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

Short Answer

Expert verified
Products: \(DA = \begin{bmatrix} 4 & 8 \\ -3 & -6 \end{bmatrix}\) and \(AD = \begin{bmatrix} 4 & -6 \\ 4 & -6 \end{bmatrix}\).

Step by step solution

01

Verify Matrix Dimensions

To multiply two matrices, check if the number of columns in the first matrix equals the number of rows in the second matrix. For matrices \(D\) and \(A\), both are 2x2, so multiplication is possible in both orders.
02

Calculate Product DA

Multiply matrix \(D\) by matrix \(A\) by taking the dot product of the rows of \(D\) with the columns of \(A\).\[ DA = \begin{bmatrix} 4 & 0 \ 0 & -3 \end{bmatrix} \begin{bmatrix} 1 & 2 \ 1 & 2 \end{bmatrix} = \begin{bmatrix} 4*1 + 0*1 & 4*2 + 0*2 \ 0*1 + (-3)*1 & 0*2 + (-3)*2 \end{bmatrix} = \begin{bmatrix} 4 & 8 \ -3 & -6 \end{bmatrix} \]
03

Calculate Product AD

Multiply matrix \(A\) by matrix \(D\) by taking the dot product of the rows of \(A\) with the columns of \(D\).\[ AD = \begin{bmatrix} 1 & 2 \ 1 & 2 \end{bmatrix} \begin{bmatrix} 4 & 0 \ 0 & -3 \end{bmatrix} = \begin{bmatrix} 1*4 + 2*0 & 1*0 + 2*(-3) \ 1*4 + 2*0 & 1*0 + 2*(-3) \end{bmatrix} = \begin{bmatrix} 4 & -6 \ 4 & -6 \end{bmatrix} \]

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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 type of square matrix where all elements outside the main diagonal are zero. The main diagonal is the collection of elements from the top-left to the bottom-right. This matrix type is special because operations involving diagonal matrices can be simplified greatly. For example, when multiplying a diagonal matrix by another matrix, each element in the result is simply the original element from the other matrix multiplied by the corresponding diagonal element. In the given example, matrix \( D \) is a diagonal matrix with values \( 4 \) and \( -3 \) on the main diagonal. This makes calculations straightforward, as it simplifies the process of matrix multiplication.
Matrix Dimensions
Understanding matrix dimensions is crucial when performing matrix operations such as multiplication. The dimension of a matrix is typically represented as \( m \times n \), which denotes the number of rows \( m \) and columns \( n \) in the matrix. For example, matrix \( D \) in our problem is a 2x2 matrix, meaning it has 2 rows and 2 columns. Similarly, matrix \( A \) is also 2x2. When multiplying matrices, it's essential that the number of columns in the first matrix equals the number of rows in the second. Since both \( D \) and \( A \) are 2x2, multiplication is possible in both the \( DA \) and \( AD \) order.
Dot Product
The dot product is a fundamental concept in matrix multiplication. It involves multiplying corresponding entries of one row by the entries of a column, then summing those products. This operation is repeated for each pair of rows and columns. In our example, when calculating \( DA \), each row of matrix \( D \) is multiplied by each column of matrix \( A \). For instance, the first element of the resulting matrix is calculated by multiplying the first element of the first row of \( D \) with the first element of the first column of \( A \), and then adding the product of the second elements. This results in "dotting" across matrices, hence the term 'dot product.'
Order of Multiplication
Matrix multiplication is not commutative, which means that the order in which matrices are multiplied affects the result. \( DA \) is generally not the same as \( AD \). In our problem, although both products are possible due to the compatible dimensions of \( D \) and \( A \), their results differ. The order of multiplication determines the arrangement of numbers and final outcome of the matrix product. Understanding this concept helps avoid errors in matrix calculations. Always verify the correct order during multiplication to ensure meaningful results.

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Most popular questions from this chapter

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 \(A \vec{u}=A \vec{v}=\vec{b}\). $$ \begin{array}{l} A=\left[\begin{array}{cccc} 0 & -3 & -1 & -3 \\ -4 & 2 & -3 & 5 \end{array}\right] \\ \vec{b}=\left[\begin{array}{c} 48 \\ 36 \end{array}\right], \vec{u}=\left[\begin{array}{c} -17 \\ -16 \\ 0 \\ 0 \end{array}\right] \\ \vec{v}=\left[\begin{array}{c} -8 \\ -28 \\ 0 \\ 12 \end{array}\right] \end{array} $$

A diagonal matrix \(D\) and a matrix \(A\) are given. Find the products \(D A\) and \(A D,\) where possible. $$ \begin{array}{l} D=\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right] \\ A=\left[\begin{array}{lll} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array}\right] \end{array} $$

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. $$ A=\left[\begin{array}{ll} 3 & 7 \\ 2 & 5 \end{array}\right] B=\left[\begin{array}{ll} 1 & -1 \\ 3 & -3 \end{array}\right] $$

Vectors \(\vec{x}\) and \(\vec{y}\) are given. Sketch \(\vec{x}, \vec{y}, \vec{x}+\vec{y},\) and \(\vec{x}-\vec{y}\) on the same Cartesian axes. $$ \vec{x}=\left[\begin{array}{c} -1 \\ 1 \end{array}\right], \vec{y}=\left[\begin{array}{c} -2 \\ 2 \end{array}\right] $$

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 \(A(\vec{u}+\vec{v})=\vec{b}\).$$ \begin{array}{l} A=\left[\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & -3 \\ 3 & 1 & -3 \end{array}\right] \\ \vec{b}=\left[\begin{array}{c} 2 \\ -4 \\ -1 \end{array}\right], \vec{u}=\left[\begin{array}{l} 0 \\ 6 \\ 2 \end{array}\right], \vec{v}=\left[\begin{array}{c} 1 \\ -1 \\ 1 \end{array}\right] \end{array} $$

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