Answer :
Of course! Let's prove that for the matrix [tex]\( P \)[/tex] given by
[tex]\[ P = \begin{pmatrix} 2 & 1 \\ -2 & -1 \end{pmatrix}, \][/tex]
it holds that [tex]\( P^2 = P \)[/tex].
To do this, we'll first calculate [tex]\( P^2 \)[/tex] by multiplying the matrix [tex]\( P \)[/tex] by itself.
Recall the matrix multiplication rule:
[tex]\[ (AB)_{ij} = \sum_{k} A_{ik} B_{kj}, \][/tex]
where the element in the [tex]\(i\)[/tex]-th row and [tex]\(j\)[/tex]-th column of the product matrix [tex]\( AB \)[/tex] is the sum of the products of the elements in the [tex]\(i\)[/tex]-th row of [tex]\( A \)[/tex] and the [tex]\(j\)[/tex]-th column of [tex]\( B \)[/tex].
Let's begin the multiplication:
[tex]\[ P^2 = P \cdot P = \begin{pmatrix} 2 & 1 \\ -2 & -1 \end{pmatrix} \cdot \begin{pmatrix} 2 & 1 \\ -2 & -1 \end{pmatrix} \][/tex]
We'll compute each element of the resulting matrix:
1. Top-left element (1,1):
[tex]\[ (P^2)_{11} = 2 \cdot 2 + 1 \cdot (-2) = 4 - 2 = 2 \][/tex]
2. Top-right element (1,2):
[tex]\[ (P^2)_{12} = 2 \cdot 1 + 1 \cdot (-1) = 2 - 1 = 1 \][/tex]
3. Bottom-left element (2,1):
[tex]\[ (P^2)_{21} = (-2) \cdot 2 + (-1) \cdot (-2) = -4 + 2 = -2 \][/tex]
4. Bottom-right element (2,2):
[tex]\[ (P^2)_{22} = (-2) \cdot 1 + (-1) \cdot (-1) = -2 +1 = -1 \][/tex]
Combining all the computed elements, we get:
[tex]\[ P^2 = \begin{pmatrix} 2 & 1 \\ -2 & -1 \end{pmatrix} \][/tex]
Notice that this is exactly the same as matrix [tex]\( P \)[/tex], which proves that:
[tex]\[ P^2 = P \][/tex]
Thus, we've demonstrated that [tex]\( P = \begin{pmatrix} 2 & 1 \\ -2 & -1 \end{pmatrix} \)[/tex] satisfies the equation [tex]\( P^2 = P \)[/tex].
[tex]\[ P = \begin{pmatrix} 2 & 1 \\ -2 & -1 \end{pmatrix}, \][/tex]
it holds that [tex]\( P^2 = P \)[/tex].
To do this, we'll first calculate [tex]\( P^2 \)[/tex] by multiplying the matrix [tex]\( P \)[/tex] by itself.
Recall the matrix multiplication rule:
[tex]\[ (AB)_{ij} = \sum_{k} A_{ik} B_{kj}, \][/tex]
where the element in the [tex]\(i\)[/tex]-th row and [tex]\(j\)[/tex]-th column of the product matrix [tex]\( AB \)[/tex] is the sum of the products of the elements in the [tex]\(i\)[/tex]-th row of [tex]\( A \)[/tex] and the [tex]\(j\)[/tex]-th column of [tex]\( B \)[/tex].
Let's begin the multiplication:
[tex]\[ P^2 = P \cdot P = \begin{pmatrix} 2 & 1 \\ -2 & -1 \end{pmatrix} \cdot \begin{pmatrix} 2 & 1 \\ -2 & -1 \end{pmatrix} \][/tex]
We'll compute each element of the resulting matrix:
1. Top-left element (1,1):
[tex]\[ (P^2)_{11} = 2 \cdot 2 + 1 \cdot (-2) = 4 - 2 = 2 \][/tex]
2. Top-right element (1,2):
[tex]\[ (P^2)_{12} = 2 \cdot 1 + 1 \cdot (-1) = 2 - 1 = 1 \][/tex]
3. Bottom-left element (2,1):
[tex]\[ (P^2)_{21} = (-2) \cdot 2 + (-1) \cdot (-2) = -4 + 2 = -2 \][/tex]
4. Bottom-right element (2,2):
[tex]\[ (P^2)_{22} = (-2) \cdot 1 + (-1) \cdot (-1) = -2 +1 = -1 \][/tex]
Combining all the computed elements, we get:
[tex]\[ P^2 = \begin{pmatrix} 2 & 1 \\ -2 & -1 \end{pmatrix} \][/tex]
Notice that this is exactly the same as matrix [tex]\( P \)[/tex], which proves that:
[tex]\[ P^2 = P \][/tex]
Thus, we've demonstrated that [tex]\( P = \begin{pmatrix} 2 & 1 \\ -2 & -1 \end{pmatrix} \)[/tex] satisfies the equation [tex]\( P^2 = P \)[/tex].
