Solved on Oct 29, 2023

Find the basis and dimension of the subspace $V \subset \mathbb{R}^{4}$ given by the homogeneous system $x_1 + 2x_2 + x_3 + 2x_4 = 0, 3x_1 + 5x_2 + 4x_3 + 7x_4 = 0$ . Then, find the homogeneous system of linear equations that define the subspace $W \subset \mathbb{R}^{4}$ spanned by the basis of $V$ and the vector $w = (1, 0, 0, 0)$ .

STEP 1

Assumptions1. The subspace $V$ is given by the homogeneous system of linear equations. The system of linear equations is $\left\{\begin{array}{c} x_{1}+ x_{}+x_{3}+ x_{4}=0 \\ 3 x_{1}+5 x_{}+4 x_{3}+7 x_{4}=0\end{array}\right.$ 3. We need to find a basis $\mathcal{A}$ and the dimension of the subspace $V$
4. We also need to find a homogeneous system of linear equations for the subspace $W$ spanned by the basis $\mathcal{A}$ and the vector $w=(1,0,0,0)$

STEP 2

First, we need to solve the given system of linear equations. We can do this by using the Gaussian elimination method. The augmented matrix for the system of equations is $\left[\begin{array}{cccc|c} 1 &2 &1 &2 &0 \\ &5 &4 &7 &0\end{array}\right]$

STEP 3

We can subtract3 times the first row from the second row to simplify the matrix. This gives us $\left[\begin{array}{cccc|c} 1 &2 &1 &2 &0 \\ 0 & -1 &1 &1 &0\end{array}\right]$

STEP 4

Next, we can multiply the second row by -1 to get $\left[\begin{array}{cccc|c} 1 &2 &1 &2 &0 \\ 0 &1 & -1 & -1 &0\end{array}\right]$

STEP 5

We can subtract2 times the second row from the first row to get $\left[\begin{array}{cccc|c} 1 &0 &3 &4 &0 \\ 0 &1 & -1 & -1 &0\end{array}\right]$

STEP 6

The system of equations corresponding to this matrix is $\left\{\begin{array}{c} x_{1}+3 x_{3}+4 x_{4}=0 \\ x_{2}-x_{3}-x_{4}=0\end{array}\right.$ We can express $x_{3}$ and $x_{4}$ in terms of $x_{1}$ and $x_{2}$ $\left\{\begin{array}{c} x_{3}=-\frac{1}{3}x_{1}+\frac{1}{3}x_{2} \\ x_{4}=-\frac{4}{3}x_{1}+\frac{1}{3}x_{2} \end{array}\right.$

STEP 7

The general solution to the system is given by $\left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}\right]=x_{1}\left[\begin{array}{c} 1 \\ 0 \\ -\frac{1}{3} \\ -\frac{4}{3} \end{array}\right]+x_{2}\left[\begin{array}{c} 0 \\ 1 \\ \frac{1}{3} \\ \frac{1}{3} \end{array}\right]$ The vectors $\left[\begin{array}{c} 1 \\ 0 \\ -\frac{1}{3} \\ -\frac{4}{3} \end{array}\right]$ and $\left[\begin{array}{c} 0 \\ 1 \\ \frac{1}{3} \\ \frac{1}{3} \end{array}\right]$ form a basis $\mathcal{A}$ for the subspace $V$ . The dimension of the subspace $V$ is2.

STEP 8

Now, let's consider the subspace $W$ spanned by the basis $\mathcal{A}$ and the vector $w=(1,0,0,0)$ . We can write down the augmented matrix for the system of equations corresponding to the subspace $W$ $\left[\begin{array}{cccc|c} 1 &0 & -\frac{1}{3} & -\frac{4}{3} &0 \\ 0 &1 & \frac{1}{3} & \frac{1}{3} &0 \\ 1 &0 &0 &0 &0\end{array}\right]$

STEP 9

We can subtract the first row from the third row to get $\left[\begin{array}{cccc|c} & & -\frac{}{3} & -\frac{4}{3} & \\ & & \frac{}{3} & \frac{}{3} & \\ & & \frac{}{3} & \frac{4}{3} &\end{array}\right]$

STEP 10

Finally, we can multiply the third row by3 to get $\left[\begin{array}{cccc|c} &0 & - & -4 &0 \\ 0 & & & &0 \\ 0 &0 & &4 &0\end{array}\right]$ The system of equations corresponding to this matrix is $\left\{\begin{array}{c} x_{}-x_{3}-4 x_{4}=0 \\ x_{2}+x_{3}+x_{4}=0 \\ x_{3}+4 x_{4}=0\end{array}\right.$ This is the homogeneous system of linear equations whose set of solutions is equal to $W$ .

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STEP 1

STEP 2

First, we need to solve the given system of linear equations. We can do this by using the Gaussian elimination method. The augmented matrix for the system of equations is $\left[\begin{array}{cccc|c} 1 &2 &1 &2 &0 \\ &5 &4 &7 &0\end{array}\right]$

STEP 3

We can subtract3 times the first row from the second row to simplify the matrix. This gives us $\left[\begin{array}{cccc|c} 1 &2 &1 &2 &0 \\ 0 & -1 &1 &1 &0\end{array}\right]$

STEP 4

Next, we can multiply the second row by -1 to get $\left[\begin{array}{cccc|c} 1 &2 &1 &2 &0 \\ 0 &1 & -1 & -1 &0\end{array}\right]$

STEP 5

We can subtract2 times the second row from the first row to get $\left[\begin{array}{cccc|c} 1 &0 &3 &4 &0 \\ 0 &1 & -1 & -1 &0\end{array}\right]$

STEP 6

STEP 7

STEP 8

STEP 9

We can subtract the first row from the third row to get $\left[\begin{array}{cccc|c} & & -\frac{}{3} & -\frac{4}{3} & \\ & & \frac{}{3} & \frac{}{3} & \\ & & \frac{}{3} & \frac{4}{3} &\end{array}\right]$

STEP 10

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STEP 1

STEP 2

First, we need to solve the given system of linear equations. We can do this by using the Gaussian elimination method. The augmented matrix for the system of equations is[121205470]\left[\begin{array}{cccc|c} 1 &2 &1 &2 &0 \\ &5 &4 &7 &0\end{array}\right] [1​25​14​27​00​]

STEP 3

We can subtract3 times the first row from the second row to simplify the matrix. This gives us[121200−1110]\left[\begin{array}{cccc|c} 1 &2 &1 &2 &0 \\ 0 & -1 &1 &1 &0\end{array}\right] [10​2−1​11​21​00​]

STEP 4

Next, we can multiply the second row by -1 to get[1212001−1−10]\left[\begin{array}{cccc|c} 1 &2 &1 &2 &0 \\ 0 &1 & -1 & -1 &0\end{array}\right] [10​21​1−1​2−1​00​]

STEP 5

We can subtract2 times the second row from the first row to get[1034001−1−10]\left[\begin{array}{cccc|c} 1 &0 &3 &4 &0 \\ 0 &1 & -1 & -1 &0\end{array}\right] [10​01​3−1​4−1​00​]

STEP 6

STEP 7

STEP 8

STEP 9

We can subtract the first row from the third row to get[−3−4333343]\left[\begin{array}{cccc|c} & & -\frac{}{3} & -\frac{4}{3} & \\ & & \frac{}{3} & \frac{}{3} & \\ & & \frac{}{3} & \frac{4}{3} &\end{array}\right] ⎣⎡​​​−3​3​3​​−34​3​34​​​⎦⎤​

STEP 10

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First, we need to solve the given system of linear equations. We can do this by using the Gaussian elimination method. The augmented matrix for the system of equations is $\left[\begin{array}{cccc|c} 1 &2 &1 &2 &0 \\ &5 &4 &7 &0\end{array}\right]$

We can subtract3 times the first row from the second row to simplify the matrix. This gives us $\left[\begin{array}{cccc|c} 1 &2 &1 &2 &0 \\ 0 & -1 &1 &1 &0\end{array}\right]$

Next, we can multiply the second row by -1 to get $\left[\begin{array}{cccc|c} 1 &2 &1 &2 &0 \\ 0 &1 & -1 & -1 &0\end{array}\right]$

We can subtract2 times the second row from the first row to get $\left[\begin{array}{cccc|c} 1 &0 &3 &4 &0 \\ 0 &1 & -1 & -1 &0\end{array}\right]$

We can subtract the first row from the third row to get $\left[\begin{array}{cccc|c} & & -\frac{}{3} & -\frac{4}{3} & \\ & & \frac{}{3} & \frac{}{3} & \\ & & \frac{}{3} & \frac{4}{3} &\end{array}\right]$