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A small shoe company took out a loan of $1,500,000 to expand their inventory. Part of the money was borrowed at 7%, part was borrowed at 8%, and part was borrowed at 10%. The amount borrowed at 10% was four times the amount borrowed at 7%, and the annual interest on all three loans was $130,500. Use matrices to find the amount borrowed at each rate.
$150,000 at 7%, $750,000 at 8%, $600,000 at 10%
Access these online resources for additional instruction and practice with solving systems of linear equations using Gaussian elimination.
Can any system of linear equations be written as an augmented matrix? Explain why or why not. Explain how to write that augmented matrix.
Yes. For each row, the coefficients of the variables are written across the corresponding row, and a vertical bar is placed; then the constants are placed to the right of the vertical bar.
Can any matrix be written as a system of linear equations? Explain why or why not. Explain how to write that system of equations.
Is there only one correct method of using row operations on a matrix? Try to explain two different row operations possible to solve the augmented matrix $\left[\begin{array}{rr}\hfill 9& \hfill 3\\ \hfill 1& \hfill -2\end{array}\text{}|\text{}\begin{array}{r}\hfill 0\\ \hfill 6\end{array}\right].$
No, there are numerous correct methods of using row operations on a matrix. Two possible ways are the following: (1) Interchange rows 1 and 2. Then $\text{\hspace{0.17em}}{R}_{2}={R}_{2}\mathrm{-9}{R}_{1}.\text{\hspace{0.17em}}$ (2) $\text{\hspace{0.17em}}{R}_{2}={R}_{1}\mathrm{-9}{R}_{2}.\text{\hspace{0.17em}}$ Then divide row 1 by 9.
Can a matrix whose entry is 0 on the diagonal be solved? Explain why or why not. What would you do to remedy the situation?
Can a matrix that has 0 entries for an entire row have one solution? Explain why or why not.
No. A matrix with 0 entries for an entire row would have either zero or infinitely many solutions.
For the following exercises, write the augmented matrix for the linear system.
$\begin{array}{l}8x\mathrm{-37}y=8\\ 2x+12y=3\end{array}$
$\begin{array}{l}\text{\hspace{0.17em}}\text{}16y=4\hfill \\ 9x-y=2\hfill \end{array}$
$\left[\begin{array}{rrrr}\hfill 0& \hfill & \hfill 16& \hfill \\ \hfill 9& \hfill & \hfill \mathrm{-1}& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 4\\ \hfill & \hfill 2\end{array}\right]$
$\begin{array}{l}\text{}3x+2y+10z=3\hfill \\ \mathrm{-6}x+2y+5z=13\hfill \\ \text{}4x+z=18\hfill \end{array}$
$\begin{array}{l}\hfill \\ \text{}x+5y+8z=19\hfill \\ \text{}12x+3y=4\hfill \\ 3x+4y+9z=\mathrm{-7}\hfill \end{array}$
$\left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill 5& \hfill & \hfill 8& \hfill \\ \hfill 12& \hfill & \hfill 3& \hfill & \hfill 0& \hfill \\ \hfill 3& \hfill & \hfill 4& \hfill & \hfill 9& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 16\\ \hfill & \hfill 4\\ \hfill & \hfill \mathrm{-7}\end{array}\right]$
$\begin{array}{l}6x+12y+16z=4\hfill \\ \text{}19x\mathrm{-5}y+3z=\mathrm{-9}\hfill \\ \text{}x+2y=\mathrm{-8}\hfill \end{array}$
For the following exercises, write the linear system from the augmented matrix.
$\left[\begin{array}{rr}\hfill \mathrm{-2}& \hfill 5\\ \hfill 6& \hfill \mathrm{-18}\end{array}\text{}|\text{}\begin{array}{r}\hfill 5\\ \hfill 26\end{array}\right]$
$\begin{array}{l}\mathrm{-2}x+5y=5\\ 6x\mathrm{-18}y=26\end{array}$
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