{b}R_4+R_5\to R_5$.

This way is not unique!


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