Question

Determinant of a Triangular matrix In Exercises 39–42, find the determinant of the triangular matrix. 39. [−2 4 −3 0 6 7 0 0 2] 40. [4 0 0 0 7 0 0 0 −2] 41. [5 0 0 0 8 0 0 0 −4 6 2 0 2 0 2 −1] 42. [ 4 −1 3 −8 0 1 2 5 7 0 0 3 0 0 0 0 −2]

Answer 1

Answer:

39) -24; 40) -56; 41) 0; 42) -126

Step-by-step explanation:

Matrices in 39 and 40 have dimensions 3x3, whether the matrices from 41 and 42 have dimensions 4x4.

To calculate determinant of the 3x3 matrix, we can use the following rule:

$\left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] =1*\left[\begin{array}{cc}5&6\\8&9\end{array}\right] -2*\left[\begin{array}{cc}4&6\\7&9\end{array}\right]+3*\left[\begin{array}{ccc}4&5\\7&8\end{array}\right]$

In this case, determinants of the small matrices 2x2, can be calculated as, for example for the first part of the equation, will be (5*9-6*8).

For the problem 39, calculations will be as following:

$\left[\begin{array}{ccc}-2&4&-3\\0&6&7\\0&0&2\end{array}\right]=-2*\left[\begin{array}{cc}6&7\\0&2\end{array}\right] -4*\left[\begin{array}{cc}0&7\\0&2\end{array}\right]+(-3)*\left[\begin{array}{ccc}0&6\\0&0\end{array}\right] \\=-2*(6*2-0*7)-4*(0*2-0*7)-3*(0*0-0*6)=-24$

So, determinant for the first problem is -24

For the second problem, the solution is:

$\left[\begin{array}{ccc}4&0&0\\0&7&0\\0&0&-2\end{array}\right]=4*\left[\begin{array}{cc}7&0\\0&-2\end{array}\right] -0*\left[\begin{array}{cc}0&0\\0&-2\end{array}\right]+0*\left[\begin{array}{ccc}0&7\\0&0\end{array}\right] \\=4*(7*-2-0*0)-0*(0*-2-0*0)-0*(0*0-0*7)=-56$

Determinant for the problem 40 is -56

In cases of the 4x4 matrices, the application of the idea is similar- we use one row or column and values from there to reduce dimensions of each matrix to 3x3 and then calculate determinants of these smaller matrices.

For the problem 41, calculations can be done, using the first row, where we have 5; 0; 0; 0- as it can be seen, based on the problem 40- once we multiply determinants of the smaller is multiplied by the coefficients of the initial matrix, multiplication with 0 will give us 0, so we can reduce calculations to the following:

$\left[\begin{array}{cccc}5&0&0&0\\8&0&0&0\\-4&6&2&0\\2&0&2&-1\end{array}\right] =5* \left[\begin{array}{ccc}0&0&0\\6&2&0\\0&2&-1\end{array}\right]=0$

The answer- 0

For the final matrix from problem 42, we can do the following:

$\left[\begin{array}{cccc}4&-13&-8&0\\1&2&5&7\\0&0&3&0\\0&0&0&-2\end{array}\right] =4* \left[\begin{array}{ccc}2&5&7\\0&3&0\\0&0&-2\end{array}\right]- (-13)* \left[\begin{array}{ccc}1&5&7\\0&3&0\\0&0&-2\end{array}\right]+(-8)* \left[\begin{array}{ccc}1&2&7\\0&0&0\\0&0&-2\end{array}\right]-0* \left[\begin{array}{ccc}1&2&5\\0&0&3\\0&0&0\end{array}\right]$

Then we can calculate determinants of the three matrices 3x3 and substituting in the equation above and removing 0 multiplier, we are getting:

$4*(-12)+13*(-6)-8*(0)=-126$

So the answer for the problem 42 is -126