-3
$\begingroup$

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.

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

$\endgroup$
  • $\begingroup$ Expand in terms of "minors" is quite quick because there are so many zeros $\endgroup$ – Mark Bennet Sep 10 '18 at 10:08
0
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.