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Chapter 1: Problem 5

a. Find values of the scalars \(a, b, c, d\) so that $$ a\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]+b\left[\begin{array}{ll} 0 & 1 \\ 2 & 3 \end{array}\right]+c\left[\begin{array}{ll} 0 & 0 \\ 1 & 2 \end{array}\right]+d\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] $$ b. Find values of the scalars \(a, b, c, d\) so that $$ a\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]+b\left[\begin{array}{ll} 0 & 1 \\ 2 & 3 \end{array}\right]+c\left[\begin{array}{ll} 0 & 0 \\ 1 & 2 \end{array}\right]+d\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{rr} 4 & 9 \\ 16 & 25 \end{array}\right] $$ c. Are there solutions to the equations in parts a and \(b\) other than the ones you found?

Short Answer

Expert verified
The solutions are \(a = b = c = d = 0\) for part a, and \(a = 4, b = 1, c = 2, d = 1\) for part b. There are no other solutions to the equations in parts a and b than the ones found.

Step by step solution

01

Set up the equations for part a

The equation in part a is \(a\left[\begin{array}{ll} 1 & 2 \ 3 & 4 \end{array}\right] + b\left[\begin{array}{ll} 0 & 1 \ 2 & 3 \end{array}\right] + c\left[\begin{array}{ll} 0 & 0 \ 1 & 2 \end{array}\right] + d\left[\begin{array}{ll} 0 & 0 \ 0 & 1 \end{array}\right] = \left[\begin{array}{ll} 0 & 0 \ 0 & 0 \end{array}\right]\). This is equivalent to four independent equations: 1. \(a = 0\) 2. \(2a + b = 0\) 3. \(3a + 2b + c = 0\) 4. \(4a + 3b + 2c + d = 0\).
02

Solve the equations for part a

From the first equation, we know that \(a = 0\). Substituting \(a = 0\) into the four equations, we have: 1. \(a = 0\) 2. \(b = 0\) 3. \(c = 0\) 4. \(d = 0\). So, the solution to part a is \(a = b = c = d = 0\).
03

Set up the equations for part b

The equation in part b is \(a\left[\begin{array}{ll} 1 & 2 \ 3 & 4 \end{array}\right] + b\left[\begin{array}{ll} 0 & 1 \ 2 & 3 \end{array}\right] + c\left[\begin{array}{ll} 0 & 0 \ 1 & 2 \end{array}\right] + d\left[\begin{array}{ll} 0 & 0 \ 0 & 1 \end{array}\right] = \left[\begin{array}{rr} 4 & 9 \ 16 & 25 \end{array}\right]\). This is equivalent to four independent equations: 1. \(a = 4\) 2. \(2a + b = 9\) 3. \(3a + 2b + c = 16\) 4. \(4a + 3b + 2c + d = 25\).
04

Solve the equations for part b

From the first equation, we know that \(a = 4\). Substituting \(a = 4\) into the remaining equations gives 2. \(b = 1\) 3. \(c = 2\) 4. \(d = 1\) and so, the solution to part b is \(a = 4, b = 1, c = 2, d = 1\).
05

Discuss Multiple Solutions

The concept of free variables applies here. Free variables allow for infinite solutions, but in both sets of equations (part a and part b), there were no free variables. In both cases, each variable could be solved in terms of constants, not other variables, so there are exactly the solutions found and no others. Therefore, other than the solutions found, there are no other solutions for the equations in parts a and b. Thus, the answer to part c is 'No, there are no other solutions.'

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Key Concepts

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

Matrix Equations
Matrix equations are a powerful tool for solving systems of linear equations. A matrix equation involves matrices, which are arrays of numbers arranged in rows and columns, used to represent and solve equations.
The basic form of a matrix equation is \( AX = B \), where \( A \) is a matrix representing the coefficients, \( X \) is a matrix containing the variables, and \( B \) is the matrix representing the constants or solutions.
Using matrix equations simplifies the process of solving complex systems of linear equations, transforming them into more manageable algebraic operations.
  • Matrices can represent multiple equations simultaneously, allowing for a compact representation of a system of equations.
  • The manipulation of these matrices, such as through row operations, makes it easier to solve for unknowns.
  • Matrix equations can be solved using various methods, including Gaussian elimination, inversion, or other computational tools.
Creating these matrix equations from a system of linear equations is an important step in linear algebra, leading to efficient problem-solving and deeper insight into mathematical relationships.
Linear Combinations
Linear combinations involve combining vector and scalar multiplication to form new vectors. In the context of equations, a linear combination refers to the sum of scalar multiples of vectors.
For instance, in an equation \( aA + bB + cC + dD = E \), each letter represents a vector or matrix, and \( a, b, c, \) and \( d \) are scalars used to multiply the respective vectors. The result is a single matrix or vector that combines these elements.
  • Linear combinations show that matrices can be scaled and added to achieve a desired outcome.
  • This concept is essential in forming and solving matrix equations, as seen when expressing one matrix in terms of others using scalars.
  • They help in understanding how different matrices relate and interact, providing a basis for more complex algebraic operations.
The ability to express a single vector or matrix as a combination of others makes linear combinations a versatile and vital concept in linear algebra.
Scalar Multiplication
Scalar multiplication is the process of multiplying each element of a matrix or vector by a scalar (a single number). This operation is fundamental in matrix algebra and a building block for more complex operations.
Consider the simple form \( a \times A \), where \( a \) is a scalar and \( A \) is a matrix or vector. Multiplying \( A \) by \( a \) results in a new matrix where each element is the product of \( a \) and the corresponding element in \( A \).
  • Scalar multiplication alters the magnitude of vectors or matrices without changing their direction or orientation.
  • It is often combined with addition in matrix equations, forming integral parts of linear combinations.
  • This operation is straightforward, yet it is crucial in scaling results or balancing equations within a system of linear equations.
Scalar multiplication extends the reach of basic arithmetic into the realm of matrices, enabling more intricate manipulations and solutions in linear algebra.

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Most popular questions from this chapter

Is it possible to add two sets? For sets of real numbers, we might define addition of sets in terms of addition of the elements in the sets. Let us introduce the following meaning to the symbol \(\oplus\) for adding a set \(A\) of real numbers to another set \(B\) of real numbers: \(A \oplus B=\\{a+b \mid a \in A\) and \(b \in B\\}\). a. List the elements in the set \(\\{1,2,3\\} \oplus\\{5,10\\}\). b. List the elements in the set \(\\{1,2,3\\} \oplus\\{5,6\\}\). c. List the elements in the set \(\\{1,2,3\\} \oplus \varnothing\). d. If a set \(A\) contams \(m\) real numbers and a set \(B\) contains \(n\) real numbers, can you predict the number of elements in \(A \oplus B\) ? If you run into difficulties, can you determine the minimum and naximum numbers of elements possible in \(A \oplus B\) ? e. Does the commutative law hold for this new addition? That is, does \(A \oplus\) \(B=B \oplus A\) ? f. Reformulate other laws of real-number addition in terms of this new addition of sets. Which of your formulas are true? Can you prove them or provide counterexamples? g. What about laws that combine set addition with union and intersection? For example, does \((A \cup B) \oplus C=(A \oplus C) \cup(B \oplus C)\) ? h. Is there any hope of extending the other operations of arithmetic to sets of real numbers? What about algebra? Limits? Power series?

Let \(f, g\), and \(h\) be functions in \(\mathbb{F}(\mathbb{R})\) defined by \(f(x)=(x+1)^{3}, g(x)=x^{3}+1\), and \(h(x)=x^{2}+x\). Let \(a\) and \(b\) denote scalars. a. Write the identity in the independent variable \(x\) that must hold if the function \(f\) is equal to the function \(a g+b h\). b. Find scalars \(a\) and \(b\) so that \(f=a g+b h\).

Find scalars \(a, b, c\), and \(d\) so that $$ \begin{aligned} &a(1,0,0,0,0)+b(1,1,0,0,0)+c(1,1,1,0,0)+d(1,1,1,1,0) \\ &=(8,5,-2,3,0) \end{aligned} $$

Determine whether we obtain a vector space from \(\mathbb{R}^{2}\) with operations defined by $$ \begin{aligned} \left(v_{1}, v_{2}\right)+\left(w_{1}, w_{2}\right) &=\left(v_{1} w_{2}+v_{2} w_{1}, v_{2} w_{2}\right) \\ r\left(v_{1}, v_{2}\right) &=\left(r_{1}, v_{2}\right) \end{aligned} $$

The following result has been attributed to Gertrude Stein: If \(S\) is a subspuce of a vector space \(V\), and \(T\) is a subspace of \(S\), then \(T\) is a subspace of \(V\). (Or, in the original formulation: A subspace of a subspace is a subspace.) Explain how this result is an immediate consequence of the definition of a subspace.
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