Chapter 4: Q64. (page 188)

Find the product, if possible.
$(\begin{matrix} 7 & - 5 & 4 \\ 6 & 1 & 3 \end{matrix}) (\begin{matrix} - 1 & 3 \\ - 2 & 8 \\ 1 & 2 \end{matrix})$

Short Answer

Expert verified

The product is $(\begin{matrix} 7 & - 11 \\ - 5 & 32 \end{matrix})$ .

Step by step solution

Step 1- Apply the concept of multiplication of matrix.

Matrix multiplication is a binary operation that produces a matrix from two matrices. For matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix.

Step 2- Step description.

Consider the matrix $(\begin{matrix} 7 & - 5 & 4 \\ 6 & 1 & 3 \end{matrix}) (\begin{matrix} - 1 & 3 \\ - 2 & 8 \\ 1 & 2 \end{matrix})$ .

Multiply the matrix as follows:

$\begin{array}{l} = (\begin{matrix} \begin{matrix} 7 (- 1) + (- 5) (- 2) + 4 \cdot 1 & 7 \cdot 3 + (- 5) \cdot 8 + 4 \cdot 2 \\ 6 (- 1) + 1 \cdot (- 2) + 3 \cdot 1 & 6 \cdot 3 + 1 \cdot 8 + 3 \cdot 2 \end{matrix} \end{matrix}) \\ = (\begin{matrix} \begin{matrix} 7 & - 11 \\ - 5 & 32 \end{matrix} \end{matrix}) \end{array}$

Step 3- Step description.

Therefore, the product of the matrix is $(\begin{matrix} 7 & - 11 \\ - 5 & 32 \end{matrix})$ .

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Find the product, if possible.
$(\begin{matrix} 7 & - 5 & 4 \\ 6 & 1 & 3 \end{matrix}) (\begin{matrix} - 1 & 3 \\ - 2 & 8 \\ 1 & 2 \end{matrix})$

Short Answer

Step by step solution

Step 1- Apply the concept of multiplication of matrix.

Step 2- Step description.

Step 3- Step description.

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Find the product, if possible.7-54 613-13 -28 12

Short Answer

Step by step solution

Step 1- Apply the concept of multiplication of matrix.

Step 2- Step description.

Step 3- Step description.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Logic and Functions

Applied Mathematics

Probability and Statistics

Discrete Mathematics

Geometry

Study anywhere. Anytime. Across all devices.

Find the product, if possible.
$(\begin{matrix} 7 & - 5 & 4 \\ 6 & 1 & 3 \end{matrix}) (\begin{matrix} - 1 & 3 \\ - 2 & 8 \\ 1 & 2 \end{matrix})$