Find the null space of the matrix \( A = \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{bmatrix} \)
Answer
Step 5: The solution to the system is a line of the form \(x = -y - z\), where y and z are free to vary. This represents the null space of the matrix A.
Step 1 ：Step 1: The null space of a matrix A is the solution to the homogeneous system \(Ax = 0\). So we need to solve the following system of equations:\[\begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}\]
Step 2 ：Step 2: This gives us the system of equations: \[ \begin{cases} x + 2y + 3z = 0 \\ 4x + 5y + 6z = 0 \\ 7x + 8y + 9z = 0 \end{cases}\]
Step 3 ：Step 3: Solve this system by choosing a variable to eliminate. Let's subtract the first equation from the second and the second from the third: \[ \begin{cases} 3x + 3y + 3z = 0 \\ 3x + 3y + 3z = 0 \end{cases}\]
Step 4 ：Step 4: Simplify the system to: \[ \begin{cases} x + y + z = 0 \\ x + y + z = 0 \end{cases}\]
Step 5 ：Step 5: The solution to the system is a line of the form \(x = -y - z\), where y and z are free to vary. This represents the null space of the matrix A.
