Find the $2 \times 2$ matrix $\bold{A}$ such that
\[\bold{A} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\]
and
\[\bold{A} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -7 \\ 4 \end{pmatrix}.\]
Regular:
Find the 2 x 2 matrix A such that:
A<6,-1> = <13,-26>
&
A<1,4> =<23,4>.
I am struggling to solve this problem, any help would be appreciated!
The numbers in your LaTeX don't seem to agree with the "regular" section, so I'll ignore them. You want A with rows (a b) and (c d) such that
6a-b = 13
6c-d = -26
a+4b = 23
c+4d = 4
I get rows (3 5) and (-4 2)
Check:
http://www.wolframalpha.com/input/?i=%7B%7B3,5%7D,%7B-4,2%7D%7D*%7B%7B6,1%7D,%7B-1,4%7D%7D
My bad! Thanks for the help! I was able to figure out the latex problem accordingly :)
Well, it seems like matrix A is a real comedian! It's trying to transform one vector into another for its own amusement. Let's see if we can decode its jokes.
So, we have two equations given by:
A<1, 0> = <1, 2>
A<0, 1> = <-7, 4>
Let's call the entries of matrix A:
A = [a, b
c, d]
Now, we can write the first equation as:
a*1 + b*0 = 1 ---> a = 1
a*0 + b*1 = 2 ---> b = 2
The second equation gives us:
c*1 + d*0 = -7 ---> c = -7
c*0 + d*1 = 4 ---> d = 4
So, it looks like we've cracked A's jokes! Matrix A is:
A = [1, 2
-7, 4]
Now, don't let matrix A's humor confuse you again!
To find the matrix $\boldsymbol{A}$, we can represent it as follows:
\[
\boldsymbol{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}
\]
We are given that $\boldsymbol{A} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $\boldsymbol{A} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -7 \\ 4\end{pmatrix}$.
Let's compute the first equation:
\[
\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} a \cdot 1 + b \cdot 0 \\ c \cdot 1 + d \cdot 0 \end{pmatrix} = \begin{pmatrix} a \\ c \end{pmatrix}
\]
Since this should equal $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$, we have:
\[
a = 1 \quad \text{and} \quad c = 2.
\]
Next, let's compute the second equation:
\[
\begin{pmatrix} a & b\\ c & d \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} a \cdot 0 + b \cdot 1 \\ c \cdot 0 + d \cdot 1 \end{pmatrix} = \begin{pmatrix} b \\ d \end{pmatrix}
\]
Since this should equal $\begin{pmatrix} -7 \\ 4 \end{pmatrix}$, we have:
\[
b = -7 \quad \text{and} \quad d = 4.
\]
Therefore, the matrix $\boldsymbol{A}$ is:
\[
\boldsymbol{A} = \begin{pmatrix} 1 & -7 \\ 2 & 4 \end{pmatrix}
\]
To find the matrix $\bold{A}$, we will set up a system of equations using the given information.
Let $\bold{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ be the $2 \times 2$ matrix we are seeking.
We are given that $\bold{A} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$.
This means that the first column of $\bold{A}$ is equal to $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$. Therefore, $a = 1$ and $c = 2$.
We are also given that $\bold{A} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -7 \\ 4 \end{pmatrix}$.
This means that the second column of $\bold{A}$ is equal to $\begin{pmatrix} -7 \\ 4 \end{pmatrix}$. Therefore, $b = -7$ and $d = 4$.
Thus, the matrix $\bold{A}$ is:
\[\bold{A} = \begin{pmatrix} 1 & -7 \\ 2 & 4 \end{pmatrix}.\]
In general, to find the matrix $\bold{A}$ given $\bold{A}\bold{v} = \bold{u}$ for two different vectors $\bold{v}$ and $\bold{u}$, you can follow these steps:
1. Let $\bold{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ be the matrix you are seeking.
2. Write down the equation $\bold{A}\bold{v} = \bold{u}$ using the given vectors $\bold{v}$ and $\bold{u}$.
3. Expand the matrix multiplication to get a system of equations. In our example, we expanded $\bold{A} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ to get $a = 1$ and $c = 2$, and $\bold{A} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -7 \\ 4 \end{pmatrix}$ to get $b = -7$ and $d = 4$.
4. Use the values of $a$, $b$, $c$, and $d$ obtained from the system of equations to form the matrix $\bold{A}$.
Remember to always verify your solution by checking if $\bold{A}\bold{v} = \bold{u}$ for both given vectors $\bold{v}$ and $\bold{u}$.