### Instruction: Compute $\begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}^3.$.### Response: I need to find the cube of a 2x2 matrix, which means multiplying it by itself three times.
I know that matrix multiplication is not commutative, so I have to keep the order of the factors the same.
I also know that matrix multiplication involves taking the dot product of the rows of the first matrix with the columns of the second matrix.
To make the calculation easier, I will use a shorthand notation where I write the entries of the matrix as $a, b, c, d$ instead of using subscripts.
So the matrix I want to cube is $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ where $a = 1, b = -1, c = 1, d = 0$.
I will start by finding the square of the matrix, which is $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.
To find the first entry of the product, I need to take the dot product of the first row of the first matrix with the first column of the second matrix.
That is, $a \cdot a + b \cdot c$.
Plugging in the values, I get $1 \cdot 1 + (-1) \cdot 1 = 0$.
To find the second entry of the product, I need to take the dot product of the first row of the first matrix with the second column of the second matrix.
That is, $a \cdot b + b \cdot d$.
Plugging in the values, I get $1 \cdot (-1) + (-1) \cdot 0 = -1$.
To find the third entry of the product, I need to take the dot product of the second row of the first matrix with the first column of the second matrix.
That is, $c \cdot a + d \cdot c$.
Plugging in the values, I get $1 \cdot 1 + 0 \cdot 1 = 1$.
To find the fourth entry of the product, I need to take the dot product of the second row of the first matrix with the second column of the second matrix.
That is, $c \cdot b + d \cdot d$.
Plugging in the values, I get $1 \cdot (-1) + 0 \cdot 0 = -1$.
So the square of the matrix is $\begin{pmatrix} 0 & -1 \\ 1 & -1 \end{pmatrix}$.
Now I need to multiply this matrix by the original matrix to get the cube.
That is, $\begin{pmatrix} 0 & -1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}$.
To find the first entry of the product, I need to take the dot product of the first row of the first matrix with the first column of the second matrix.
That is, $0 \cdot 1 + (-1) \cdot 1 = -1$.
To find the second entry of the product, I need to take the dot product of the first row of the first matrix with the second column of the second matrix.
That is, $0 \cdot (-1) + (-1) \cdot 0 = 0$.
To find the third entry of the product, I need to take the dot product of the second row of the first matrix with the first column of the second matrix.
That is, $1 \cdot 1 + (-1) \cdot 1 = 0$.
To find the fourth entry of the product, I need to take the dot product of the second row of the first matrix with the second column of the second matrix.
That is, $1 \cdot (-1) + (-1) \cdot 0 = -1$.