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A diagonal matrix \(D\) and a matrix \(A\) are given. Find the products \(D A\) and \(A D,\) where possible. $$ \begin{array}{l} D=\left[\begin{array}{ll} 3 & 0 \\ 0 & -1 \end{array}\right] \\ A=\left[\begin{array}{ll} 2 & 4 \\ 6 & 8 \end{array}\right] \end{array} $$

Short Answer

Expert verified
DA = \( \begin{bmatrix} 6 & 12 \\ -6 & -8 \end{bmatrix} \), AD = \( \begin{bmatrix} 6 & -4 \\ 18 & -8 \end{bmatrix} \)

Step by step solution

01

Confirm Matrix Dimensions

To multiply two matrices, check the dimensions. Matrix \( D \) is 2x2 and matrix \( A \) is also 2x2. Since they are both square matrices of the same dimension, both products \( DA \) and \( AD \) are possible.
02

Compute the Product DA

Calculate \( DA \) by multiplying matrix \( D \) with matrix \( A \). The calculation follows the rules of matrix multiplication: for each element in the resultant matrix, multiply the corresponding row from \( D \) by the corresponding column from \( A \). Thus, the resulting matrix \( DA \) is: \[ DA = \begin{bmatrix} (3 \cdot 2 + 0 \cdot 6) & (3 \cdot 4 + 0 \cdot 8) \ (0 \cdot 2 + -1 \cdot 6) & (0 \cdot 4 + -1 \cdot 8) \end{bmatrix} = \begin{bmatrix} 6 & 12 \ -6 & -8 \end{bmatrix} \]
03

Compute the Product AD

Now calculate \( AD \) by multiplying matrix \( A \) with matrix \( D \). Follow the matrix multiplication rules: for each element in the resultant matrix, multiply the corresponding row from \( A \) by the corresponding column from \( D \). Thus, the resulting matrix \( AD \) is: \[ AD = \begin{bmatrix} (2 \cdot 3 + 4 \cdot 0) & (2 \cdot 0 + 4 \cdot -1) \ (6 \cdot 3 + 8 \cdot 0) & (6 \cdot 0 + 8 \cdot -1) \end{bmatrix} = \begin{bmatrix} 6 & -4 \ 18 & -8 \end{bmatrix} \]

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Key Concepts

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

Diagonal Matrix
A diagonal matrix is a special kind of matrix where all the non-diagonal elements are zero. This means that only the elements on the leading diagonal (the diagonal from the top left to the bottom right) are non-zero. Diagonal matrices are significant in mathematics due to their simple structure, which makes them easy to work with.
Consider the matrix \( D = \begin{bmatrix} 3 & 0 \ 0 & -1 \end{bmatrix} \) from the exercise. In this case, the non-zero elements are 3 and -1, located on the diagonal, while the off-diagonal elements are all zeroes.
  • Diagonal matrices are often used in simplifying matrix operations, like multiplication.
  • They are easy to invert, as long as all the diagonal elements are non-zero.
  • Multiplying another matrix by a diagonal matrix often results in scaling the rows or columns of the other matrix.
Understanding diagonal matrices helps simplify many complex operations in linear algebra.
Resultant Matrix
When you multiply two matrices, the outcome is called a resultant matrix. Each element of the resultant matrix is calculated as the sum of the products of corresponding entries from the rows of the first matrix and the columns of the second matrix.
For the matrices \( D \) and \( A \) in our example, the resultant matrix from multiplying \( D \) by \( A \) (represented as \( DA \)) is \( \begin{bmatrix} 6 & 12 \ -6 & -8 \end{bmatrix} \).
  • The position of each element in the resultant matrix is important and reflects the influence of both input matrices.
  • Matrix multiplication is not commutative, meaning \( DA \) and \( AD \) are generally not the same.
  • This specific characteristic also results in different resultant matrices, as shown where \( DA eq AD \) in the exercise.
The concept of the resultant matrix is essential for understanding how matrices interact through multiplication.
Matrix Dimensions
Matrix dimensions fundamentally dictate if two matrices can be multiplied. The dimensions are described by the numbers of rows and columns a matrix has, usually in the format 'rows x columns'.
In our exercise, both matrices \( D \) and \( A \) are 2x2 matrices, meaning they each have 2 rows and 2 columns. This allows them to be multiplied in either order since for matrix multiplication, the number of columns in the first matrix must match the number of rows in the second matrix.
  • If the matrices are not of compatible dimensions, the multiplication is undefined.
  • The resulting product of two matrices will have dimensions defined by the outer dimensions of the multiplied matrices. Here, a product of two 2x2 matrices results in another 2x2 matrix.
  • Always check dimensions first to ensure the matrix multiplication is possible.
Recognizing matrix dimensions simplifies the process of working with matrices, allowing for smoother calculations.

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Most popular questions from this chapter

Matrices \(A\) and \(B\) are defined. (a) Give the dimensions of \(A\) and \(B\). If the dimensions properly match, give the dimensions of \(A B\) and \(B A\). (b) Find the products \(A B\) and \(B A\), if possible. $$ \begin{array}{l} A=\left[\begin{array}{ccc} -4 & -1 & 3 \\ 2 & -3 & 5 \\ 1 & 5 & 3 \end{array}\right] \\ B=\left[\begin{array}{ccc} -2 & 4 & 3 \\ -1 & 1 & -1 \\ 4 & 0 & 2 \end{array}\right] \end{array} $$

Row and column vectors \(\vec{u}\) and \(\vec{v}\) are defined. Find the product \(\vec{u} \vec{v},\) where possible. $$ \vec{u}=\left[\begin{array}{lll} 1 & 2 & 3 \end{array}\right] \quad \vec{v}=\left[\begin{array}{l} 3 \\ 2 \end{array}\right] $$

A matrix \(A\) and vectors \(\vec{b}, \bar{u}\) and \(\vec{v}\) are given. Verify that \(A \vec{u}=\overrightarrow{0}, A \vec{v}=\vec{b}\) and \(A(\vec{u}+\vec{v})=\vec{b}\).$$ A=\left[\begin{array}{ccc} 1 & -1 & 3 \\ 3 & -3 & -3 \\ -1 & 1 & 1 \end{array}\right] $$ $$ \vec{b}=\left[\begin{array}{c} -1 \\ -3 \\ 1 \end{array}\right], \vec{u}=\left[\begin{array}{l} 2 \\ 2 \\ 0 \end{array}\right], \vec{v}=\left[\begin{array}{l} 2 \\ 3 \\ 0 \end{array}\right] $$

Row and column vectors \(\vec{u}\) and \(\vec{v}\) are defined. Find the product \(\vec{u} \vec{v},\) where possible. $$ \vec{u}=\left[\begin{array}{ll} 0.6 & 0.8 \end{array}\right] \quad \vec{v}=\left[\begin{array}{l} 0.6 \\ 0.8 \end{array}\right] $$

A matrix \(A\) and vectors \(\vec{b}, \vec{u}\) and \(\vec{v}\) are given. Verify that \(\vec{u}\) and \(\vec{v}\) are both solutions to the equation \(A \vec{x}=\vec{b} ;\) that is, show that \(A \vec{u}=A \vec{v}=\vec{b}\). $$ \begin{array}{l} A=\left[\begin{array}{cc} 1 & -2 \\ -3 & 6 \end{array}\right] \\ \vec{b}=\left[\begin{array}{c} 2 \\ -6 \end{array}\right], \vec{u}=\left[\begin{array}{c} 0 \\ -1 \end{array}\right], \vec{v}=\left[\begin{array}{l} 2 \\ 0 \end{array}\right] \end{array} $$

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