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Chapter 2: Problem 43

Choose the number \(q\) so that (if possible) the ranks are (a) 1, (b) 2 , (c) \(3:\) $$ A=\left[\begin{array}{rrr} 6 & 4 & 2 \\ -3 & -2 & -1 \\ 9 & 6 & q \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{lll} 3 & 1 & 3 \\ q & 2 & q \end{array}\right] $$

Short Answer

Expert verified
For (a) A: q=3, B: q=6; (b) A: non-q multiple, B: q=3; (c) A: q=3. Rank 3 is not possible for B.

Step by step solution

01

Understanding Rank and Matrix

The rank of a matrix is the maximum number of linearly independent row vectors in it. To find the rank, reduce the matrix to its row echelon form and count the number of non-zero rows.
02

Rank 1 for Matrix A

For Matrix A to have rank 1, every row must be a scalar multiple of each other. Assume the base row is the first one, \(6, 4, 2\), and verify whether other rows can be written as \( k \times (6, 4, 2) \). The second row \(-3, -2, -1\) is \(-0.5\times(6, 4, 2)\). Substituting q in the third row as \(k(6, 4, 2)\), we get \(q=3\).
03

Rank 2 for Matrix A

For rank 2, at least one row should not be a multiple of the others. We already have two rows of rank 1. So we let third row \(9, 6, q\) be linearly independent. Use a different value from 3 for q (like any non-scaling multiple relation). For instance, taking q = 5 or any other non-related value keeps it rank 2 as there won't be any dependency relations between rows.
04

Rank 3 for Matrix A

For rank 3, no row should be expressible as a linear combination of others. Substitute \(q\) so that the determinant of the matrix is non-zero, leading to rank 3 condition. Selecting \(q = 3\), as tried with rank-1, returns \(6 \times (-12) + 4 \times (-9) + 2\times (-27) = 3[18]\) a non-zero determinant, leading to rank 3.
05

Rank 1 for Matrix B

For rank 1, the second row must be a multiple of the first. If \(q = 6\), then the second row \(q, 2, q = 6 \times (3, 1, 3) \).
06

Rank 2 for Matrix B

We need any non-zero row with another row not being multiples or parallels. Choosing \(q = 3\) makes rows linearly independent \((3, 1, 3)\) and \((3, 2, 3)\).
07

Rank 3 for Matrix B

Rank 3 is not possible as B is a 2x3 matrix implying the maximum possible rank is 2.

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Key Concepts

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

Linear Independence
Linear independence is a fundamental concept in linear algebra. It describes a situation where no vector in a set can be expressed as a linear combination of the others. In simpler terms, if vectors are linearly independent, none of them is redundant. Each vector brings something unique to the table.
To break it down:
  • Consider vectors as arrows with both direction and magnitude.
  • If you can't write any vector as a sum of multiples of others, they're independent.
  • For example, in a 2D plane, the vectors (1,0) and (0,1) are independent because you can't make one out of the other using multiplication and addition.
  • In contrast, vectors like (2,2) and (4,4) are dependent, because (4,4) is just 2 times (2,2).
Grasping linear independence helps in understanding why some matrices have higher rank than others, as rank is determined by the number of linearly independent rows or columns.
Row Echelon Form
The row echelon form is a key tool for solving systems of linear equations and computing the rank of a matrix. In this form, the matrix has a staircase shape, with leading coefficients (the first non-zero number from the left, in each row) creating a pattern that progresses down and right across the matrix.
Features include:
  • Every leading entry (or pivot) is strictly to the right of the pivot in the row above.
  • All zero rows, if any, are at the bottom of the matrix.
  • The elements below each pivot are zeros.
To achieve row echelon form:
  • Use row operations like swapping rows, multiplying rows by non-zero numbers, and adding multiples of one row to another.
  • The goal is to simplify the matrix while keeping the solution to the system of equations unchanged.
Understanding how to transform a matrix to its row echelon form allows you to identify how many rows are independent, and thus, determine its rank.
Determinant
A determinant is a powerful scalar value associated with matrices. It provides important properties and insights about a matrix, such as whether it is invertible and the volume transformation it describes.
Key aspects:
  • Determinants only apply to square matrices (e.g., 2x2, 3x3).
  • It can be calculated using various methods, like cofactor expansion.
  • A determinant of zero means the matrix is singular (non-invertible) and its rows (or columns) are linearly dependent.
  • For a 3x3 matrix \( A \) with elements \( a, b, c, d, e, f, g, h, i \), the determinant is calculated as:
    \[ \text{det}(A) = a(ei鈭抐h)鈭抌(di鈭抐g)+c(dh鈭抏g) \]
In matrix rank finding, a non-zero determinant implies full rank, meaning all rows (and columns) are independent. This makes the determinant a quick check for linear independence and the maximum rank of a matrix.

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Most popular questions from this chapter

From the cubics \(\mathbf{P}_{3}\) to the fourth-degree polynomials \(\mathbf{P}_{4}\), what matrix represents multiplication by \(2+3 t\) ? The columns of the 5 by 4 matrix \(A\) come from applying the transformation to \(1, t, t^{2}, t^{3}\).

The transformation \(T\) that transposes every matrix is definitely linear. Which of these extra properties are true? (a) \(T^{2}=\) identity transformation. (b) The kernel of \(T\) is the zero matrix. (c) Every matrix is in the range of \(T\). (d) \(T(M)=-M\) is impossible.

By locating the pivots, find a basis for the column space of $$ U=\left[\begin{array}{llll} 0 & 5 & 4 & 3 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ Express each column that is not in the basis as a combination of the basic columns. Find also a matrix \(A\) with this echelon form \(U\), but a different column space.

Suppose \(\mathbf{S}\) is a five-dimensional subspace of \(\mathbf{R}^{6}\). True or false? (a) Every basis for \(\mathbf{S}\) can be extended to a basis for \(\mathbf{R}^{6}\) by adding one more vector. (b) Every basis for \(\mathbf{R}^{6}\) can be reduced to a basis for \(\mathbf{S}\) by removing one vector.

If the entries of a 3 by 3 matrix are chosen randomly between 0 and 1 , what are the most likely dimensions of the four subspaces? What if the matrix is 3 by \(5 ?\)
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