/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Calculate the products \(A B\) a... [FREE SOLUTION] | 91Ó°ÊÓ

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Calculate the products \(A B\) and \(B A\) to verify that \(B\) is the inverse of \(A .\) $$ A=\left[\begin{array}{ll}{4} & {1} \\ {7} & {2}\end{array}\right] \quad B=\left[\begin{array}{rr}{2} & {-1} \\ {-7} & {4}\end{array}\right] $$

Short Answer

Expert verified
Matrix \( B \) is the inverse of matrix \( A \).

Step by step solution

01

Understand the Condition for Matrix Inverses

For a matrix \( B \) to be the inverse of matrix \( A \), the products \( AB \) and \( BA \) must both equal the identity matrix \( I \), which for 2x2 matrices is \( I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \).
02

Calculate the Product \( AB \)

Multiply the matrices \( A \) and \( B \):\[AB = \begin{bmatrix} 4 & 1 \ 7 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 \ -7 & 4 \end{bmatrix} = \begin{bmatrix} (4 \cdot 2 + 1 \cdot -7) & (4 \cdot -1 + 1 \cdot 4) \ (7 \cdot 2 + 2 \cdot -7) & (7 \cdot -1 + 2 \cdot 4) \end{bmatrix}\]Calculating each element:- \((4 \cdot 2) + (1 \cdot -7) = 8 - 7 = 1\)- \((4 \cdot -1) + (1 \cdot 4) = -4 + 4 = 0\)- \((7 \cdot 2) + (2 \cdot -7) = 14 - 14 = 0\)- \((7 \cdot -1) + (2 \cdot 4) = -7 + 8 = 1\)Thus, \( AB = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \).
03

Calculate the Product \( BA \)

Multiply the matrices \( B \) and \( A \):\[BA = \begin{bmatrix} 2 & -1 \ -7 & 4 \end{bmatrix} \begin{bmatrix} 4 & 1 \ 7 & 2 \end{bmatrix} = \begin{bmatrix} (2 \cdot 4 + (-1) \cdot 7) & (2 \cdot 1 + (-1) \cdot 2) \ (-7 \cdot 4 + 4 \cdot 7) & (-7 \cdot 1 + 4 \cdot 2) \end{bmatrix}\]Calculating each element:- \((2 \cdot 4) + (-1 \cdot 7) = 8 - 7 = 1\)- \((2 \cdot 1) + (-1 \cdot 2) = 2 - 2 = 0\)- \((-7 \cdot 4) + (4 \cdot 7) = -28 + 28 = 0\)- \((-7 \cdot 1) + (4 \cdot 2) = -7 + 8 = 1\)Thus, \( BA = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \).
04

Verify the Inverse

Since both \( AB = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \) and \( BA = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \), matrix \( B \) is indeed the inverse of matrix \( A \).

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Key Concepts

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

Matrix Multiplication
Matrix multiplication is a critical operation in linear algebra, and it involves combining rows from one matrix with columns from another. When multiplying matrices, each element in the resulting matrix is obtained by taking the dot product of corresponding rows and columns.
For example, given two matrices, say matrix \( A \) with dimensions \( m \times n \) and matrix \( B \) with dimensions \( n \times p \), the resultant matrix \( AB \) will have dimensions \( m \times p \). Each element \( c_{ij} \) of the product matrix is calculated as follows:
  • Take the \( i \)-th row of matrix \( A \) and the \( j \)-th column of matrix \( B \).
  • Multiply corresponding elements and sum them up: \( c_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + \, ... \, + a_{in}b_{nj} \).
Remember, matrix multiplication is not commutative. This means \( AB \) is not generally equal to \( BA \). Explore this operation on different sizes of matrices to see its behavior.
Identity Matrix
The identity matrix is a cornerstone concept in the realm of matrices. It functions as the multiplicative identity in matrix algebra, similar to the number 1 in regular arithmetic.
For a square matrix of size \( n \times n \), the identity matrix is denoted by \( I_n \) and holds the following properties:
  • All the diagonal elements are 1.
  • All the off-diagonal elements are 0.
For instance, the identity matrix for a 2x2 matrix is:\[I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\] When any matrix \( A \) is multiplied by an identity matrix \( I \) of compatible dimensions, the result is matrix \( A \) itself:
\[ AI = A \quad \text{and} \quad IA = A \]This property is vital when verifying whether one matrix is the inverse of another.
2x2 Matrices
2x2 matrices are among the simplest types of matrices and are often used as a foundation for understanding more complex matrix operations. Each 2x2 matrix consists of four elements, laid out in two rows and two columns.

Consider a 2x2 matrix \( A \):\[A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\] Operations on such matrices, such as addition, subtraction, and multiplication, follow specific rules that make use of the individual elements.
The determinant of \( A \), a scalar value, is calculated as \( ad - bc \). This value is particularly important when discussing inverses; for example, a 2x2 matrix will only have an inverse if its determinant is non-zero. 2x2 matrices are an excellent starting point for exploring the properties of linear transformations and matrix equations.
Verification of Inverse
Verifying that one matrix is the inverse of another involves a particular procedure. An inverse matrix \( B \) of a matrix \( A \) should satisfy the condition that the products \( AB \) and \( BA \) yield the identity matrix.

To check this for 2x2 matrices, you can follow these steps:
  • Compute the product \( AB \) and ensure it equals the identity matrix.
  • Compute the product \( BA \) to confirm it also equals the identity matrix.
If both conditions hold true, then \( B \) is indeed the inverse of \( A \). For instance, if \( A = \begin{bmatrix} 4 & 1 \ 7 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 2 & -1 \ -7 & 4 \end{bmatrix} \), and both products result in the identity matrix:\[AB = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \quad \text{and} \quad BA = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\] Then \( B \) is confirmed as the inverse of \( A \), reinforcing the concept of matrix inverses in linear algebra.

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Most popular questions from this chapter

\(47-50\) . Use a graphing calculator to graph the solution of the system of inequalities. Find the coordinates of all vertices, rounded to one decimal place. $$ \left\\{\begin{array}{c}{y \leq 6 x-x^{2}} \\ {x+y \geq 4}\end{array}\right. $$

Shading Unwanted Regions To graph the solution of a system of inequalities, we have shaded the solution of each inequality in a different color; the solution of the system is the region where all the shaded parts overlap. Here is a different method: For each inequality, shade the region that does not satisfy the inequality. Explain why the part of the plane that is left unshaded is the solution of the system. Solve the following system by both methods. Which do you prefer? Why? $$ \left\\{\begin{aligned} x+2 y &>4 \\\\-x+y &<1 \\ x+3 y &<9 \\ x &<3 \end{aligned}\right. $$

Evaluate the determinant, using row or column operations whenever possible to simplify your work. $$ \left|\begin{array}{rrrr}{-2} & {3} & {-1} & {7} \\ {4} & {6} & {-2} & {3} \\\ {7} & {7} & {0} & {5} \\ {3} & {-12} & {4} & {0}\end{array}\right| $$

Find the determinant of the matrix. Determine whether the matrix has an inverse, but don’t calculate the inverse. $$ \left[\begin{array}{lll}{1} & {3} & {7} \\ {2} & {0} & {8} \\ {0} & {2} & {2}\end{array}\right] $$

Classroom Use A small school has 100 students who occupy three classrooms: \(A, B,\) and \(C .\) After the first period of the school day, half the students in room A move to room B, one-fifth of the students in room B move to room \(C,\) and one-third of the students in room \(C\) move to room A. Nevertheless, the total number of students in each room is the same for both periods. How many students occupy each room?

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