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Explain why we can always evaluate the determinant of a square matrix.
A determinant is the sum and products of the entries in the matrix, so you can always evaluate that product—even if it does end up being 0.
Examining Cramer’s Rule, explain why there is no unique solution to the system when the determinant of your matrix is 0. For simplicity, use a $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ matrix.
Explain what it means in terms of an inverse for a matrix to have a 0 determinant.
The inverse does not exist.
The determinant of $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ matrix $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ is 3. If you switch the rows and multiply the first row by 6 and the second row by 2, explain how to find the determinant and provide the answer.
For the following exercises, find the determinant.
$\left|\begin{array}{rr}\hfill -1& \hfill 2\\ \hfill 3& \hfill -4\end{array}\right|$
$$\left|\begin{array}{rr}\hfill 2& \hfill -5\\ \hfill -1& \hfill 6\end{array}\right|$$
$7$
$\left|\begin{array}{cc}-8& 4\\ -1& 5\end{array}\right|$
$\left|\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 3& \hfill -4\end{array}\right|$
$-4$
$\left|\begin{array}{rr}\hfill 10& \hfill 20\\ \hfill 0& \hfill -10\end{array}\right|$
$\left|\begin{array}{cc}10& 0.2\\ 5& 0.1\end{array}\right|$
$0$
$\left|\begin{array}{rr}\hfill 6& \hfill -3\\ \hfill 8& \hfill 4\end{array}\right|$
$\left|\begin{array}{rr}\hfill -2& \hfill -3\\ \hfill 3.1& \hfill 4,000\end{array}\right|$
$-7,990.7$
$\left|\begin{array}{rr}\hfill -1.1& \hfill 0.6\\ \hfill 7.2& \hfill -0.5\end{array}\right|$
$\left|\begin{array}{rrr}\hfill -1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0\\ \hfill 0& \hfill 0& \hfill -3\end{array}\right|$
$3$
$\left|\begin{array}{rrr}\hfill -1& \hfill 4& \hfill 0\\ \hfill 0& \hfill 2& \hfill 3\\ \hfill 0& \hfill 0& \hfill -3\end{array}\right|$
$\left|\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right|$
$-1$
$\left|\begin{array}{rrr}\hfill 2& \hfill -3& \hfill 1\\ \hfill 3& \hfill -4& \hfill 1\\ \hfill -5& \hfill 6& \hfill 1\end{array}\right|$
$\left|\begin{array}{rrr}\hfill -2& \hfill 1& \hfill 4\\ \hfill -4& \hfill 2& \hfill -8\\ \hfill 2& \hfill -8& \hfill -3\end{array}\right|$
$224$
$\left|\begin{array}{rrr}\hfill 6& \hfill -1& \hfill 2\\ \hfill -4& \hfill -3& \hfill 5\\ \hfill 1& \hfill 9& \hfill -1\end{array}\right|$
$\left|\begin{array}{rrr}\hfill 5& \hfill 1& \hfill -1\\ \hfill 2& \hfill 3& \hfill 1\\ \hfill 3& \hfill -6& \hfill -3\end{array}\right|$
$15$
$\left|\begin{array}{rrr}\hfill 1.1& \hfill 2& \hfill -1\\ \hfill -4& \hfill 0& \hfill 0\\ \hfill 4.1& \hfill -0.4& \hfill 2.5\end{array}\right|$
$\left|\begin{array}{rrr}\hfill 2& \hfill -1.6& \hfill 3.1\\ \hfill 1.1& \hfill 3& \hfill -8\\ \hfill -9.3& \hfill 0& \hfill 2\end{array}\right|$
$-17.03$
$\left|\begin{array}{ccc}-\frac{1}{2}& \frac{1}{3}& \frac{1}{4}\\ \frac{1}{5}& -\frac{1}{6}& \frac{1}{7}\\ 0& 0& \frac{1}{8}\end{array}\right|$
For the following exercises, solve the system of linear equations using Cramer’s Rule.
$\begin{array}{l}2x-3y=\mathrm{-1}\\ 4x+5y=9\end{array}$
$\left(1,1\right)$
$\begin{array}{r}5x-4y=2\\ -4x+7y=6\end{array}$
$\begin{array}{l}\text{}6x-3y=2\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hfill \\ -8x+9y=\mathrm{-1}\hfill \end{array}$
$\left(\frac{1}{2},\frac{1}{3}\right)$
$\begin{array}{l}2x+6y=12\\ 5x-2y=13\end{array}$
$\begin{array}{l}4x+3y=23\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hfill \\ \text{}2x-y=\mathrm{-1}\hfill \end{array}$
$\left(2,5\right)$
$\begin{array}{l}10x-6y=2\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hfill \\ -5x+8y=\mathrm{-1}\hfill \end{array}$
$\begin{array}{l}4x-3y=\mathrm{-3}\\ 2x+6y=\mathrm{-4}\end{array}$
$\left(-1,-\frac{1}{3}\right)$
$\begin{array}{r}4x-5y=7\\ -3x+9y=0\end{array}$
$\begin{array}{l}4x+10y=180\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hfill \\ -3x-5y=\mathrm{-105}\hfill \end{array}$
$\left(15,12\right)$
$\begin{array}{l}\text{}8x-2y=\mathrm{-3}\hfill \\ -4x+6y=4\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hfill \end{array}$
For the following exercises, solve the system of linear equations using Cramer’s Rule.
$\begin{array}{l}\text{}x+2y-4z=-1\hfill \\ \text{}7x+3y+5z=26\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hfill \\ -2x-6y+7z=-6\hfill \end{array}$
$\left(1,3,2\right)$
$\begin{array}{l}-5x+2y-4z=-47\hfill \\ \text{}4x-3y-z=-94\hfill \\ \text{}3x-3y+2z=94\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hfill \end{array}$
$\begin{array}{l}\text{}4x+5y-z=\mathrm{-7}\hfill \\ \mathrm{-2}x-9y+2z=8\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hfill \\ \text{}5y+7z=21\text{\hspace{0.17em}}\hfill \end{array}$
$\left(-1,0,3\right)$
$\begin{array}{r}4x-3y+4z=10\\ 5x-2z=-2\\ 3x+2y-5z=-9\end{array}$
$\begin{array}{l}4x-2y+3z=6\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hfill \\ \text{}-6x+y=-2\hfill \\ 2x+7y+8z=24\hfill \end{array}$
$\left(\frac{1}{2},1,2\right)$
$\begin{array}{r}\hfill 5x+2y-z=1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \hfill -7x-8y+3z=1.5\\ \hfill 6x-12y+z=7\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$
$\begin{array}{l}\text{}13x-17y+16z=73\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hfill \\ -11x+15y+17z=61\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hfill \\ \text{}46x+10y-30z=-18\hfill \end{array}$
$\left(2,1,4\right)$
$\begin{array}{l}\begin{array}{l}\hfill \\ -4x-3y-8z=-7\hfill \end{array}\hfill \\ \text{}2x-9y+5z=0.5\hfill \\ \text{}5x-6y-5z=-2\hfill \end{array}$
$\begin{array}{l}\text{}4x-6y+8z=10\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hfill \\ -2x+3y-4z=-5\hfill \\ \text{}x+y+z=1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hfill \end{array}$
Infinite solutions
$\begin{array}{r}\hfill 4x-6y+8z=10\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \hfill -2x+3y-4z=-5\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \hfill 12x+18y-24z=-30\end{array}$
For the following exercises, use the determinant function on a graphing utility.
$\left|\begin{array}{rrrr}\hfill 1& \hfill 0& \hfill 8& \hfill 9\\ \hfill 0& \hfill 2& \hfill 1& \hfill 0\\ \hfill 1& \hfill 0& \hfill 3& \hfill 0\\ \hfill 0& \hfill 2& \hfill 4& \hfill 3\end{array}\right|$
$24$
$\left|\begin{array}{rrrr}\hfill 1& \hfill 0& \hfill 2& \hfill 1\\ \hfill 0& \hfill \mathrm{-9}& \hfill 1& \hfill 3\\ \hfill 3& \hfill 0& \hfill \mathrm{-2}& \hfill \mathrm{-1}\\ \hfill 0& \hfill 1& \hfill 1& \hfill \mathrm{-2}\end{array}\right|$
$\left|\begin{array}{rrrr}\hfill \frac{1}{2}& \hfill 1& \hfill 7& \hfill 4\\ \hfill 0& \hfill \frac{1}{2}& \hfill 100& \hfill 5\\ \hfill 0& \hfill 0& \hfill 2& \hfill \mathrm{2,000}\\ \hfill 0& \hfill 0& \hfill 0& \hfill 2\end{array}\right|$
$1$
$\left|\begin{array}{rrrr}\hfill 1& \hfill 0& \hfill 0& \hfill 0\\ \hfill 2& \hfill 3& \hfill 0& \hfill 0\\ \hfill 4& \hfill 5& \hfill 6& \hfill 0\\ \hfill 7& \hfill 8& \hfill 9& \hfill 0\end{array}\right|$
For the following exercises, create a system of linear equations to describe the behavior. Then, calculate the determinant. Will there be a unique solution? If so, find the unique solution.
Two numbers add up to 56. One number is 20 less than the other.
Yes; 18, 38
Two numbers add up to 104. If you add two times the first number plus two times the second number, your total is 208
Three numbers add up to 106. The first number is 3 less than the second number. The third number is 4 more than the first number.
Yes; 33, 36, 37
Three numbers add to 216. The sum of the first two numbers is 112. The third number is 8 less than the first two numbers combined.
For the following exercises, create a system of linear equations to describe the behavior. Then, solve the system for all solutions using Cramer’s Rule.
You invest $10,000 into two accounts, which receive 8% interest and 5% interest. At the end of a year, you had $10,710 in your combined accounts. How much was invested in each account?
$7,000 in first account, $3,000 in second account.
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