Math

QuestionFind all x\mathbf{x} in R4\mathbb{R}^{4} such that Ax=0A \mathbf{x} = \mathbf{0} for the matrix A=[1271103401231574]A=\begin{bmatrix} 1 & 2 & 7 & -1 \\ 1 & 0 & 3 & -4 \\ 0 & 1 & 2 & 3 \\ -1 & 5 & 7 & 4 \end{bmatrix}. Choose A, B, C, or D.

Studdy Solution

STEP 1

Assumptions1. The vector x\mathbf{x} is in R4\mathbb{R}^{4}. . The transformation is xAx\mathbf{x} \mapsto A \mathbf{x}.
3. The matrix AA is given asA=[17110340131574]A=\left[\begin{array}{rrrr} 1 & &7 & -1 \\ 1 &0 &3 & -4 \\ 0 &1 & &3 \\ -1 &5 &7 &4\end{array}\right] 4. We are looking for vectors that are mapped into the zero vector by the transformation.

STEP 2

The problem is asking for the null space of the matrix AA. The null space of a matrix AA is the set of all vectors x\mathbf{x} such that Ax=0A\mathbf{x} = \mathbf{0}.

STEP 3

Let's denote the vector x\mathbf{x} as x=[x1x2x3x]\mathbf{x} = \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{} \end{array}\right]. Then, the equation Ax=0A\mathbf{x} = \mathbf{0} can be written as a system of linear equations[12711030123157][x1x2x3x]=[0000]\left[\begin{array}{rrrr} 1 &2 &7 & -1 \\ 1 &0 &3 & - \\ 0 &1 &2 &3 \\ -1 &5 &7 &\end{array}\right] \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{} \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \\ 0 \\ 0\end{array}\right]

STEP 4

This system of equations can be written as1. x1+2x2+7x3x4=0x_{1} +2x_{2} +7x_{3} - x_{4} =0
2. x1+3x34x4=0x_{1} +3x_{3} -4x_{4} =0
3. x2+2x3+3x4=0x_{2} +2x_{3} +3x_{4} =0
4. x1+x2+7x3+4x4=0-x_{1} +x_{2} +7x_{3} +4x_{4} =0

STEP 5

We can solve this system of equations using the Gaussian elimination method. First, we form the augmented matrix[12710103400123015740]\left[\begin{array}{rrrr|r} 1 &2 &7 & -1 &0 \\ 1 &0 &3 & -4 &0 \\ 0 &1 &2 &3 &0 \\ -1 &5 &7 &4 &0\end{array}\right]

STEP 6

Subtract the first row from the second and add the first row to the fourth to get[1210024300123001430]\left[\begin{array}{rrrr|r} 1 &2 & & -1 &0 \\ 0 & -2 & -4 & -3 &0 \\ 0 &1 &2 &3 &0 \\ 0 & &14 &3 &0\end{array}\right]

STEP 7

Multiply the second row by -1/2, the third row by2, and subtract the first row multiplied by7 from the fourth row to get[127100121.500246000040]\left[\begin{array}{rrrr|r} 1 &2 &7 & -1 &0 \\ 0 &1 &2 &1.5 &0 \\ 0 &2 &4 &6 &0 \\ 0 &0 &0 & -4 &0\end{array}\right]

STEP 8

Subtract the second row multiplied by2 from the third row, and divide the fourth row by -4 to get[127100121.500003000010]\left[\begin{array}{rrrr|r} 1 &2 &7 & -1 &0 \\ 0 &1 &2 &1.5 &0 \\ 0 &0 &0 &3 &0 \\ 0 &0 &0 &1 &0\end{array}\right]

STEP 9

Subtract the fourth row multiplied by3 from the third row, and subtract the fourth row multiplied by.5 from the second row to get[272]\left[\begin{array}{rrrr|r} &2 &7 & - & \\ & &2 & & \\ & & & & \\ & & & &\end{array}\right]

STEP 10

Subtract the second row multiplied by2 from the first row to get[0300200000000000]\left[\begin{array}{rrrr|r} &0 &3 & - &0 \\ 0 & &2 &0 &0 \\ 0 &0 &0 &0 &0 \\ 0 &0 &0 & &0\end{array}\right]

STEP 11

Add the fourth row to the first row to get[0300000000000000]\left[\begin{array}{rrrr|r} &0 &3 &0 &0 \\ 0 & & &0 &0 \\ 0 &0 &0 &0 &0 \\ 0 &0 &0 & &0\end{array}\right]

STEP 12

The solution to the system of equations is then. x=xx_{} = -x_{}
2. x2=2xx_{2} = -2x_{} . xx_{} is free4. x4=0x_{4} =0

So, the solution is x=x[20]\mathbf{x} = x_{} \left[\begin{array}{c} - \\ -2 \\ \\0 \end{array}\right], where xx_{} is any real number.
Therefore, the correct choice is B. x+x+x4x_{} \square+x_{} \square+x_{4} \square, where x=xx_{} = -x_{}, x2=2xx_{2} = -2x_{}, xx_{} is free, and x4=0x_{4} =0.

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