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Problem 178

Give an example to illustrate the following theorem: The system of n homogeneous linear equations in \(\mathrm{n}\) unknowns, \(\mathrm{AX}=0\), has a nontrivial solution if and only if rank \(\mathrm{A}<\mathrm{n}\)

Short Answer

Expert verified
Consider a system of 3 homogeneous linear equations in 3 unknowns: \[ x_1 + 2x_2 + 3x_3 = 0 \] \[ 2x_1 + 4x_2 + 6x_3 = 0 \] \[ 3x_1 + 6x_2 + 9x_3 = 0 \] We represent the system in matrix form AX=0 and row reduce A to find its rank: \[ \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{pmatrix} \sim \begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \] The rank of A is 1, which is less than the number of unknowns (n=3). Since rank(A) < n, there exists a nontrivial solution for the system. For example, the nontrivial solution is \(x_1=-3\), \(x_2=0\), and \(x_3=1\). This illustrates the theorem for a system of 3 homogeneous linear equations in 3 unknowns.
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Step 1: Define the system of linear equations

Consider a system of 3 homogeneous linear equations in 3 unknowns as follows: \[ x_1 + 2x_2 + 3x_3 = 0 \] \[ 2x_1 + 4x_2 + 6x_3 = 0 \] \[ 3x_1 + 6x_2 + 9x_3 = 0 \] #Step 2: Write the system in matrix form#

Step 2: Write the system in matrix form

We can represent the system of linear equations in the form AX=0, where A is the coefficient matrix and X is the column vector of unknowns: \[ \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0 \] #Step 3: Find the rank of the matrix A#

Step 3: Find the rank of the matrix A

We determine the rank of the matrix A by row reducing it to its row echelon form: \[ \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{pmatrix} \sim \begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \] The rank of the matrix A is the number of non-zero rows, which is 1 in this case. #Step 4: Determine if the rank(A) < n#

Step 4: Determine if the rank(A) < n

Now, we need to compare the rank of A with the number of unknowns (n) in the equation. In this case, n=3, and since rank(A)=1, it means rank(A) < n. #Step 5: Show that there is a nontrivial solution#

Step 5: Show that there is a nontrivial solution

Since rank(A) < n (1 < 3), there exists a nontrivial solution for the system of linear equations. For example, let's take \(x_3 = 1\), then we would have: \[x_1 = -3x_3 = -3\] \[x_2 = (4x_1 + 6x_3)/2 = (4*(-3) + 6)/2 = 0\] The nontrivial solution is \(x_1=-3\), \(x_2=0\), and \(x_3=1\). This shows that the given theorem holds true for the provided example: The system of 3 homogeneous linear equations in 3 unknowns, AX=0, has a nontrivial solution because the rank of A is less than 3 (number of unknowns).

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Most popular questions from this chapter

Chapter 7

Find the rank of matrix A where $$ \mathrm{A}=\begin{array}{cccc} 1 & 0 & 2 & 3 \mid \\ 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 \end{array} $$

Chapter 7

Show that the matrix $$ \mathrm{A}=\begin{array}{cll} 0 & 1 & 2 \\ 2 & 3 & 4 \\ 14 & 7 & 10 \end{array} $$ is equivalent to \(\mathrm{D}^{33}{ }_{2}\) where $\mathrm{D}^{\mathrm{m}, \mathrm{n}} \mathrm{r}$ denotes the canonical form under equivalence of A. D \(^{\mathrm{m}, \mathrm{n}}_{\mathrm{r}}\) is the echelon form that has one's along the diagonal and zeros elsewhere, and where all the zero rows are consigned to the depths of the matrix.

Chapter 7

Let \(\mathrm{V}\) be the subspace of \(\mathrm{R}^{4}\) spanned by $\mathrm{S}=\left\\{\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}, \alpha_{5}\right\\}\( where \)\alpha_{1}=[1,2,1,2], \alpha_{2}=[2,1,2,1]$ \(\alpha_{3}=[3,2,3,2], \alpha_{4}=[3,3,3,3], \alpha_{5}=[5,3,5,3]\). Find a basis for \(\mathrm{V}\).

Chapter 7

Let \(\mathrm{A}\) be the matrix $$ \begin{array}{rrrrrr} \mid 0 & 1 & 3 & -2 & -1 & 2 \\ 10 & 2 & 6 & -4 & -2 & 4 \\ \mid 0 & 1 & 3 & -2 & 1 & 4 \\ \mid 0 & 2 & 6 & 1 & -1 & 0 \end{array} $$ Find the determinant rank of \(\mathrm{A}\)

Chapter 7

Let the homogeneous linear system \(\mathrm{AX}=\mathrm{B}\) be given by \(\begin{array}{llll}1 & 2 & 0 \mid & \left|x_{1}\right| & |0|\end{array}\) \(|0 \quad 1 \quad 3| \quad\left|\mathrm{x}_{2}\right|=|0|\) $\mid \begin{array}{llll}2 & 1 & 3 \mid & \left|\mathrm{x}_{3}\right| & |0| \text { . }\end{array}$ Show that \(\mathrm{A}\) has only the trivial solution, $\left(\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}\right)=(0,0,0)$

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