Question

If the system below was written as a matrix equation, by which matrix could you multiply both sides to obtain a soultion?
System: 4x+6y=24
5x+8y=40

Answer Choices:
A) [8 -6
-5 4]
B) [4 -3
-2.5 2]
C) [-4 5
6 -8]
D) [-2 2.5
3 -4]

Answer 1

\begin{array}l\color{#FF0000}{\text{f}}\color{#FF7F00}{\text{i}}\color{#FFFF00}{\text{n}}\color{#00FF00}{\text{d}}\color{#0000FF}{\text{ }}\color{#0000FF}{\text{t}}\color{#6600FF}{\text{h}}\color{#8B00FF}{\text{e}}\color{#FF0000}{\text{ }}\color{#FF0000}{\text{i}}\color{#FF7F00}{\text{n}}\color{#FFFF00}{\text{v}}\color{#00FF00}{\text{e}}\color{#0000FF}{\text{r}}\color{#6600FF}{\text{s}}\color{#8B00FF}{\text{e}}\color{#FF0000}{\text{ }}\color{#FF0000}{\text{m}}\color{#FF7F00}{\text{a}}\color{#FFFF00}{\text{t}}\color{#00FF00}{\text{r}}\color{#0000FF}{\text{i}}\color{#6600FF}{\text{x}}\color{#8B00FF}{\text{ }}\color{#FF0000}{\text{o}}\color{#FF7F00}{\text{f}}\color{#FFFF00}{\text{ }}\color{#FFFF00}{\text{t}}\color{#00FF00}{\text{h}}\color{#0000FF}{\text{e}}\color{#6600FF}{\text{ }}\color{#6600FF}{\text{c}}\color{#8B00FF}{\text{o}}\color{#FF0000}{\text{e}}\color{#FF7F00}{\text{f}}\color{#FFFF00}{\text{f}}\color{#00FF00}{\text{i}}\color{#0000FF}{\text{c}}\color{#6600FF}{\text{i}}\color{#8B00FF}{\text{e}}\color{#FF0000}{\text{n}}\color{#FF7F00}{\text{t}}\color{#FFFF00}{\text{ }}\color{#FFFF00}{\text{m}}\color{#00FF00}{\text{a}}\color{#0000FF}{\text{t}}\color{#6600FF}{\text{r}}\color{#8B00FF}{\text{i}}\color{#FF0000}{\text{x}}\end{array}

Answer 2

How?

Answer 3

first write the coefficient matrix:

[ 4 6]
[5 8]

then augment it to an identity matrix of the same size

[ 4 6 | 1 0 ]
[ 5 8 | 0 1 ]

and write it in reduced row echelon form

the right matrix will be its inverse and the answer to your question

Answer 4

you should get B) as the solution

Answer 5

thank you!

Answer 6

\begin{array}l\color{#FF0000}{\text{D}}\color{#FF7F00}{\text{o}}\color{#FFFF00}{\text{ }}\color{#FFFF00}{\text{y}}\color{#00FF00}{\text{o}}\color{#0000FF}{\text{u}}\color{#6600FF}{\text{ }}\color{#6600FF}{\text{n}}\color{#8B00FF}{\text{o}}\color{#FF0000}{\text{w}}\color{#FF7F00}{\text{ }}\color{#FF7F00}{\text{k}}\color{#FFFF00}{\text{n}}\color{#00FF00}{\text{o}}\color{#0000FF}{\text{w}}\color{#6600FF}{\text{ }}\color{#6600FF}{\text{h}}\color{#8B00FF}{\text{o}}\color{#FF0000}{\text{w}}\color{#FF7F00}{\text{ }}\color{#FF7F00}{\text{t}}\color{#FFFF00}{\text{o}}\color{#00FF00}{\text{ }}\color{#00FF00}{\text{f}}\color{#0000FF}{\text{i}}\color{#6600FF}{\text{n}}\color{#8B00FF}{\text{d}}\color{#FF0000}{\text{ }}\color{#FF0000}{\text{m}}\color{#FF7F00}{\text{a}}\color{#FFFF00}{\text{t}}\color{#00FF00}{\text{r}}\color{#0000FF}{\text{i}}\color{#6600FF}{\text{x}}\color{#8B00FF}{\text{ }}\color{#FF0000}{\text{i}}\color{#FF7F00}{\text{n}}\color{#FFFF00}{\text{v}}\color{#00FF00}{\text{e}}\color{#0000FF}{\text{r}}\color{#6600FF}{\text{s}}\color{#8B00FF}{\text{e}}\color{#FF0000}{\text{s}}\color{#FF7F00}{\text{?}}\end{array}

Answer 7

Yes

Answer 8

there's another way where you find the determinant and take it's reciprocal, and switch elements in the original matrix while multiplying and taking negatives etc...