Chapter 4: Q29. (page 164)

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

Short Answer

Expert verified

The final matrix after performing the indicated matrix operations is

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

Step by step solution

- Substitute the matrix

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

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

- Perform the scalar multiplication

Scalar multiplication - 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.

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

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

- Perform the matrix addition

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

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

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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=57−163−9, B=835144, C=04−257−1, D=6290−3029. 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

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