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
\[\bold{A} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -7 \\ 4 \end{pmatrix}.\]
I do not know how to solve this, I am struggling, any help would be appreciated!
To find the matrix $\bold{A}$, we need to determine the values of its entries.
Let $\bold{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, where $a$, $b$, $c$, and $d$ are the unknown entries.
Now, we can use the given equations to find $a$, $b$, $c$, and $d$.
From the first equation, we have:
\[\bold{A} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}.\]
This translates to the equation:
\[\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}.\]
Multiplying the matrices, we get:
\[\begin{pmatrix} a \\ c \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}.\]
Similarly, from the second equation, we have:
\[\bold{A} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -7 \\ 4 \end{pmatrix}.\]
This translates to the equation:
\[\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -7 \\ 4 \end{pmatrix}.\]
Multiplying the matrices, we get:
\[\begin{pmatrix} b \\ d \end{pmatrix} = \begin{pmatrix} -7 \\ 4 \end{pmatrix}.\]
Therefore, we have the following system of equations:
\[\begin{aligned} a &= 1, \\ c &= 2, \\ b &= -7, \\ d &= 4. \end{aligned}\]
Thus, the matrix $\bold{A}$ is given by:
\[\bold{A} = \begin{pmatrix} 1 & -7 \\ 2 & 4 \end{pmatrix}.\]
To find the matrix $\mathbf{A}$, we need to determine what happens to the standard basis vectors $\mathbf{e}_1$ and $\mathbf{e}_2$ when they are multiplied by $\mathbf{A}$.
The standard basis vector $\mathbf{e}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ gets mapped to $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ under the transformation $\mathbf{A}$.
Similarly, the standard basis vector $\mathbf{e}_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ gets mapped to $\begin{pmatrix} -7 \\ 4 \end{pmatrix}$ under the transformation $\mathbf{A}$.
Since the columns of a matrix represent the images of the standard basis vectors under the matrix transformation, we can write:
\[\mathbf{A} = \begin{pmatrix} \mathbf{A} \mathbf{e}_1 & \mathbf{A} \mathbf{e}_2 \end{pmatrix}.\]
Substituting the given values, we get:
\[\mathbf{A} = \begin{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} & \begin{pmatrix} -7 \\ 4 \end{pmatrix} \end{pmatrix}.\]
Therefore, the matrix $\mathbf{A}$ is:
\[\mathbf{A} = \begin{pmatrix} 1 & -7 \\ 2 & 4 \end{pmatrix}.\]