Chapter 8, Problem 13PS

To determine

The value of $4\times 4$ matrix.

Answer to Problem 13PS

A $4\times 4$ matrix whose determinant is equal to $a{x}^3+b{x}^2+cx+d$

Explanation of Solution

Given information:

Find a $4\times 4$ matrix whose determinant is equal to $a{x}^3+b{x}^2+cx+d$

Formula used:

Matrix multiplication

Calculation:

Let A be a $4\times 4$ matrix $A=\left[\begin{array}{cccc}x& 0& 0& d\\ 0& x& 0& c\\ -1& -1& x& b\\ 0& 0& -1& a\end{array}\right]$

Using the first column expand it to find its determinant.

  $\begin{array}{l}\left[\begin{array}{cccc}x& 0& 0& d\\ 0& x& 0& c\\ -1& -1& x& b\\ 0& 0& -1& a\end{array}\right]=x\left[\begin{array}{ccc}x& 0& c\\ -1& x& b\\ 0& -1& a\end{array}\right]-0+0-d\left[\begin{array}{ccc}0& x& 0\\ -1& -1& x\\ 0& 0& -1\end{array}\right]\\ =x\left\{x\left[\begin{array}{cc}x& b\\ -1& a\end{array}\right]-\left(-\left[\begin{array}{cc}0& c\\ -1& a\end{array}\right]\right)\right\}-d\left(-1\right)\\ =x\left[x\left(ax+b\right)+\left(0+c\right)\right]+d\\ \left[\begin{array}{cccc}x& 0& 0& d\\ 0& x& 0& c\\ -1& -1& x& b\\ 0& 0& -1& a\end{array}\right]=a{x}^3+b{x}^2+cx+d\end{array}$

Thus, a $4\times 4$ matrix whose determinant is equal to $a{x}^3+b{x}^2+cx+d$

Conclusion:

A $4\times 4$ matrix whose determinant is equal to $a{x}^3+b{x}^2+cx+d$