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Problem 173

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

Short Answer

Expert verified
The given matrix A is already in Row Echelon Form: $$ \mathrm{A}=\begin{array}{cccc} 1 & 0 & 2 & 3 \\\ 0 & 0 & 5 & 1 \\\ 0 & 0 & 0 & 0 \end{array} $$ The number of non-zero rows is 2, so the rank of matrix A is 2. Therefore, \( \text{rank}(A) = 2 \).
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Step by step solution

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Step 1: Identify the linearly independent rows

Since the given matrix A is already in Row Echelon Form, we can simply count the number of non-zero rows to find the rank. The matrix A looks like this: $$ \mathrm{A}=\begin{array}{cccc} 1 & 0 & 2 & 3 \\\ 0 & 0 & 5 & 1 \\\ 0 & 0 & 0 & 0 \end{array} $$ We can see that there are two non-zero rows in the matrix.

Step 2: Determine the rank of the matrix

Since there are two linearly independent rows, the rank of matrix A is 2. Therefore, rank(A) = 2.

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Most popular questions from this chapter

Chapter 7

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 \(\mathrm{A}<\mathrm{n}\)

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, $\left(\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}\right)=(0,0,0)$

Chapter 7

Let \(\mathrm{P}_{4}\), denote the vector space of all polynomials of degree at most equal to four. Let \(\mathrm{V}\) be the subspace of \(\mathrm{P}_{4}\). spanned by $\mathrm{S}=\left\\{\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}\right\\}\( where \)\alpha_{1}=\mathrm{t}^{4}+\mathrm{t}^{2}+2 \mathrm{t}+1$ \(\alpha_{2}=t^{4}+t^{2}+2 t+2, \alpha_{3}=2 t^{4}+t^{3}+t+2\) and \(\alpha_{4}=t^{4}+t^{3}-t^{2}\) \(-\mathrm{t}\). Find a basis for \(\mathrm{V}\).

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 \(\mathrm{V}\).

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