Q29. Use matrices聽A, B, C聽and Dto f... [FREE SOLUTION]

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Chapter 4: Q29. (page 164)

Use matrices $A$ , $B$ , $C$ and $D$ to find the following.
$A = [\begin{matrix} 5 & 7 \\ 鈭� 1 & 6 \\ 3 & 鈭� 9 \end{matrix}], 鈥夆赌� B = [\begin{matrix} 8 & 3 \\ 5 & 1 \\ 4 & 4 \end{matrix}], 鈥夆赌� C = [\begin{matrix} 0 & 4 \\ 鈭� 2 & 5 \\ 7 & 鈭� 1 \end{matrix}], 鈥夆赌� D = [\begin{matrix} 6 & 2 \\ 9 & 0 \\ 鈭� 3 & 0 \end{matrix}]$
29. $C + \frac{1}{3} D$

Short Answer

Expert verified

The final matrix after performing the indicated matrix operations is

$C + \frac{1}{3} D = [\begin{matrix} 2 & \frac{14}{3} \\ 1 & 5 \\ 6 & 鈭� 1 \end{matrix}]$ .

Step by step solution

- Substitute the matrix

Substitute $[\begin{matrix} 0 & 4 \\ 鈭� 2 & 5 \\ 7 & 鈭� 1 \end{matrix}]$ for C and $[\begin{matrix} 6 & 2 \\ 9 & 0 \\ 鈭� 3 & 0 \end{matrix}]$ for D into $C + \frac{1}{3} D$ .

$C + \frac{1}{3} D = [\begin{matrix} 0 & 4 \\ 鈭� 2 & 5 \\ 7 & 鈭� 1 \end{matrix}] + \frac{1}{3} [\begin{matrix} 6 & 2 \\ 9 & 0 \\ 鈭� 3 & 0 \end{matrix}]$

- Perform the scalar multiplication

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

Multiply each element of the second matrix $[\begin{matrix} 6 & 2 \\ 9 & 0 \\ 鈭� 3 & 0 \end{matrix}]$ by $\frac{1}{3}$ .

$\begin{matrix} C + \frac{1}{3} D = [\begin{matrix} 0 & 4 \\ 鈭� 2 & 5 \\ 7 & 鈭� 1 \end{matrix}] + \frac{1}{3} [\begin{matrix} 6 & 2 \\ 9 & 0 \\ 鈭� 3 & 0 \end{matrix}] \\ = [\begin{matrix} 0 & 4 \\ 鈭� 2 & 5 \\ 7 & 鈭� 1 \end{matrix}] + [\begin{matrix} \frac{1}{3} (6) & \frac{1}{3} (2) \\ \frac{1}{3} (9) & \frac{1}{3} (0) \\ \frac{1}{3} (鈭� 3) & \frac{1}{3} (0) \end{matrix}] \\ = [\begin{matrix} 0 & 4 \\ 鈭� 2 & 5 \\ 7 & 鈭� 1 \end{matrix}] + [\begin{matrix} 2 & \frac{2}{3} \\ 3 & 0 \\ 鈭� 1 & 0 \end{matrix}] \end{matrix}$

- Perform the matrix addition

Matrix addition - If A and B are two matrices of order $m 脳 n$ , then the addition of the two matrices will also yield a matrix of order $m 脳 n$ wherein each element is the sum of the corresponding elements.

$\begin{matrix} C + \frac{1}{3} D = [\begin{matrix} 0 & 4 \\ 鈭� 2 & 5 \\ 7 & 鈭� 1 \end{matrix}] + [\begin{matrix} 2 & \frac{2}{3} \\ 3 & 0 \\ 鈭� 1 & 0 \end{matrix}] \\ = [\begin{matrix} 0 + 2 & 4 + \frac{2}{3} \\ 鈭� 2 + 3 & 5 + 0 \\ 7 鈭� 1 & 鈭� 1 + 0 \end{matrix}] \\ = [\begin{matrix} 2 & \frac{14}{3} \\ 1 & 5 \\ 6 & 鈭� 1 \end{matrix}] \end{matrix}$

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Short Answer

Step by step solution

- Substitute the matrix

- Perform the scalar multiplication

- Perform the matrix addition

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Use matricesA, B, Cand Dto find the following.A=57鈭�163鈭�9,鈥夆赌�B=835144,鈥夆赌�C=04鈭�257鈭�1,鈥夆赌�D=6290鈭�3029. C+13D

Short Answer

Step by step solution

- Substitute the matrix

- Perform the scalar multiplication

- Perform the matrix addition

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Discrete Mathematics

Geometry

Probability and Statistics

Pure Maths

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.