21. Determine the rank of the following matrices:3 -1 4 -20 26 -1 10 -142-1 3(a)-1 212 -4 - 5(b)4|Ans. (a) 2, (b) 2.

Answered: Determine the rank of the following…

College Algebra

7th Edition

ISBN:9781305115545

Author:James Stewart, Lothar Redlin, Saleem Watson

Publisher:James Stewart, Lothar Redlin, Saleem Watson

Chapter6: Matrices And Determinants

Section: Chapter Questions

Problem 7CC: If A and B are matrices with the same dimension and k is a real number, how do you find A+B and kA ?

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Question

21. Determine the rank of the following matrices:
3 -1 4 -2
0 2
6 -1 10 -1
4
2
-1 3
(a)
-1 2
12 -4 - 5
(b)
4
|
Ans. (a) 2, (b) 2.

Expert Solution

Step 1

The number of non-zero rows in the row echelon form of a matrix gives the rank of that matrix.

The given matrix is $(\begin{matrix} 4 & 2 & - 1 & 3 \\ 0 & 5 & - 1 & 2 \\ 12 & - 4 & - 1 & 5 \end{matrix})$ .

Find the row echelon form of $(\begin{matrix} 4 & 2 & - 1 & 3 \\ 0 & 5 & - 1 & 2 \\ 12 & - 4 & - 1 & 5 \end{matrix})$ as shown below.

Step 2

$\begin{array}{rcl} (\begin{matrix} 4 & 2 & - 1 & 3 \\ 0 & 5 & - 1 & 2 \\ 12 & - 4 & - 1 & 5 \end{matrix}) ~ (\begin{matrix} 1 & \frac{1}{2} & - \frac{1}{4} & \frac{3}{4} \\ 0 & 5 & - 1 & 2 \\ 12 & - 4 & - 1 & 5 \end{matrix}) (R_{1} \to \frac{1}{4} R_{1}) \\ ~ (\begin{matrix} 1 & \frac{1}{2} & - \frac{1}{4} & \frac{3}{4} \\ 0 & 5 & - 1 & 2 \\ 0 & - 10 & 2 & - 4 \end{matrix}) (R_{3} \to R_{3} - 12 R_{1}) \\ ~ (\begin{matrix} 1 & \frac{1}{2} & - \frac{1}{4} & \frac{3}{4} \\ 0 & 1 & - \frac{1}{5} & \frac{2}{5} \\ 0 & - 10 & 2 & - 4 \end{matrix}) (R_{2} \to \frac{1}{5} R_{2}) \\ ~ (\begin{matrix} 1 & \frac{1}{2} & - \frac{1}{4} & \frac{3}{4} \\ 0 & 1 & - \frac{1}{5} & \frac{2}{5} \\ 0 & 0 & 0 & 0 \end{matrix}) (R_{3} \to 10 R_{2} + R_{3}) \end{array}$

Step 3

Thus, the row echelon form of $(\begin{matrix} 4 & 2 & - 1 & 3 \\ 0 & 5 & - 1 & 2 \\ 12 & - 4 & - 1 & 5 \end{matrix})$ is $\begin{array}{rcl} (\begin{matrix} 1 & \frac{1}{2} & - \frac{1}{4} & \frac{3}{4} \\ 0 & 1 & - \frac{1}{5} & \frac{2}{5} \\ 0 & 0 & 0 & 0 \end{matrix}) \end{array}$ .

The number of non-zero rows in row echelon form $\begin{array}{rcl} (\begin{matrix} 1 & \frac{1}{2} & - \frac{1}{4} & \frac{3}{4} \\ 0 & 1 & - \frac{1}{5} & \frac{2}{5} \\ 0 & 0 & 0 & 0 \end{matrix}) \end{array}$ is 2.

Hence, the rank of the matrix $(\begin{matrix} 4 & 2 & - 1 & 3 \\ 0 & 5 & - 1 & 2 \\ 12 & - 4 & - 1 & 5 \end{matrix})$ is $2$ .

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