# Difference between revisions of "Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 1 (a)/Solution 1"

From UBC Wiki

HyunjuKwon (Talk | contribs) |
NicholasHu (Talk | contribs) m |
||

Line 1: | Line 1: | ||

− | Note that <math>x_1, x_2</math> and <math>x_3</math> denote the amount of money Uno, Duo, and Traea owe, respectively. | + | Note that <math>x_1</math>, <math>x_2</math>, and <math>x_3</math> denote the amount of money Uno, Duo, and Traea owe, respectively. |

− | We first express three | + | We first express the three statements (1) "All together they owe $600", (2) "Duo owes $200 more than Uno", and (3) "Uno and Duo combined owe as much as Traea" as linear equations in <math>x_1</math>, <math>x_2</math>, and <math>x_3</math>: |

− | <math display="block">\begin{cases}x_1+x_2+x_3=600 | + | <math display="block">\begin{cases}x_1+x_2+x_3=600 & (1)\\x_2-x_1=200 & (2)\\ x_1+x_2=x_3 & (3)\end{cases}.</math> |

This system can be rewritten as | This system can be rewritten as | ||

− | <math display="block"> | + | <math display="block">\begin{cases}x_1 +x_2+x_3=600 \\-x_1 +x_2+0x_3=200 \\ x_1 +x_2-x_3=0\end{cases}.</math> |

− | Then, the coefficient matrix <math>A</math> can be formed by | + | Then, the coefficient matrix <math>A</math> can be formed by reading off the coefficients of each variable in each row. For the constant vector <math>\mathbf{b}</math>, we keep the same values: |

− | <math display="block">\begin{ | + | <math display="block">\begin{bmatrix}1&1&1\\-1&1&0\\1&1&-1 \end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}=\begin{bmatrix}600\\200\\0 \end{bmatrix}.</math> |

− | Therefore, we find | + | Therefore, we find that |

− | <math display="block">A = \begin{ | + | <math display="block">\color{blue} A = \begin{bmatrix}1&1&1\\-1&1&0\\1&1&-1 \end{bmatrix},\quad \mathbf{b}=\begin{bmatrix}600\\200\\0 \end{bmatrix}.</math> |

+ | Note that the rows of the matrix <math>A</math> and vector <math>\mathbf{b}</math> may be freely permuted without affecting the system; thus, for instance, | ||

− | + | <math display="block">A = \begin{bmatrix}-1&1&0\\1&1&1\\1&1&-1 \end{bmatrix},\quad \mathbf{b}=\begin{bmatrix}200\\600\\0 \end{bmatrix}.</math> | |

− | + | is also a valid answer. | |

− | + | ||

− | + | ||

− | + | ||

− | + | ||

− | + | ||

− | + | ||

− | + | ||

− | + |

## Latest revision as of 20:04, 8 March 2018

Note that , , and denote the amount of money Uno, Duo, and Traea owe, respectively.

We first express the three statements (1) "All together they owe $600", (2) "Duo owes $200 more than Uno", and (3) "Uno and Duo combined owe as much as Traea" as linear equations in , , and :

This system can be rewritten as

Then, the coefficient matrix can be formed by reading off the coefficients of each variable in each row. For the constant vector , we keep the same values:

Therefore, we find that

Note that the rows of the matrix and vector may be freely permuted without affecting the system; thus, for instance,

is also a valid answer.