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Given \(A=\left[\begin{array}{cc}4 & -2 \\ 3 & 1\end{array}\right], B=\left[\begin{array}{cc}5 & 1 \\ 0 & -2 \\ 3 & 7\end{array}\right],\) and \(C=\left[\begin{array}{ccc}-5 & 4 & 1 \\ 0 & 3 & 6\end{array}\right],\) find each product if possible. $$C B$$

Short Answer

Expert verified
The product of matrices \(C\) and \(B\) is \( \begin{bmatrix} -22 & -6 \\ 18 & 36 \end{bmatrix} \).

Step by step solution

01

Check Matrix Dimensions for Multiplication

To multiply two matrices, the number of columns in the first matrix must equal the number of rows in the second matrix. Matrix \(C\) has dimensions \(2 \times 3\) and matrix \(B\) has dimensions \(3 \times 2\). Since the number of columns in \(C\) (which is 3) matches the number of rows in \(B\) (also 3), the product \(CB\) is possible.
02

Determine the Dimensions of Resulting Matrix

When multiplying two matrices, the resulting matrix will have the same number of rows as the first matrix and the same number of columns as the second matrix. Therefore, matrix \(CB\) will have dimensions \(2 \times 2\).
03

Multiply the First Row by Each Column

Compute the elements of the first row of \(CB\) by multiplying the first row of \(C\) by each column of \(B\). For the first column: \((-5) \times 5 + 4 \times 0 + 1 \times 3 = -25 + 0 + 3 = -22\). For the second column: \((-5) \times 1 + 4 \times (-2) + 1 \times 7 = -5 - 8 + 7 = -6\).
04

Multiply the Second Row by Each Column

Compute the elements of the second row of \(CB\) by multiplying the second row of \(C\) by each column of \(B\). For the first column: \(0 \times 5 + 3 \times 0 + 6 \times 3 = 0 + 0 + 18 = 18\). For the second column: \(0 \times 1 + 3 \times (-2) + 6 \times 7 = 0 - 6 + 42 = 36\).
05

Construct the Resulting Matrix

Combine the calculated elements to form the resulting matrix \(CB\):\[CB = \begin{bmatrix} -22 & -6 \ 18 & 36 \end{bmatrix}\]

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Key Concepts

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

Matrix Dimensions
When you're diving into matrix multiplication, understanding matrix dimensions is your starting point. For two matrices to be multiplied, the number of columns in the first matrix must be equal to the number of rows in the second matrix. This requirement is essential as it determines whether the multiplication is possible or not.

For example, consider matrix \(C\) with dimensions \(2 \times 3\) and matrix \(B\) with dimensions \(3 \times 2\). The number of columns in \(C\) (which is 3) perfectly matches the number of rows in \(B\) (also 3), making the product \(CB\) feasible. If these dimensions do not align, the matrices cannot be multiplied. This rule is crucial when dealing with matrix operations in linear algebra.

In summary, always check:
  • The number of columns in the first matrix.
  • The number of rows in the second matrix.
  • If they are equal, multiplication is possible!
Resulting Matrix
Once you've confirmed that matrix multiplication is possible, the next step is to determine the dimensions of the resulting matrix. This resulting matrix will have a distinct set of dimensions derived from the original matrices.

The dimensions of the new matrix are determined as follows:
  • The number of rows in the resulting matrix equals the number of rows of the first matrix.
  • The number of columns equals the number of columns of the second matrix.
For the matrices \(C\) and \(B\), which are \(2 \times 3\) and \(3 \times 2\) respectively, the resulting matrix \(CB\) will thus be \(2 \times 2\). This is a direct result of having two rows from \(C\) and two columns from \(B\).

Understanding the resulting matrix dimensions is crucial as it sets up the structure for filling in the elements during matrix multiplication.
Matrix Operations
Matrix operations, especially multiplication, involve a methodical approach to filling elements in the resulting matrix. With the previous steps confirming the feasibility and dimensions of the product \(CB\), you're ready to perform the actual operations.

Here's how multiplication works in this context:1. Multiply rows by columns:
- For each element in the resulting matrix, take a row from the first matrix and a column from the second matrix.
- Perform a dot product; that is, multiply corresponding elements and sum them up.
2. Fill each position in the resulting matrix:
- For example, to find the top-left element of \(CB\), multiply the first row of \(C\) by the first column of \(B\): \((-5 \times 5) + (4 \times 0) + (1 \times 3) = -25 + 0 + 3 = -22\).
- Continue this pattern for each position in the resulting matrix.

This systematic approach helps in achieving accurate calculations, ensuring each element in the resulting matrix is computed correctly. These operations are at the heart of working with matrices and are widely used in various mathematical and real-world applications.

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Most popular questions from this chapter

Solve each system graphically. Give \(x\) - and y-coordinates correct to the nearest hundredth. $$\begin{aligned}y &=5^{x} \\\x y &=1\end{aligned}$$

Graph the solution set of each system of inequalities by hand. $$\begin{aligned} &y > x^{2}+4 x+4\\\ &y < -x^{2} \end{aligned}$$

In certain parts of the Rocky Mountains, deer are the main food source for mountain lions. When the deer population \(d\) is large, the mountain lions ( \(m\) ) thrive. However, a large mountain lion population drives down the size of the deer population. Suppose the fluctuations of the two populations from year to year can be modeled with the matrix equation $$\left[\begin{array}{c} m_{n+1} \\ d_{n+1} \end{array}\right]=\left[\begin{array}{cc} 0.51 & 0.4 \\ -0.05 & 1.05 \end{array}\right]\left[\begin{array}{l} m_{n} \\ d_{n} \end{array}\right]$$ The numbers in the column matrices give the numbers of animals in the two populations after \(n\) years and \(n+1\) years, where the number of deer is measured in hundreds. (a) Give the equation for \(d_{n+1}\) obtained from the second row of the square matrix. Use this equation to determine the rate the deer population will grow from year to year if there are no mountain lions. (b) Suppose we start with a mountain lion population of 2000 and a deer population of \(500,000\) (that is, 5000 hundred deer). How large would each population be after 1 year? 2 years? (c) Consider part (b), but change the initial mountain lion population to \(4000 .\) Show that the populations would both grow at a steady annual rate of \(1 \$ 6\).

Because variables appear in denominators, the system $$\begin{aligned}&\frac{5}{x}+\frac{15}{y}=16\\\&\frac{5}{x}+\frac{4}{y}=5\end{aligned}$$ is not a linear system. However, we can solve it in a manner similar to the method for solving a linear system by using a substitution-of-variable technique. Let \(t=\frac{1}{x}\) and let \(u=\frac{1}{y} \). Write a system of equations in \(t\) and \(u\) by making the appropriate substitutions.

Graph the solution set of each system of inequalities by hand. $$\begin{array}{r} -2 < x < 2 \\ y > 1 \\ x-y > 0 \end{array}$$

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