QuestionIs the vector $u=\begin{bmatrix}7 \\ 4 \\ 8\end{bmatrix}$ in the plane spanned by the columns of $A=\begin{bmatrix}4 & -2 \\ -2 & 6 \\ 2 & 2\end{bmatrix}$ ? Explain.

Studdy Solution

STEP 1

Assumptions1. The vector $u$ is \left[\begin{array}{l}7 \\4 \\8\end{}\right] . The matrix $A$ is $\left[\begin{array}{rr}4 & - \\ - &6 \\ &\end{array}\right]$
3. We are checking if $u$ lies in the plane spanned by the columns of $A$ in $\mathbb{R}^{3}$
4. If $u$ can be written as a linear combination of the columns of $A$ , then $u$ is in the plane spanned by the columns of $A$

STEP 2

To determine if $u$ is in the plane spanned by the columns of $A$ , we need to check if there exists a solution to the equation $Ax = u$ , where $x$ is a vector of coefficients. If a solution exists, then $u$ can be written as a linear combination of the columns of $A$ .

STEP 3

We can write the equation $Ax = u$ in the form of an augmented matrix.
$\left[\begin{array}{cc|c} & -2 &7 \\ -2 &6 & \\2 &2 &8\end{array}\right]$

STEP 4

We will now perform Gaussian elimination to bring the augmented matrix to row echelon form. Let's start by swapping row1 and row2 to bring the largest leading coefficient to the top.
$\left[\begin{array}{cc|c}-2 &6 &4 \\4 & -2 &7 \\2 &2 &8\end{array}\right]$

STEP 5

Now, we multiply the first row by -1/2 to make the leading coefficient1.
$\left[\begin{array}{cc|c}1 & -3 & -2 \\4 & -2 &7 \\2 &2 &8\end{array}\right]$

STEP 6

Next, we subtract4 times the first row from the second row and2 times the first row from the third row to eliminate the leading coefficients in the second and third rows.
$\left[\begin{array}{cc|c}1 & -3 & -2 \\0 &10 &15 \\0 &8 &12\end{array}\right]$

STEP 7

We divide the second row by10 and the third row by to make the leading coefficients1.
$\left[\begin{array}{cc|c}1 & -3 & -2 \\0 &1 &1.5 \\0 &1 &1.5\end{array}\right]$

STEP 8

Finally, we subtract the third row from the second row to eliminate the leading coefficient in the third row.
$\left[\begin{array}{cc|c}1 & -3 & -2 \\0 &0 &0 \\0 &1 &1.5\end{array}\right]$

STEP 9

The resulting augmented matrix indicates that there is an inconsistent system, meaning there are no solutions to the equation $Ax = u$ . Therefore, the vector $u$ cannot be written as a linear combination of the columns of $A$ .
Hence, the answer is B. No, the reduced echelon form of the augmented matrix is which is an inconsistent system.

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QuestionIs the vector $u=\begin{bmatrix}7 \\ 4 \\ 8\end{bmatrix}$ in the plane spanned by the columns of $A=\begin{bmatrix}4 & -2 \\ -2 & 6 \\ 2 & 2\end{bmatrix}$ ? Explain.

Studdy Solution

STEP 1

STEP 2

To determine if $u$ is in the plane spanned by the columns of $A$ , we need to check if there exists a solution to the equation $Ax = u$ , where $x$ is a vector of coefficients. If a solution exists, then $u$ can be written as a linear combination of the columns of $A$ .

STEP 3

We can write the equation $Ax = u$ in the form of an augmented matrix.
$\left[\begin{array}{cc|c} & -2 &7 \\ -2 &6 & \\2 &2 &8\end{array}\right]$

STEP 4

We will now perform Gaussian elimination to bring the augmented matrix to row echelon form. Let's start by swapping row1 and row2 to bring the largest leading coefficient to the top.
$\left[\begin{array}{cc|c}-2 &6 &4 \\4 & -2 &7 \\2 &2 &8\end{array}\right]$

STEP 5

Now, we multiply the first row by -1/2 to make the leading coefficient1.
$\left[\begin{array}{cc|c}1 & -3 & -2 \\4 & -2 &7 \\2 &2 &8\end{array}\right]$

STEP 6

Next, we subtract4 times the first row from the second row and2 times the first row from the third row to eliminate the leading coefficients in the second and third rows.
$\left[\begin{array}{cc|c}1 & -3 & -2 \\0 &10 &15 \\0 &8 &12\end{array}\right]$

STEP 7

We divide the second row by10 and the third row by to make the leading coefficients1.
$\left[\begin{array}{cc|c}1 & -3 & -2 \\0 &1 &1.5 \\0 &1 &1.5\end{array}\right]$

STEP 8

Finally, we subtract the third row from the second row to eliminate the leading coefficient in the third row.
$\left[\begin{array}{cc|c}1 & -3 & -2 \\0 &0 &0 \\0 &1 &1.5\end{array}\right]$

STEP 9

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QuestionIs the vector u=[748]u=\begin{bmatrix}7 \\ 4 \\ 8\end{bmatrix}u=⎣⎡​748​⎦⎤​ in the plane spanned by the columns of A=[4−2−2622]A=\begin{bmatrix}4 & -2 \\ -2 & 6 \\ 2 & 2\end{bmatrix}A=⎣⎡​4−22​−262​⎦⎤​? Explain.

Studdy Solution

STEP 1

STEP 2

To determine if uuu is in the plane spanned by the columns of AAA, we need to check if there exists a solution to the equation Ax=uAx = uAx=u, where xxx is a vector of coefficients. If a solution exists, then uuu can be written as a linear combination of the columns of AAA.

STEP 3

We can write the equation Ax=uAx = uAx=u in the form of an augmented matrix.[−27−26228]\left[\begin{array}{cc|c} & -2 &7 \\ -2 &6 & \\2 &2 &8\end{array}\right]⎣⎡​−22​−262​78​⎦⎤​

STEP 4

STEP 5

Now, we multiply the first row by -1/2 to make the leading coefficient1.[1−3−24−27228]\left[\begin{array}{cc|c}1 & -3 & -2 \\4 & -2 &7 \\2 &2 &8\end{array}\right]⎣⎡​142​−3−22​−278​⎦⎤​

STEP 6

STEP 7

We divide the second row by10 and the third row by to make the leading coefficients1.[1−3−2011.5011.5]\left[\begin{array}{cc|c}1 & -3 & -2 \\0 &1 &1.5 \\0 &1 &1.5\end{array}\right]⎣⎡​100​−311​−21.51.5​⎦⎤​

STEP 8

Finally, we subtract the third row from the second row to eliminate the leading coefficient in the third row.[1−3−2000011.5]\left[\begin{array}{cc|c}1 & -3 & -2 \\0 &0 &0 \\0 &1 &1.5\end{array}\right]⎣⎡​100​−301​−201.5​⎦⎤​

STEP 9

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QuestionIs the vector $u=\begin{bmatrix}7 \\ 4 \\ 8\end{bmatrix}$ in the plane spanned by the columns of $A=\begin{bmatrix}4 & -2 \\ -2 & 6 \\ 2 & 2\end{bmatrix}$ ? Explain.

To determine if $u$ is in the plane spanned by the columns of $A$ , we need to check if there exists a solution to the equation $Ax = u$ , where $x$ is a vector of coefficients. If a solution exists, then $u$ can be written as a linear combination of the columns of $A$ .

We can write the equation $Ax = u$ in the form of an augmented matrix.
$\left[\begin{array}{cc|c} & -2 &7 \\ -2 &6 & \\2 &2 &8\end{array}\right]$

Now, we multiply the first row by -1/2 to make the leading coefficient1.
$\left[\begin{array}{cc|c}1 & -3 & -2 \\4 & -2 &7 \\2 &2 &8\end{array}\right]$

We divide the second row by10 and the third row by to make the leading coefficients1.
$\left[\begin{array}{cc|c}1 & -3 & -2 \\0 &1 &1.5 \\0 &1 &1.5\end{array}\right]$

Finally, we subtract the third row from the second row to eliminate the leading coefficient in the third row.
$\left[\begin{array}{cc|c}1 & -3 & -2 \\0 &0 &0 \\0 &1 &1.5\end{array}\right]$