Problem 33 Find the nullspace of the matrix... [FREE SOLUTION]

Chapter 4: Problem 33

Find the nullspace of the matrix. $$ A=\left[\begin{array}{rrr} 1 & 2 & -3 \\ 2 & -1 & 4 \\ 4 & 3 & -2 \end{array}\right] $$

Short Answer

Expert verified

The null space of the matrix is all scalar multiples of the vector $[-1.2, 2, 1]^T$.

Step by step solution

Set the matrix-vector multiplication equal to the null vector

Start by setting up the matrix multiplication $Ax = 0$, where $x = \{x_1, x_2, x_3\}$ is a vector, which gives us the following system of linear equations: \[ \begin{align*} 1*x_1 + 2*x_2 - 3*x_3 = 0 \ 2*x_1 - 1*x_2 + 4*x_3 = 0 \ 4*x_1 + 3*x_2 - 2*x_3 = 0 \end{align*} \]

Reduce the augmented matrix to row-echelon form

Next we transform the system of linear equations into an augmented matrix, and then perform row operations (multiplication, swapping, or addition) to bring the matrix to row-echelon form. The row-echelon form of the augmented matrix - representing the same system of linear equations - is: \[ \begin{align*} 1 & 2 & -3 \ 0 & -5 & 10 \ 0 & 0 & 0 \end{align*} \]

Reduce to reduced row-echelon form and solve

Now we'll further reduce the row-echelon matrix to reduced row-echelon form: \[ \begin{align*} 1 & 0 & 1.2 \ 0 & 1 & -2 \ 0 & 0 & 0 \end{align*} \] This gives us the solutions:\[ x_1 = -1.2x_3 \x_2 = 2x_3 \x_3 = x_3 \] So the general solution vector becomes $[-1.2, 2, 1]^T * x_3$. The nullspace of the matrix is all scalar multiples of the vector $[-1.2, 2, 1]^T$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

matrix-vector multiplication

The process of matrix-vector multiplication involves taking a matrix and multiplying it by a vector. This is a fundamental operation in linear algebra. In simple terms, you take each row of the matrix and compute a dot product with the vector. The result is a new vector. In our exercise, we multiplied the given matrix $ A $ by the vector $ x = \{x_1, x_2, x_3\} $ and set it equal to the null vector. This operation helps us form a system of linear equations:

Each equation corresponds to a row in the matrix.
Each equation is formed by the dot product of the matrix row and the vector.

For example, the first row of the matrix formed the equation $1*x_1 + 2*x_2 - 3*x_3 = 0$. Multiplying matrices by vectors is key in transforming geometric data and solving complex problems.

row-echelon form

The row-echelon form of a matrix is a type of matrix simplification. This process involves converting a matrix into a simpler form using a set of rules. The rules often involve performing row operations such as:

Swapping rows.
Multiplying a row by a non-zero scalar.
Adding or subtracting the multiples of one row from another.

In our problem, we started with the augmented matrix from the system of equations and brought it to a form that displays a staircase-like pattern. Each leading entry in a non-zero row is to the right of the leading entry in the row above it. The matrix eventually becomes:\[\begin{align*}1 & 2 & -3 \0 & -5 & 10 \0 & 0 & 0\end{align*}\]This form makes it much easier to solve the corresponding system of linear equations.

reduced row-echelon form

Further simplifying a matrix, the reduced row-echelon form (RREF) provides greater insight into solving linear equations. In RREF:

The leading entry in each non-zero row is 1.
Each leading 1 is the only non-zero value in its column.

This means the RREF is even more simplified than the row-echelon form. In the exercise, we continued from the row-echelon form and carried additional operations to achieve:\[\begin{align*}1 & 0 & 1.2 \0 & 1 & -2 \0 & 0 & 0\end{align*}\]By doing this, the solution to the system becomes apparent, which shows dependencies between variables. It also helps in expressing some variables in terms of others, providing the nullspace of the matrix.

system of linear equations

A system of linear equations is a set of equations where each one is linear. Each equation in a linear system describes a hyperplane in space. In the context of matrices, solving the system means finding the intersection of these hyperplanes. To do this for the given matrix:

We translated the problem into matrix form and solved using row and reduction operations.
The system initially provided equations like $1*x_1 + 2*x_2 - 3*x_3 = 0$.

Solving the system boils down to finding variable values that satisfy all equations simultaneously. Our transformed matrix approach aided in neatly breaking down and revealing solutions that pointed to the nullspace. The discovery of the nullspace offers all solutions to the homogeneous system where the matrix times the solution vector equals the zero vector.

91影视

Find the nullspace of the matrix. $$ A=\left[\begin{array}{rrr} 1 & 2 & -3 \\ 2 & -1 & 4 \\ 4 & 3 & -2 \end{array}\right] $$

Short Answer

Step by step solution

Set the matrix-vector multiplication equal to the null vector

Reduce the augmented matrix to row-echelon form

Reduce to reduced row-echelon form and solve

Key Concepts

matrix-vector multiplication

row-echelon form

reduced row-echelon form

system of linear equations

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