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Verify that the matrices are inverses of each other. $$\left[\begin{array}{rr}-6 & 5 \\\4 & -3\end{array}\right],\left[\begin{array}{ll}\frac{3}{2} & \frac{5}{2} \\\2 & 3\end{array}\right]$$

Short Answer

Expert verified
By multiplying the matrices, if the resulting matrix is the identity matrix, then it's known that these two matrices are inverses of each other. In this case, the matrices are indeed inverses of one another as the product is the 2x2 identity matrix.

Step by step solution

01

Write the matrices

Write down the given matrices.
02

Multiply the two matrices

We will multiply these matrices. Matrix multiplication can be completed element by element. For instance, to find the element in the first row, first column of the resulting matrix, you would multiply the first row of the first matrix element-by-element with the first column of the second matrix and add up those products.
03

Perform the multiplication

After doing the multiplication, you should have a new 2x2 matrix. Now, do these calculations for every combination of rows from the first matrix and columns from the second matrix.
04

Check the result against the identity matrix

After performing the multiplication, compare the resulting matrix with the identity matrix. The identity matrix you are looking for has 1's along the diagonal and 0's everywhere else.

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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 fundamental operation in linear algebra that combines two matrices to produce a third matrix. However, unlike regular number multiplication, matrix multiplication is not element-wise and has specific rules. You calculate the element in the i-th row and j-th column of the product matrix by taking the dot product of the i-th row of the first matrix with the j-th column of the second matrix.
In our exercise, to calculate the element in the first row and first column of the result, you would take the dot product of the first row of the first matrix with the first column of the second matrix. Repeat this process for each element of the resulting matrix by cycling through all combinations of rows from the first matrix and columns from the second matrix.
It's important to remember that matrix multiplication is only possible when the number of columns in the first matrix equals the number of rows in the second matrix. Also, matrix multiplication is not commutative, meaning that the order in which you multiply the matrices matters. The final product will differ if the order is changed.
Identity Matrix
The identity matrix plays a role similar to the number 1 in regular multiplication. It is a square matrix, meaning it has the same number of rows as columns, and it has 1s on its main diagonal (from the top left to the bottom right) and 0s in all other positions.
When you multiply any matrix by its corresponding identity matrix (which must be the same size), the result is the original matrix. This property is akin to multiplying any number by 1; it doesn't change the number.
In the context of our exercise, after multiplying the two matrices, we check to see if the result is an identity matrix with 1s on its diagonal and 0s elsewhere. If it is, we can say that the matrices are indeed inverses of each other, because just like 1 is the multiplicative identity for numbers, the identity matrix serves as the multiplicative identity for matrices.
Inverse of a Matrix
The inverse of a matrix is akin to the reciprocal of a number. It is a matrix that, when multiplied by the original matrix, yields the identity matrix. Finding the inverse is crucial for solving systems of linear equations and for understanding linear transformations.
However, not all matrices have inverses. A matrix must be square and have a property called 'non-singularity' to have an inverse. To illustrate, as in our exercise, if multiplying matrix A by matrix B gives us the identity matrix, it implies that B is A's inverse, and vice versa.
Understanding matrix inverses is not only about finding them but also recognizing their significance in solving linear equations. When we multiply both sides of a matrix equation by the inverse of the coefficient matrix, we isolate the variable matrix, much like dividing both sides of a regular equation by the same non-zero number.

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Most popular questions from this chapter

Find the decoding matrix for each encoding matrix. $$\left[\begin{array}{ll}5 & 7 \\\2 & 3\end{array}\right]$$

Find \(A^{2}\) (the product \(A A\) ) and \(A^{3}\) (the prod\(\left.u c t\left(A^{2}\right) A\right)\). $$A=\left[\begin{array}{rr}-4 & 0 \\\0 & 3\end{array}\right]$$

A farmer has 90 acres available for planting corn and soybeans. The cost of seed per acre is \(\$ 4\) for corn and \(\$ 6\) for soybeans. To harvest the crops, the farmer will need to hire some temporary help. It will cost the farmer \(\$ 20\) per acre to harvest the corn and \(\$ 10\) per acre to harvest the soybeans. The farmer has \(\$ 480\) available for seed and \(\$ 1400\) available for labor. His profit is \(\$ 120\) per acre of corn and \(\$ 150\) per acre of soybeans. How many acres of each crop should the farmer plant to maximize the profit?

This set of exercises will draw on the ideas presented in this section and your general math background. Compute \(A(B C)\) and \((A B) C,\) where \(A=\left[\begin{array}{rr}3 & -1 \\ 0 & 2\end{array}\right], \quad B=\left[\begin{array}{ll}1 & 4 \\ 0 & 1\end{array}\right], \quad\) and \(\quad C=\left[\begin{array}{rr}-1 & 0 \\ 3 & 1\end{array}\right]\) What do you observe?

$$\begin{aligned}&\text { If } A=\left[\begin{array}{cc}4 a+5 & -1 \\\\-4 & -7\end{array}\right] \text { and } B=\left[\begin{array}{rr}7 & 0 \\\\-4 & -8\end{array}\right], \text { for what }\\\&\text { value(s) of } a \text { does } 2 B-3 A=\left[\begin{array}{ll}2 & 3 \\\4 & 5\end{array}\right] ?\end{aligned}$$

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