11.8 Solving systems with cramer's rule  (Page 5/11)

Explain why we can always evaluate the determinant of a square matrix.

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 $2\times 2$ matrix.

Explain what it means in terms of an inverse for a matrix to have a 0 determinant.

The determinant of $2\times 2$ matrix $A$ 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.

$\left|\begin{array}{cc}1& 2\\ 3& 4\end{array}\right|$

$\left|\begin{array}{rr}\hfill -1& \hfill 2\\ \hfill 3& \hfill -4\end{array}\right|$

$$\left|\begin{array}{rr}\hfill 2& \hfill -5\\ \hfill -1& \hfill 6\end{array}\right|$$

$\left|\begin{array}{cc}-8& 4\\ -1& 5\end{array}\right|$

$\left|\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 3& \hfill -4\end{array}\right|$

$\left|\begin{array}{rr}\hfill 10& \hfill 20\\ \hfill 0& \hfill -10\end{array}\right|$

$\left|\begin{array}{cc}10& 0.2\\ 5& 0.1\end{array}\right|$

$\left|\begin{array}{rr}\hfill 6& \hfill -3\\ \hfill 8& \hfill 4\end{array}\right|$

$\left|\begin{array}{rr}\hfill -2& \hfill -3\\ \hfill 3.1& \hfill 4,000\end{array}\right|$

$\left|\begin{array}{rr}\hfill -1.1& \hfill 0.6\\ \hfill 7.2& \hfill -0.5\end{array}\right|$

$\left|\begin{array}{rrr}\hfill -1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0\\ \hfill 0& \hfill 0& \hfill -3\end{array}\right|$

$\left|\begin{array}{rrr}\hfill -1& \hfill 4& \hfill 0\\ \hfill 0& \hfill 2& \hfill 3\\ \hfill 0& \hfill 0& \hfill -3\end{array}\right|$

$\left|\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right|$

$\left|\begin{array}{rrr}\hfill 2& \hfill -3& \hfill 1\\ \hfill 3& \hfill -4& \hfill 1\\ \hfill -5& \hfill 6& \hfill 1\end{array}\right|$

$\left|\begin{array}{rrr}\hfill -2& \hfill 1& \hfill 4\\ \hfill -4& \hfill 2& \hfill -8\\ \hfill 2& \hfill -8& \hfill -3\end{array}\right|$

$\left|\begin{array}{rrr}\hfill 6& \hfill -1& \hfill 2\\ \hfill -4& \hfill -3& \hfill 5\\ \hfill 1& \hfill 9& \hfill -1\end{array}\right|$

$\left|\begin{array}{rrr}\hfill 5& \hfill 1& \hfill -1\\ \hfill 2& \hfill 3& \hfill 1\\ \hfill 3& \hfill -6& \hfill -3\end{array}\right|$

$\left|\begin{array}{rrr}\hfill 1.1& \hfill 2& \hfill -1\\ \hfill -4& \hfill 0& \hfill 0\\ \hfill 4.1& \hfill -0.4& \hfill 2.5\end{array}\right|$

$\left|\begin{array}{rrr}\hfill 2& \hfill -1.6& \hfill 3.1\\ \hfill 1.1& \hfill 3& \hfill -8\\ \hfill -9.3& \hfill 0& \hfill 2\end{array}\right|$

$\left|\begin{array}{ccc}-\frac{1}{2}& \frac{1}{3}& \frac{1}{4}\\ \frac{1}{5}& -\frac{1}{6}& \frac{1}{7}\\ 0& 0& \frac{1}{8}\end{array}\right|$

$\begin{array}{l}2x-3y=\mathrm{-1}\\ 4x+5y=9\end{array}$

$\begin{array}{r}5x-4y=2\\ -4x+7y=6\end{array}$

$\begin{array}{l}6x-3y=2\hfill \\ -8x+9y=\mathrm{-1}\hfill \end{array}$

$\begin{array}{l}2x+6y=12\\ 5x-2y=13\end{array}$

$\begin{array}{l}4x+3y=23\hfill \\ 2x-y=\mathrm{-1}\hfill \end{array}$

$\begin{array}{l}10x-6y=2\hfill \\ -5x+8y=\mathrm{-1}\hfill \end{array}$

$\begin{array}{l}4x-3y=\mathrm{-3}\\ 2x+6y=\mathrm{-4}\end{array}$

$\begin{array}{r}4x-5y=7\\ -3x+9y=0\end{array}$

$\begin{array}{l}4x+10y=180\hfill \\ -3x-5y=\mathrm{-105}\hfill \end{array}$

$\begin{array}{l}8x-2y=\mathrm{-3}\hfill \\ -4x+6y=4\hfill \end{array}$

$\begin{array}{l}x+2y-4z=-1\hfill \\ 7x+3y+5z=26\hfill \\ -2x-6y+7z=-6\hfill \end{array}$

$\begin{array}{l}-5x+2y-4z=-47\hfill \\ 4x-3y-z=-94\hfill \\ 3x-3y+2z=94\hfill \end{array}$

$\begin{array}{l}4x+5y-z=\mathrm{-7}\hfill \\ \mathrm{-2}x-9y+2z=8\hfill \\ 5y+7z=21\hfill \end{array}$

$\begin{array}{r}4x-3y+4z=10\\ 5x-2z=-2\\ 3x+2y-5z=-9\end{array}$

$\begin{array}{l}4x-2y+3z=6\hfill \\ -6x+y=-2\hfill \\ 2x+7y+8z=24\hfill \end{array}$

$\begin{array}{r}\hfill 5x+2y-z=1\\ \hfill -7x-8y+3z=1.5\\ \hfill 6x-12y+z=7\end{array}$

$\begin{array}{l}13x-17y+16z=73\hfill \\ -11x+15y+17z=61\hfill \\ 46x+10y-30z=-18\hfill \end{array}$

$\begin{array}{l}\begin{array}{l}\hfill \\ -4x-3y-8z=-7\hfill \end{array}\hfill \\ 2x-9y+5z=0.5\hfill \\ 5x-6y-5z=-2\hfill \end{array}$

$\begin{array}{l}4x-6y+8z=10\hfill \\ -2x+3y-4z=-5\hfill \\ x+y+z=1\hfill \end{array}$

$\begin{array}{r}\hfill 4x-6y+8z=10\\ \hfill -2x+3y-4z=-5\\ \hfill 12x+18y-24z=-30\end{array}$

$\left|\begin{array}{rrrr}\hfill 1& \hfill 0& \hfill 8& \hfill 9\\ \hfill 0& \hfill 2& \hfill 1& \hfill 0\\ \hfill 1& \hfill 0& \hfill 3& \hfill 0\\ \hfill 0& \hfill 2& \hfill 4& \hfill 3\end{array}\right|$

$\left|\begin{array}{rrrr}\hfill 1& \hfill 0& \hfill 2& \hfill 1\\ \hfill 0& \hfill \mathrm{-9}& \hfill 1& \hfill 3\\ \hfill 3& \hfill 0& \hfill \mathrm{-2}& \hfill \mathrm{-1}\\ \hfill 0& \hfill 1& \hfill 1& \hfill \mathrm{-2}\end{array}\right|$

$\left|\begin{array}{rrrr}\hfill \frac{1}{2}& \hfill 1& \hfill 7& \hfill 4\\ \hfill 0& \hfill \frac{1}{2}& \hfill 100& \hfill 5\\ \hfill 0& \hfill 0& \hfill 2& \hfill \mathrm{2,000}\\ \hfill 0& \hfill 0& \hfill 0& \hfill 2\end{array}\right|$

$\left|\begin{array}{rrrr}\hfill 1& \hfill 0& \hfill 0& \hfill 0\\ \hfill 2& \hfill 3& \hfill 0& \hfill 0\\ \hfill 4& \hfill 5& \hfill 6& \hfill 0\\ \hfill 7& \hfill 8& \hfill 9& \hfill 0\end{array}\right|$

Two numbers add up to 56. One number is 20 less than the other.

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.

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.

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?