Math Problem Statement

The Fundamental Theorem of Linear Algebra express the general solution of Ax = b where A = [-3 -6; -9 -18] and b = [-6; -18] as x = x_r + cx_n where x_r is the row space solution and x_n is a basis for null(A)

Solution

To express the general solution of the system Ax=bAx = b in the form x=xr+cxnx = x_r + cx_n, where xrx_r is the row space solution and xnx_n is a basis for the null space of AA, we proceed with the following steps.

1. Matrix and Vector Definition

Given:

A=(36918),b=(618)A = \begin{pmatrix} -3 & -6 \\ -9 & -18 \end{pmatrix}, \quad b = \begin{pmatrix} -6 \\ -18 \end{pmatrix}

2. Check Consistency of the System Ax=bAx = b

First, let's verify if the system Ax=bAx = b is consistent. This involves checking if the vector bb lies in the column space of AA.

Rank(A)=1andRank([Ab])=1\text{Rank}(A) = 1 \quad \text{and} \quad \text{Rank}([A|b]) = 1

Since the ranks are equal, the system is consistent, and bb lies in the column space of AA.

3. Row Reduction to Row Echelon Form (REF)

Let's reduce the augmented matrix [Ab][A|b] to its row echelon form (REF):

(36691818)(122000)\begin{pmatrix} -3 & -6 & -6 \\ -9 & -18 & -18 \end{pmatrix} \sim \begin{pmatrix} 1 & 2 & 2 \\ 0 & 0 & 0 \end{pmatrix}

4. Particular Solution xrx_r (Row Space Solution)

From the row-reduced form, we obtain the following equation:

x1+2x2=2x_1 + 2x_2 = 2

We can express the particular solution xrx_r as:

xr=(20)(set x2=0)x_r = \begin{pmatrix} 2 \\ 0 \end{pmatrix} \quad \text{(set } x_2 = 0 \text{)}

5. Null Space Solution xnx_n

To find the null space null(A)\text{null}(A), solve Ax=0Ax = 0:

(36918)(x1x2)=(00)\begin{pmatrix} -3 & -6 \\ -9 & -18 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}

This leads to the equation:

x1+2x2=0orx1=2x2x_1 + 2x_2 = 0 \quad \text{or} \quad x_1 = -2x_2

Thus, the null space solution xnx_n can be written as:

xn=(21)x_n = \begin{pmatrix} -2 \\ 1 \end{pmatrix}

6. General Solution

The general solution to Ax=bAx = b is given by:

x=xr+cxnx = x_r + c x_n

Substituting the values of xrx_r and xnx_n:

x=(20)+c(21)x = \begin{pmatrix} 2 \\ 0 \end{pmatrix} + c \begin{pmatrix} -2 \\ 1 \end{pmatrix}

Where cc is any scalar.

Final Answer

The general solution is:

x=(20)+c(21)x = \begin{pmatrix} 2 \\ 0 \end{pmatrix} + c \begin{pmatrix} -2 \\ 1 \end{pmatrix}

Would you like any more details or have any questions?

Here are 5 related questions to consider:

  1. How do you find the null space of a matrix?
  2. What is the significance of the row space and null space in linear algebra?
  3. How can you verify if a solution is indeed a solution to a given system?
  4. What happens when bb does not lie in the column space of AA?
  5. Can a matrix have a null space with more than one dimension?

Tip: Always check the consistency of a system Ax=bAx = b by comparing the ranks of AA and [Ab][A|b] before proceeding with solutions.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Operations
Null Space
Row Space

Formulas

Matrix-vector multiplication
Row reduction

Theorems

Fundamental Theorem of Linear Algebra

Suitable Grade Level

Undergraduate