Answer
See the explanation
Work Step by Step
Let's compute each product separately:
1. **Product of Matrices (1)**:
\[ \begin{bmatrix} 4 & 0 & 1 \\ 0 & 1 & 0 \\ 4 & 0 & 1 \end{bmatrix} \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix} \]
To compute this product, we take the dot product of each row of the first matrix with the column vector.
For the first row:
\[ \begin{bmatrix} 4 & 0 & 1 \end{bmatrix} \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix} = (4 \times 3) + (0 \times 4) + (1 \times 5) = 12 + 0 + 5 = 17 \]
For the second row:
\[ \begin{bmatrix} 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix} = (0 \times 3) + (1 \times 4) + (0 \times 5) = 0 + 4 + 0 = 4 \]
For the third row:
\[ \begin{bmatrix} 4 & 0 & 1 \end{bmatrix} \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix} = (4 \times 3) + (0 \times 4) + (1 \times 5) = 12 + 0 + 5 = 17 \]
So, the resulting vector is:
\[ \begin{bmatrix} 17 \\ 4 \\ 17 \end{bmatrix} \]
2. **Product of Matrices (2)**:
\[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 5 \\ -2 \\ 3 \end{bmatrix} \]
Since the first matrix is the identity matrix, the resulting vector is just the same as the given column vector:
\[ \begin{bmatrix} 5 \\ -2 \\ 3 \end{bmatrix} \]
3. **Product of Column Vectors**:
\[ \begin{bmatrix} 2 \\ 1 \end{bmatrix} \begin{bmatrix} 0 \\ 3 \end{bmatrix} \]
Graphically, when we multiply (2,1) by (0,3), it means scaling the vector (2,1) by the scalar 0 and 3, respectively. When you scale by 0, you end up with the zero vector (0,0). When you scale by 3, you triple the length of the vector in the direction of (0,3), ending up with (0,9).
4. **Product of Matrices (3)**:
\[ \begin{bmatrix} 2 & 0 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]
\[ \begin{bmatrix} 2 & 0 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \cdot 1 + 0 \cdot 1 \\ 1 \cdot 1 + 3 \cdot 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \end{bmatrix} \]
So, the resulting vector is:
\[ \begin{bmatrix} 2 \\ 4 \end{bmatrix} \]
