Chapter 4: Q22. (page 164)

Perform the indicated matrix operation. If the matrix doesn’t exist, write impossible.
22. $\frac{1}{2} [\begin{matrix} 4 & 6 \\ 3 & 0 \end{matrix}] - \frac{2}{3} [\begin{matrix} 9 & 27 \\ 0 & 3 \end{matrix}]$

Short Answer

Expert verified

The final matrix after performing the indicated matrix operations is

$\frac{1}{2} [\begin{matrix} 4 & 6 \\ 3 & 0 \end{matrix}] - \frac{2}{3} [\begin{matrix} 9 & 27 \\ 0 & 3 \end{matrix}] = [\begin{matrix} - 4 & - 15 \\ \frac{3}{2} & - 2 \end{matrix}]$ .

Step by step solution

- Define the concept used

Matrix subtraction and scalar multiplication:

If A and B are two matrices of order $m \times n$ , then the subtraction of the two matrices will also yield a matrix of order $m \times n$ wherein each element is the difference of the corresponding elements.

The product of a scalar k and an $m \times n$ matrix is an $m \times n$ matrix in which each element equals k times the corresponding elements of the original matrix.

- Perform the scalar multiplication first

Multiply each element in the first matrix $[\begin{matrix} 4 & 6 \\ 3 & 0 \end{matrix}]$ by $\frac{1}{2}$ and multiply each element in the second matrix $[\begin{matrix} 9 & 27 \\ 0 & 3 \end{matrix}]$ by $\frac{2}{3}$ .

$\begin{matrix} \frac{1}{2} [\begin{matrix} 4 & 6 \\ 3 & 0 \end{matrix}] - \frac{2}{3} [\begin{matrix} 9 & 27 \\ 0 & 3 \end{matrix}] = [\begin{matrix} \frac{1}{2} (4) & \frac{1}{2} (6) \\ \frac{1}{2} (3) & \frac{1}{2} (0) \end{matrix}] - [\begin{matrix} \frac{2}{3} (9) & \frac{2}{3} (27) \\ \frac{2}{3} (0) & \frac{2}{3} (3) \end{matrix}] \\ = [\begin{matrix} 2 & 3 \\ \frac{3}{2} & 0 \end{matrix}] - [\begin{matrix} 6 & 18 \\ 0 & 2 \end{matrix}] \end{matrix}$

- Subtract corresponding elements

Subtract corresponding elements of both the matrices $[\begin{matrix} 2 & 3 \\ \frac{3}{2} & 0 \end{matrix}] - [\begin{matrix} 6 & 18 \\ 0 & 2 \end{matrix}]$ and simplify.

$\begin{matrix} \frac{1}{2} [\begin{matrix} 4 & 6 \\ 3 & 0 \end{matrix}] - \frac{2}{3} [\begin{matrix} 9 & 27 \\ 0 & 3 \end{matrix}] = [\begin{matrix} 2 & 3 \\ \frac{3}{2} & 0 \end{matrix}] - [\begin{matrix} 6 & 18 \\ 0 & 2 \end{matrix}] \\ = [\begin{matrix} 2 - 6 & 3 - 18 \\ \frac{3}{2} - 0 & 0 - 2 \end{matrix}] \\ = [\begin{matrix} - 4 & - 15 \\ \frac{3}{2} & - 2 \end{matrix}] \end{matrix}$

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Perform the indicated matrix operation. If the matrix doesn’t exist, write impossible.
22. $\frac{1}{2} [\begin{matrix} 4 & 6 \\ 3 & 0 \end{matrix}] - \frac{2}{3} [\begin{matrix} 9 & 27 \\ 0 & 3 \end{matrix}]$

Short Answer

Step by step solution

- Define the concept used

- Perform the scalar multiplication first

- Subtract corresponding elements

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Perform the indicated matrix operation. If the matrix doesn’t exist, write impossible.22.124630−2392703

Short Answer

Step by step solution

- Define the concept used

- Perform the scalar multiplication first

- Subtract corresponding elements

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Probability and Statistics

Logic and Functions

Discrete Mathematics

Applied Mathematics

Statistics

Study anywhere. Anytime. Across all devices.

Perform the indicated matrix operation. If the matrix doesn’t exist, write impossible.
22. $\frac{1}{2} [\begin{matrix} 4 & 6 \\ 3 & 0 \end{matrix}] - \frac{2}{3} [\begin{matrix} 9 & 27 \\ 0 & 3 \end{matrix}]$