To demonstrate that the product of matrices [tex]\( P \)[/tex] and [tex]\( y \)[/tex] results in a null matrix, follow these steps:
Given:
[tex]\[ P = \begin{pmatrix} 1 & -1 \\ 3 & -3 \end{pmatrix} \][/tex]
[tex]\[ y = \begin{pmatrix} 2 & -5 \\ 2 & -5 \end{pmatrix} \][/tex]
We need to calculate the matrix product [tex]\( P y \)[/tex].
Step 1: Write down the dimensions of the matrices.
- [tex]\( P \)[/tex] is a 2x2 matrix.
- [tex]\( y \)[/tex] is also a 2x2 matrix.
Step 2: Check the feasibility of the matrix multiplication.
- Since both [tex]\( P \)[/tex] and [tex]\( y \)[/tex] are 2x2 matrices, the multiplication [tex]\( P y \)[/tex] is valid and will result in a 2x2 matrix.
Step 3: Perform the matrix multiplication.
[tex]\[ P y = \begin{pmatrix} 1 & -1 \\ 3 & -3 \end{pmatrix} \begin{pmatrix} 2 & -5 \\ 2 & -5 \end{pmatrix} \][/tex]
To find each element of the resulting matrix, we use the rule of matrix multiplication:
[tex]\[ (P y)_{ij} = \sum_{k=1}^n p_{ik} \cdot y_{kj} \][/tex]
For the element in the first row and first column, [tex]\((P y)_{11}\)[/tex]:
[tex]\[ (P y)_{11} = (1 \cdot 2) + (-1 \cdot 2) = 2 - 2 = 0 \][/tex]
For the element in the first row and second column, [tex]\((P y)_{12}\)[/tex]:
[tex]\[ (P y)_{12} = (1 \cdot -5) + (-1 \cdot -5) = -5 + 5 = 0 \][/tex]
For the element in the second row and first column, [tex]\((P y)_{21}\)[/tex]:
[tex]\[ (P y)_{21} = (3 \cdot 2) + (-3 \cdot 2) = 6 - 6 = 0 \][/tex]
For the element in the second row and second column, [tex]\((P y)_{22}\)[/tex]:
[tex]\[ (P y)_{22} = (3 \cdot -5) + (-3 \cdot -5) = -15 + 15 = 0 \][/tex]
Step 4: Combine all the calculated elements into the resulting matrix:
[tex]\[ P y = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \][/tex]
This confirms that the product of matrices [tex]\( P \)[/tex] and [tex]\( y \)[/tex] is indeed a null matrix.
