QuestionDo the columns of $A$ span $\mathbb{R}^{4}$ ? Does $A \mathbf{x}=\mathbf{b}$ have a solution for all $\mathbf{b}$ ?
$A=\begin{bmatrix} 1 & 3 & 5 & -9 \\ 2 & 0 & -8 & 6 \\ 3 & 4 & 0 & -7 \\ -2 & -7 & -13 & 18 \end{bmatrix}$

Studdy Solution

STEP 1

Assumptions1. Matrix $A$ is a4x4 matrix with real number entries. . We are asked to determine if the columns of $A$ span $\mathbb{R}^{4}$ .
3. We are also asked to verify if the equation $A\mathbf{x}=\mathbf{b}$ has a solution for all $\mathbf{b}$ in $\mathbb{R}^{4}$ .

STEP 2

To determine if the columns of $A$ span $\mathbb{R}^{4}$ , we need to check if the rank of the matrix is4. This is equivalent to checking if the matrix is full rank. We can do this by reducing the matrix to its row echelon form (REF) or reduced row echelon form (RREF) and checking if there are any rows of zeros.

STEP 3

Let's start by reducing the matrix $A$ to its row echelon form (REF). This can be achieved by performing a series of row operations. The goal is to create a leading1 (also known as a pivot) in each row that is to the right of the pivot in the row above it, and to make all other entries in the column containing the pivot zero.

STEP 4

We can start by swapping the first and second rows to get a leading1 in the first row.
$A=\left[\begin{array}{rrrr} 2 &0 & -8 &6 \\ 1 &3 & & -9 \\ 3 &4 &0 & -7 \\ -2 & -7 & -13 &18\end{array}\right]$

STEP 5

Next, we can multiply the first row by1/2 to get a leading1.
$A=\left[\begin{array}{rrrr} 1 &0 & -4 &3 \\ 1 &3 &5 & -9 \\ 3 &4 &0 & -7 \\ -2 & -7 & -13 &18\end{array}\right]$

STEP 6

We can then subtract the first row from the second and third rows to get zeros below the leading1 in the first row.
$A=\left[\begin{array}{rrrr} 1 &0 & -4 &3 \\ 0 &3 &9 & -12 \\ 2 &4 &4 & -10 \\ -2 & - & -13 &18\end{array}\right]$

STEP 7

We can multiply the second row by1/3 to get a leading1 in the second row.
$A=\left[\begin{array}{rrrr} 1 &0 & -4 &3 \\ 0 &1 &3 & -4 \\ 2 &4 &4 & -10 \\ -2 & -7 & -13 &18\end{array}\right]$

STEP 8

We can then subtract4 times the second row from the third row and add7 times the second row to the fourth row to get zeros below the leading1 in the second row.
$A=\left[\begin{array}{rrrr} 1 &0 & -4 &3 \\ 0 &1 &3 & -4 \\ 2 &0 & -8 & -6 \\ 0 &0 &6 & -10\end{array}\right]$

STEP 9

We can then subtract2 times the first row from the third row to get a zero below the leading in the first row.
$A=\left[\begin{array}{rrrr} & & -4 &3 \\ & &3 & -4 \\ & & & \\ & &6 & -\end{array}\right]$

STEP 10

We can then divide the fourth row by6 to get a leading in the fourth row.
$A=\left[\begin{array}{rrrr} &0 & -4 &3 \\ 0 & &3 & -4 \\ 0 &0 &0 &0 \\ 0 &0 & & -5/3\end{array}\right]$

STEP 11

Finally, we can add4 times the fourth row to the first row and subtract3 times the fourth row from the second row to get zeros above the leading in the fourth row.
$A=\left[\begin{array}{rrrr} &0 &0 &/3 \\ 0 & &0 &/3 \\ 0 &0 &0 &0 \\ 0 &0 & & -5/3\end{array}\right]$

STEP 12

The matrix is now in its reduced row echelon form (RREF). We can see that there is a row of zeros, which means that the rank of the matrix is less than4. Therefore, the columns of $A$ do not span $\mathbb{R}^{4}$ .

STEP 13

Since the columns of $A$ do not span $\mathbb{R}^{}$ , the equation $A\mathbf{x}=\mathbf{b}$ does not have a solution for each $\mathbf{b}$ in $\mathbb{R}^{}$ . This is because there are vectors in $\mathbb{R}^{}$ that cannot be expressed as a linear combination of the columns of $A$ .
The correct choice is A. No, because the reduced echelon form of $A$ is $\left[\begin{array}{rrrr} &0 &0 &/3 \\ 0 & &0 &/3 \\ 0 &0 &0 &0 \\ 0 &0 & & -5/3\end{array}\right]$

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