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Let $$ A=\left[\begin{array}{rr} 1 & 3 \\ -2 & -1 \end{array}\right] \text { and } B=\left[\begin{array}{ll} 3 & -4 \\ 2 & -2 \end{array}\right] $$ a. Find \(A^{T}\) and show that \(\left(A^{T}\right)^{T}=A\). b. Show that \((A+B)^{T}=A^{T}+B^{T}\). c. Show that \((A B)^{T}=B^{T} A^{T}\).

Short Answer

Expert verified
a. \(A^T = \begin{bmatrix} 1 & -2 \\ 3 & -1 \end{bmatrix}\), and \((A^T)^T = A\). b. \((A+B)^T = A^T + B^T\), where \((A+B)^T = \begin{bmatrix} 4 & 0 \\ -1 & -3 \end{bmatrix}\). c. \((AB)^T = B^T A^T\), where \((AB)^T = \begin{bmatrix} 9 & -4 \\ -10 & 2 \end{bmatrix}\).

Step by step solution

01

Finding A^T

To find the transpose of matrix A, we interchange its rows and columns. In other words, the element at row i and column j in the original matrix A will now be placed at row j and column i in the transpose. So, we have: \(A =\begin{bmatrix} 1 & 3 \\ -2 & -1 \end{bmatrix}\) \(A^T = \begin{bmatrix} 1 & -2 \\ 3 & -1 \end{bmatrix}\)
02

Find (A^T)^T and compare with A

Now, we take the transpose of A^T: \((A^T)^T = \begin{bmatrix} 1 & 3 \\ -2 & -1 \end{bmatrix}\) As we can see, \((A^T)^T = A\). #b. Show (A+B)^T = A^T + B^T#
03

Compute A+B

First, we need to compute A+B: \(A + B = \begin{bmatrix} 1+3 & 3+(-4) \\ -2+2 & -1+(-2) \end{bmatrix} = \begin{bmatrix} 4 & -1 \\ 0 & -3 \end{bmatrix}\)
04

Compute (A+B)^T, A^T and B^T

Now, we need to find the transpose of (A+B), A^T, and B^T: \((A+B)^T = \begin{bmatrix} 4 & 0 \\ -1 & -3 \end{bmatrix}\) \(A^T = \begin{bmatrix} 1 & -2 \\ 3 & -1 \end{bmatrix}\) \(B = \begin{bmatrix} 3 & -4 \\ 2 & -2 \end{bmatrix}\) \(B^T = \begin{bmatrix} 3 & 2 \\ -4 & -2 \end{bmatrix}\)
05

Compare (A+B)^T and A^T + B^T

Finally, let's compute A^T + B^T and compare it with (A+B)^T: \(A^T + B^T = \begin{bmatrix} 4 & 0 \\ -1 & -3 \end{bmatrix}\) We can see that \((A+B)^T = A^T + B^T\). #c. Show (AB)^T = B^T A^T#
06

Compute the product AB

First, we need to compute the product of the matrices A and B: \(AB = \begin{bmatrix} 1 & 3 \\ -2 & -1 \end{bmatrix}\begin{bmatrix} 3 & -4 \\ 2 & -2 \end{bmatrix} = \begin{bmatrix} 9 & -10 \\ -4 & 2\end{bmatrix}\)
07

Compute (AB)^T, B^T, and A^T

Now, we need to find the transpose of (AB), B^T, and A^T: \((AB)^T = \begin{bmatrix} 9 & -4 \\ -10 & 2 \end{bmatrix}\) \(B^T = \begin{bmatrix} 3 & 2 \\ -4 & -2 \end{bmatrix}\) \(A^T = \begin{bmatrix} 1 & -2 \\ 3 & -1 \end{bmatrix}\)
08

Compute B^T A^T and compare with (AB)^T

Finally, let's compute B^T A^T and compare it with (AB)^T: \(B^T A^T = \begin{bmatrix} 3 & 2 \\ -4 & -2 \end{bmatrix}\begin{bmatrix} 1 & -2 \\ 3 & -1 \end{bmatrix} = \begin{bmatrix} 9 & -4 \\ -10 & 2 \end{bmatrix}\) We can see that \((AB)^T = B^T A^T\).

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Key Concepts

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

Matrix Operations
Understanding matrix operations is crucial for anyone delving into linear algebra or related fields. Matrix operations include a variety of procedures that can be performed on matrices, such as addition, subtraction, multiplication, and transposition. Each operation has specific rules that must be followed for the operation to be valid.

For instance, when adding or subtracting matrices, they must have the same dimensions; the operations are executed element-wise. Multiplication is a bit more complex; the number of columns in the first matrix must match the number of rows in the second matrix to multiply them.

Another vital operation is the transpose of a matrix, where the matrix is flipped over its diagonal, meaning the column and row indices of each element are swapped. The transpose of a matrix is not merely a reorganization of elements; it often holds special significance in linear transformations and inner product spaces.
Transpose of a Matrix
The transpose of a matrix is an operation that flips a matrix over its diagonal. In mathematical terms, the transpose of a matrix \(A\), denoted \(A^T\), is obtained by interchanging rows and columns. This means that the element in the i-th row and j-th column of \(A\) becomes the element in the j-th row and i-th column of \(A^T\).

One interesting property of the transpose operation is that transposing a matrix twice will return the original matrix, i.e., \( (A^T)^T = A \). Understanding this property can be particularly helpful in simplifying expressions involving transposes.
Matrix Addition
Moving on to matrix addition, this operation involves adding corresponding elements from two matrices of the same size. The result is a new matrix also of the same size, where each element is the sum of the corresponding elements. The rule of thumb is that if \(A\) and \(B\) are matrices of the same dimension \(m \times n\), their sum \(A+B\) is also an \(m \times n\) matrix where the element at position \(i, j\) is the sum of elements \(A_{ij}\) and \(B_{ij}\).

Considering the transpose operation in this context, if you first sum two matrices and then take the transpose, it's equivalent to transposing the two matrices individually and then summing them: \( (A+B)^T = A^T + B^T \). This property can simplify the cumbersome task of transposing each matrix individually before addition.
Matrix Multiplication
Lastly, matrix multiplication is a fundamental operation that is not as straightforward as addition or subtraction. For two matrices \(A\) and \(B\) to be multiplicable, the number of columns in \(A\) must be equal to the number of rows in \(B\). The resulting matrix has dimensions that are determined by the rows of \(A\) and the columns of \(B\).

The product of \(A \times B\) is computed by taking the dot product of the rows of \(A\) with the columns of \(B\). An important property associated with matrix multiplication and transposition is that the transpose of a product is equal to the product of the transposes in reverse order: \( (AB)^T = B^T A^T \). This is a non-intuitive property because it does not align with the commutative property of scalar multiplication but is crucial in matrix theory.

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Most popular questions from this chapter

Solve the system of linear equations using the Gauss-Jordan elimination method. $$ \begin{array}{l} 2 x+3 y-2 z=10 \\ 3 x-2 y+2 z=0 \\ 4 x-y+3 z=-1 \end{array} $$

The problems in exercise correspond to those in exercises 15-27, Section 2.1. Use the results of your previous work to help you solve these problems. A dietitian wishes to plan a meal around three foods. The percent of the daily requirements of proteins, carbohydrates, and iron contained in each ounce of the three foods is summarized in the following table: $$\begin{array}{lccc} \hline & \text { Food I } & \text { Food II } & \text { Food III } \\ \hline \text { Proteins }(\%) & 10 & 6 & 8 \\ \hline \text { Carbohydrates }(\%) & 10 & 12 & 6 \\ \hline \text { Iron }(\%) & 5 & 4 & 12 \\ \hline \end{array}$$ Determine how many ounces of each food the dietitian should include in the meal to meet exactly the daily requirement of proteins, carbohydrates, and iron \((100 \%\) of each).

Solve the system of linear equations using the Gauss-Jordan elimination method. $$ \begin{array}{r} 2 x+2 y+z=9 \\ x+z=4 \\ 4 y-3 z=17 \end{array} $$

Hartman Lumber Company has two branches in the city. The sales of four of its products for the last year (in thousands of dollars) are represented by the matrix $$ B=\begin{array}{l} \text { Branch I } \\ \text { Branch II } \end{array} \quad\left[\begin{array}{rrrr} 5 & 3 & C & D \\ 3 & 4 & 6 & 8 \end{array}\right] $$ For the present year, management has projected that the sales of the four products in branch I will be \(10 \%\) more than the corresponding sales for last year and the sales of the four products in branch II will be \(15 \%\) more than the corresponding sales for last year. a. Show that the sales of the four products in the two branches for the current year are given by the matrix \(A B\), where $$ A=\left[\begin{array}{ll} 1.1 & 0 \\ 0 & 1.15 \end{array}\right] $$ Compute \(A B\). b. Hartman has \(m\) branches nationwide, and the sales of \(n\) of its products (in thousands of dollars) last year are represented by the matrix Product $$ \left.B=\begin{array}{c} 1 & 2 & 3 & \cdots & n \\ \text { Branch } 1 \\ \text { Branch 2 } \\ \vdots & a_{11} & a_{12} & a_{13} & \cdots & a_{1 n} \\ a_{21} & a_{22} & a_{23} & \cdots & a_{2 n} \\ \vdots & \vdots & \vdots & & \vdots \\ a_{m 1} & a_{m 2} & a_{m 3} & \cdots & a_{m n} \end{array}\right] $$ Also, management has projected that the sales of the \(n\) products in branch 1 , branch \(2, \ldots\), branch \(m\) will be \(r_{1} \%, r_{2} \%, \ldots, r_{m} \%\), respectively, more than the corresponding sales for last year. Write the matrix \(A\) such that \(A B\) gives the sales of the \(n\) products in the \(m\) branches for the current year.

Let $$ A=\left[\begin{array}{rr} 2 & 3 \\ -4 & -5 \end{array}\right] $$ a. Find \(A^{-1}\). b. Show that \(\left(A^{-1}\right)^{-1}=A\).

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