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If \(A=\left[\begin{array}{rrr}1 & 0 & 1 \\ 0 & 0 & 1 \\ 2 & -1 & 0\end{array}\right]\) and \(B=\left[\begin{array}{rrr}0 & 3 & -1 \\ -1 & 2 & 0 \\\ 0 & 0 & 1\end{array}\right],\) for what values of \(a\) and \(b\) does \(A B=\left[\begin{array}{rrr}0 & 2 a+2 b+1 & 0 \\ 3 a+4 b & 0 & 1 \\ 1 & 4 & -2\end{array}\right] ?\)

Short Answer

Expert verified
The values that make the condition satisfied are \( a = 0 \) and \( b = 4 \)

Step by step solution

01

Multiply Matrices A and B

Firstly, we need to perform the multiplication of matrices A and B. The resulting matrix AB will have elements that are the sums of the products of corresponding entries from rows of A and columns of B. Using this rule for multiplying matrices, the resulting matrix AB is: \( AB = \left[ \begin{array}{ccc} 0 & 3 & 0 \ 0 & 0 & 1 \ -1 & 6 & 0 \end{array} \right] \)
02

Compare Matrix AB to the Given Matrix

Next, we compare each element in the matrix AB to the corresponding element in the matrix we are trying to equal to. For example, the elements in the first row and second column of these two matrices should be equal. So, we present the following equations according to the equality of the corresponding points in the matrices: \(2a + 2b + 1 = 3\) and \(3a + 4b = -1\)
03

Solve the System of Equations

Finally, we solve the system of equations obtained in step 2. This can be achieved by substitution. From the first equation, we find that \(a = 1 - b\), substituting a value in the second equation we get \( 3(1 - b) + 4b = -1 \), which yields \(3 - 3b + 4b = -1 \) and therefore \(b = 4 \). Substitute \( b = 4 \) back in the first equation gives \(a + 1 = 1 \), hence \(a = 0\).

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Key Concepts

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

Understanding Systems of Equations
A system of equations is essentially a collection of two or more equations with the same set of unknowns. In the context of our matrix multiplication exercise, we encounter a system when we equate the product of two matrices to a third matrix. Each element in the resulting matrix from multiplying matrices A and B provides us with an equation based on the position within the matrix.

For instance, when we multiply matrices A and B, we form a new matrix AB. If we want AB to match a given third matrix, we compare the corresponding elements and set them as a system of equations. Specifically in this exercise, the comparison yields equations such as \(2a + 2b + 1 = 3\) and \(3a + 4b = -1\). Solving this system helps us find the unknowns 'a' and 'b' that make the equality true.

We can solve this system using various methods, such as substitution, elimination, or matrix methods like row reduction. In our example, substitution proves handy. Such systems crop up all the time not only in algebra but also in real-world scenarios where multiple conditions have to be satisfied simultaneously.
Matrix Equality For Problem-Solving
Matrix equality states that two matrices are equal if and only if their corresponding elements are equal. In terms of the exercise, this means that we can only say matrix AB is equal to the given matrix if all the entries match up perfectly.

When we multiply matrix A by matrix B, we must ensure each element of the product matrix matches the corresponding element in the given matrix. This check for matrix equality is what suggests the system of equations we need to solve. One essential point to note is that matrix multiplication is not commutative—meaning AB does not necessarily equal BA—which is why careful alignment of the matrix elements is key.

Matrix equality also serves as a bedrock for more complex concepts like matrix inversion and eigenvalues, which are both pivotal in higher-level mathematics and applied disciplines such as physics and engineering.
Algebraic Manipulation in Matrix Equations
Algebraic manipulation involves rearranging and simplifying mathematical expressions and equations to find the values of unknowns. In the context of our problem, once we have a system of equations, we use algebraic manipulation to solve for the variables 'a' and 'b'.

We often start by isolating one of the variables in one of the equations—like isolating 'a' in the equation \(2a + 2b + 1 = 3\) to get \(a = 1 - b\). Then, we substitute this expression for 'a' into the other equations, a process that typifies algebraic manipulation. The goal is to simplify the equations to a point where the variables can be easily identified.

Successful algebraic manipulation not only requires understanding how to operate with the terms and coefficients but also a strategic approach to selecting which manipulations will lead to a simpler equation more quickly. It's a critical skill in mathematics that underpins problem-solving across topics, from basic algebra to calculus.

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Most popular questions from this chapter

A family owns and operates three businesses. On their income-tax return, they have to report the depreciation deductions for the three businesses separately. In \(2004,\) their depreciation deductions consisted of use of a car, plus depreciation on 5 -year equipment (on which onc-fifth of the original value is deductible per year) and 10-year equipment (on which one- tenth of the original valuc is deductible per year). The car use (in miles) for cach business in 2004 is given in the following table, along with the original value of the depreciable 5 - and 10 year equipment used in each business that year. $$\begin{array}{|c|c|c|c|}\hline & \begin{array}{c}\text { Car } \\\\\text { Use }\end{array} & \begin{array}{c}\text { Original } \\\\\text { Value, 5-Year }\end{array} & \begin{array}{c}\text { Original } \\\\\text { Value, 10-Year }\end{array} \\\\\text { Business } & \text { (miles) } & \text { Equipment (S) } & \text { Equipment (S) } \\\\\hline 1 & 3200 & 9850 & 435 \\\2 & 8800 & 12,730 & 980 \\\3 & 6880 & 2240 & 615\\\\\hline\end{array}$$ The depreciation deduction for car use in 2004 was 37.5 cents per mile. Use matrix multiplication to determine the total depreciation deduction for each business in 2004.

A telephone company manufactures two different models of phones: Model 120 is cordless and Model 140 is not cordless. It takes 1 hour to manufacture the cordless model and 1 hour and 30 minutes to manufacture the traditional phone. At least 300 of the cordless models are to be produced. The manufacturer realizes a profit per phone of \(\$ 12\) for Model 120 and \(\$ 10\) for Model \(140 .\) If at most 1000 hours are to be allocated to the manufacture of the two models combined, how many of each model should be produced to maximize the total profit?

The following table lists the caloric content of a typical fast-food meal. Food (single serving) Calories Cheeseburger Medium order of fries Medium cola (21 oz) \(\begin{array}{lc}\text { Food (single serving) } & \text { Calories } \\\ \text { Cheeseburger } & 330 \\ \text { Medium order of fries } & 450 \\\ \text { Medium cola }(210 z) & 220\end{array}\) (a) After a lunch that consists of a cheeseburger, a medium order of fries, and a medium cola, you decide to burn off a quarter of the total calories in the meal by some combination of running and walking. You know that running burns 8 calories per minute and walking burns 3 calories per minute. If you exercise for a total of 40 minutes, how many minutes should you spend on each activity? (b) Rework part (a) for the case in which you exercise for a total of only 20 minutes. Do you get a realistic solution? Explain your answer.

In this set of exercises, you will use the method of solving linear systems using matrices to study real-world problems. A grocery store carries two brands of diapers. For a certain week, the number of boxes of Brand \(A\) diapers sold was 4 more than the number of boxes of Brand B diapers sold. Brand A diapers cost \(\$ 10\) per box and Brand B diapers cost \(\$ 12\) per box. If the total revenue generated that week from the sale of diapers was \(\$ 172,\) how many of each brand did the store sell?

Answer the question pertaining to the matrices. $$A=\left[\begin{array}{ll}a & b \\\c & d \\\e & f\end{array}\right] \text { and } B=\left[\begin{array}{lll}g & h & i \\\j & k & l\end{array}\right]$$ Let \(Q=B A,\) and find \(q_{11}\) and \(q_{22}\) without performing the entire multiplication of matrix \(B\) by matrix \(A\).

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