Math / Algebra

QuestionLet $A=\left[\begin{array}{ccc} -2 & 2 & 0 \\ 2 & -1 & 1 \\ 3 & -2 & 1 \end{array}\right]$ a. A basis for the row space of $A$ is $\{\square\}$ . You should be able to explain and justify your answer. Enter a coordinate vector, such as $\langle 1,2,3>$ , or a comma separated list of coordinate vectors, such as $\langle 1,2,3\rangle,\langle 4,5,6\rangle$ . b. The dimension of the row space of $A$ is $\square$ because (select all correct answers -- there may be more than one correct answer): A. $\operatorname{rref}(A)$ is the identity matrix. B. Two of the three rows in $\operatorname{rref}(A)$ have pivots. C. The basis we found for the row space of $A$ has two vectors. D. Two of the three columns in $\operatorname{rref}(A)$ are free variable columns. E. $\operatorname{rref}(A)$ has a pivot in every row. F. Two of the threê rows in $\operatorname{rref}(A)$ do not have a pivot. c. The row space of $A$ is a subspace of $\square$ because choose $\square$ d. The geometry of the row space of $A$ is choose $\square$

Studdy Solution

STEP 1

What is this asking? We need to find a basis for the row space of matrix $A$ , figure out its dimension, what larger space it lives in, and what it looks like geometrically. Watch out! Don't mix up row operations with column operations!
Also, remember the basis vectors come from the original matrix, not the reduced one.

STEP 2

1. Row reduce the matrix
2. Identify the pivot rows
3. Determine the basis and dimension
4. Describe the subspace and geometry

STEP 3

We're starting with matrix $A$ : $A = \begin{bmatrix} -2 & 2 & 0 \\ 2 & -1 & 1 \\ 3 & -2 & 1 \end{bmatrix}$ Our goal here is to get this matrix into row-echelon form, which will make it easier to find a basis for the row space.

STEP 4

Let's add the first row to the second row to eliminate the 2 in the second row, first column.
This is a valid row operation, and it helps us move towards row-echelon form. $\begin{bmatrix} -2 & 2 & 0 \\ 0 & 1 & 1 \\ 3 & -2 & 1 \end{bmatrix}$

STEP 5

Now, let's multiply the first row by $\frac{3}{2}$ and add it to the third row to eliminate the 3 in the third row, first column.
We're doing this to continue our pursuit of row-echelon form. $\begin{matrix} -2 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{matrix}$

STEP 6

Finally, let's subtract the second row from the third row to eliminate the 1 in the third row, second column.
This gets us to row-echelon form! $\begin{pmatrix} -2 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}$

STEP 7

The pivot rows are the rows with the leading non-zero entries.
In our row-reduced matrix, the first and second rows have pivots (the $-2$ in the first row and the $1$ in the second row).
The third row is all zeros, so it doesn't have a pivot.

STEP 8

The basis for the row space is formed by the corresponding rows from the original matrix $A$ .
So, our basis is $\langle -2, 2, 0 \rangle, \langle 2, -1, 1 \rangle$ .
Remember, we use the rows from the original matrix $A$ , not the row-reduced form!

STEP 9

The dimension of the row space is the number of vectors in the basis.
Since we have two basis vectors, the dimension is 2.

STEP 10

The row space of $A$ is a subspace of $\mathbb{R}^3$ , because each row of $A$ has three entries.

STEP 11

Since the row space is a two-dimensional subspace of $\mathbb{R}^3$ , it represents a plane in three-dimensional space.

STEP 12

a. A basis for the row space of $A$ is $\langle -2, 2, 0 \rangle, \langle 2, -1, 1 \rangle$ . b. The dimension of the row space of $A$ is 2 because B and C are correct. c. The row space of $A$ is a subspace of $\mathbb{R}^3$ . d. The geometry of the row space of $A$ is a plane.

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Row reduce the matrix
2. Identify the pivot rows
3. Determine the basis and dimension
4. Describe the subspace and geometry

STEP 3

We're starting with matrix $A$ : $A = \begin{bmatrix} -2 & 2 & 0 \\ 2 & -1 & 1 \\ 3 & -2 & 1 \end{bmatrix}$ Our goal here is to get this matrix into row-echelon form, which will make it easier to find a basis for the row space.

STEP 4

Let's add the first row to the second row to eliminate the 2 in the second row, first column.
This is a valid row operation, and it helps us move towards row-echelon form. $\begin{bmatrix} -2 & 2 & 0 \\ 0 & 1 & 1 \\ 3 & -2 & 1 \end{bmatrix}$

STEP 5

Now, let's multiply the first row by $\frac{3}{2}$ and add it to the third row to eliminate the 3 in the third row, first column.
We're doing this to continue our pursuit of row-echelon form. $\begin{matrix} -2 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{matrix}$

STEP 6

Finally, let's subtract the second row from the third row to eliminate the 1 in the third row, second column.
This gets us to row-echelon form! $\begin{pmatrix} -2 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}$

STEP 7

The pivot rows are the rows with the leading non-zero entries.
In our row-reduced matrix, the first and second rows have pivots (the $-2$ in the first row and the $1$ in the second row).
The third row is all zeros, so it doesn't have a pivot.

STEP 8

The basis for the row space is formed by the corresponding rows from the original matrix $A$ .
So, our basis is $\langle -2, 2, 0 \rangle, \langle 2, -1, 1 \rangle$ .
Remember, we use the rows from the original matrix $A$ , not the row-reduced form!

STEP 9

The dimension of the row space is the number of vectors in the basis.
Since we have two basis vectors, the dimension is 2.

STEP 10

The row space of $A$ is a subspace of $\mathbb{R}^3$ , because each row of $A$ has three entries.

STEP 11

Since the row space is a two-dimensional subspace of $\mathbb{R}^3$ , it represents a plane in three-dimensional space.

STEP 12

a. A basis for the row space of $A$ is $\langle -2, 2, 0 \rangle, \langle 2, -1, 1 \rangle$ . b. The dimension of the row space of $A$ is 2 because B and C are correct. c. The row space of $A$ is a subspace of $\mathbb{R}^3$ . d. The geometry of the row space of $A$ is a plane.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

1. Row reduce the matrix2. Identify the pivot rows3. Determine the basis and dimension4. Describe the subspace and geometry

STEP 3

We're starting with matrix AAA: A=[−2202−113−21] A = \begin{bmatrix} -2 & 2 & 0 \\ 2 & -1 & 1 \\ 3 & -2 & 1 \end{bmatrix} A=⎣⎡​−223​2−1−2​011​⎦⎤​ Our goal here is to get this matrix into row-echelon form, which will make it easier to find a basis for the row space.

STEP 4

STEP 5

Now, let's multiply the first row by 32 \frac{3}{2} 23​ and add it to the third row to eliminate the **3** in the third row, first column.We're doing this to continue our pursuit of row-echelon form. −220011011\begin{matrix} -2 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{matrix}−200​211​011​

STEP 6

Finally, let's subtract the second row from the third row to eliminate the **1** in the third row, second column.This gets us to row-echelon form! (−220011000) \begin{pmatrix} -2 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix} ⎝⎛​−200​210​010​⎠⎞​

STEP 7

The **pivot rows** are the rows with the leading non-zero entries.In our row-reduced matrix, the first and second rows have pivots (the −2-2−2 in the first row and the 111 in the second row).The third row is all zeros, so it doesn't have a pivot.

STEP 8

STEP 9

The dimension of the row space is the number of vectors in the basis.Since we have **two** basis vectors, the dimension is **2**.

STEP 10

The row space of AAA is a subspace of R3\mathbb{R}^3R3, because each row of AAA has **three** entries.

STEP 11

Since the row space is a two-dimensional subspace of R3\mathbb{R}^3R3, it represents a *plane* in three-dimensional space.

STEP 12

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Row reduce the matrix
2. Identify the pivot rows
3. Determine the basis and dimension
4. Describe the subspace and geometry

We're starting with matrix $A$ : $A = \begin{bmatrix} -2 & 2 & 0 \\ 2 & -1 & 1 \\ 3 & -2 & 1 \end{bmatrix}$ Our goal here is to get this matrix into row-echelon form, which will make it easier to find a basis for the row space.

Now, let's multiply the first row by $\frac{3}{2}$ and add it to the third row to eliminate the 3 in the third row, first column.
We're doing this to continue our pursuit of row-echelon form. $\begin{matrix} -2 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{matrix}$

Finally, let's subtract the second row from the third row to eliminate the 1 in the third row, second column.
This gets us to row-echelon form! $\begin{pmatrix} -2 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}$

The pivot rows are the rows with the leading non-zero entries.
In our row-reduced matrix, the first and second rows have pivots (the $-2$ in the first row and the $1$ in the second row).
The third row is all zeros, so it doesn't have a pivot.

The dimension of the row space is the number of vectors in the basis.
Since we have two basis vectors, the dimension is 2.

The row space of $A$ is a subspace of $\mathbb{R}^3$ , because each row of $A$ has three entries.

Since the row space is a two-dimensional subspace of $\mathbb{R}^3$ , it represents a plane in three-dimensional space.