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Compute the rank and nullity of the given matrices over the indicated \(\mathbb{Z}_{p}\)? $$\left[\begin{array}{lll} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{array}\right] \text { over } \mathbb{Z}_{2}$$

Short Answer

Expert verified
The rank is 2, and the nullity is 1.

Step by step solution

01

Understand Matrix over \(\mathbb{Z}_2\)

The given matrix is over the field \(\mathbb{Z}_2\), which means each entry is either 0 or 1 and calculations are done modulo 2. The matrix given is \([\begin{array}{ccc} 1 & 1 & 0 \ 0 & 1 & 1 \ 1 & 0 & 1 \end{array}]\).
02

Perform Row Reduction

Apply row operations to convert the matrix to its row-reduced echelon form (RREF). Subtract the first row from the third row: \([1, 0, 1] - [1, 1, 0] = [0, 1, 1]\), resulting in the matrix: \[[\begin{array}{ccc} 1 & 1 & 0 \ 0 & 1 & 1 \ 0 & 1 & 1 \end{array}]\].
03

Continue Row Reduction

To further simplify, subtract the second row from the third row: \([0, 1, 1] - [0, 1, 1] = [0, 0, 0]\), achieving the RREF: \([\begin{array}{ccc} 1 & 1 & 0 \ 0 & 1 & 1 \ 0 & 0 & 0 \end{array}]\).
04

Determine Rank

Count the number of non-zero rows in the RREF. There are 2 non-zero rows, so the rank of the matrix is 2.
05

Calculate Nullity

The nullity of a matrix is defined as the number of columns minus the rank. With 3 columns and a rank of 2, the nullity is \(3 - 2 = 1\).

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Key Concepts

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

Row Reduction
Row reduction is a powerful methodological tool in linear algebra that transforms a matrix into a simpler form by applying a series of row operations. These operations include swapping rows, scaling rows by non-zero scalars, and adding multiples of one row to another. This process helps solve systems of linear equations and find key matrix properties, such as its rank.

In our case, we need to work with a matrix over \(\mathbb{Z}_2\), where calculations are performed modulo 2. Here, '1 - 1' equals 0 while performing operations.

This simplification process is crucial because it leads us to the row-reduced echelon form (RREF), a more digestible form of any matrix that can quickly reveal information about the matrix's solutions and properties. It helps in systematically evaluating a matrix by eliminating variables and reducing complexity.
Row-Reduced Echelon Form
The row-reduced echelon form (RREF) is a special kind of matrix form achieved through row reduction. In RREF, each leading entry in a row is 1, and all other elements in the column containing the leading entry are zeros.

The RREF for a matrix can quickly tell us about the system of equations it represents. For instance, when looking at our processed matrix:\[\begin{array}{ccc}1 & 1 & 0 \0 & 1 & 1 \0 & 0 & 0 \\end{array}\]we can see a clear pattern indicating that it has two pivot positions, suggesting two basic variables.

Achieving this form is essential for easily identifying the rank, which is simply the number of non-zero rows, as it represents the count of linearly independent rows.
Nullity of a Matrix
Nullity is a concept in linear algebra that measures the dimension of the solution space to the homogeneous equation corresponding to a matrix. It's calculated by subtracting the rank from the total number of columns in the matrix.

For example, with our 3-column matrix that has a rank of 2, the nullity is computed as 1.This means there is a one-dimensional space of solutions to the equation \(A\mathbf{x} = \mathbf{0}\).

  • The rank provides a picture of the solutions' span.
  • Nullity informs us about the number of free variables, corresponding here to the one zero row in the RREF.
Understanding nullity is crucial for grasping the kernel of a matrix, which is the set of all possible solutions to this equation.

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Most popular questions from this chapter

By considering the matrix with the given vectors as its columns. $$\text { Do }\left[\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right],\left[\begin{array}{l} 0 \\ 1 \\ 1 \end{array}\right] \text { form a basis for } \mathbb{R}^{3} \text { ? }$$

Use the Gauss-Jordan method to find the inverse of the given matrix (if it exists). $$\left[\begin{array}{rrr}1 & -1 & 2 \\\3 & 1 & 2 \\\2 & 3 & -1\end{array}\right]$$

Verify Theorem 3.32 by finding the matrix of \(S\) o \(T\) (a) by direct substitution and (b) by matrix multiplication of \([\mathrm{S}][\mathrm{T}]\). $$T\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right]=\left[\begin{array}{l} x_{1}+2 x_{2} \\ 2 x_{2}-x_{3} \end{array}\right], S\left[\begin{array}{l} y_{1} \\ y_{2} \end{array}\right]=\left[\begin{array}{c} y_{1}-y_{2} \\ y_{1}+y_{2} \\ -y_{1}+y_{2} \end{array}\right]$$

A graph is called bipartite if its vertices can be subdivided into two sets U and V such that every edge has one endpoint in \(U\) and the other endpoint in \(V\). For example, the graph in Exercise 46 is bipartite with \(U=\left\\{v_{1}, v_{2}, v_{3}\right\\}\) and \(V=\left\\{v_{4}, v_{5}\right\\}\). Determine whether \(a\) graph with the given adjacency matrix is bipartite. $$\left[\begin{array}{cccccc} 0 & 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 0 \end{array}\right]$$

Compute the rank and nullity of the given matrices over the indicated \(\mathbb{Z}_{p}\)? $$\left[\begin{array}{llll} 1 & 3 & 1 & 4 \\ 2 & 3 & 0 & 1 \\ 1 & 0 & 4 & 0 \end{array}\right] \text { over } \mathbb{Z}_{5}$$

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