Question

The value of \( \begin{pmatrix} 7 & 1 & 5 \\ 8 & 0 & 0 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} + 5 \begin{pmatrix} 1 \\ 0 \end{pmatrix} \) is equal to:

• A. \( \begin{pmatrix} 16 \\ 27 \end{pmatrix} \)
• B. \( \begin{pmatrix} 27 \\ 16 \end{pmatrix} \)
• C. \( \begin{pmatrix} 15 \\ 16 \end{pmatrix} \)
• D. \( \begin{pmatrix} 16 \\ 15 \end{pmatrix} \)
• E. \( \begin{pmatrix} 16 \\ 16 \end{pmatrix} \)

Hint:

Always double-check the final vector addition after a long multiplication. Zeros in the first matrix (like the second row here) act as a "filter," allowing you to skip parts of the multiplication for that specific row.

Answer 1

The Correct Option is B

Concept: Matrix multiplication involves the dot product of rows from the first matrix and columns from the second. For the resulting vector, each entry is the sum of the products of corresponding elements. Scalar multiplication requires multiplying every element in the matrix by the given constant.

Step 1: Performing the initial matrix-vector multiplication.
We multiply the \( 2 \times 3 \) matrix by the \( 3 \times 1 \) vector: \[ \text{Product}_1 = \begin{pmatrix} (7 \times 2) + (1 \times 3) + (5 \times 1) \\ (8 \times 2) + (0 \times 3) + (0 \times 1) \end{pmatrix} \] Simplifying the arithmetic for each row: \[ \text{Row 1:} \quad 14 + 3 + 5 = 22 \] \[ \text{Row 2:} \quad 16 + 0 + 0 = 16 \] So, the first part of the expression results in the vector \( \begin{pmatrix} 22 16 \end{pmatrix} \).

Step 2: Applying scalar multiplication and vector addition.
Next, calculate the scalar product for the second term: \[ 5 \begin{pmatrix} 1 0 \end{pmatrix} = \begin{pmatrix} 5 \times 1 \\ 5 \times 0 \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \end{pmatrix} \] Finally, add the two vectors together element-wise: \[ \text{Result} = \begin{pmatrix} 22 \\ 16 \end{pmatrix} + \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} 22 + 5 16 + 0 \end{pmatrix} = \begin{pmatrix} 27 \\ 16 \end{pmatrix} \]