Problem 91 - Precalculus with Limits

Question

Let \(A\) and \(B\) be unequal diagonal matrices of the same order. (A diagonal matrix is a square matrix in which each entry not on the main diagonal is zero.) Determine the products \(A B\) for several pairs of such matrices. Make a conjecture about a quick rule for such products.

Step-by-Step Solution

Answer

If \(A\) and \(B\) are diagonal matrices given by \(A =\begin{bmatrix}a & 0 & 0 \\0 & b & 0 \\0 & 0 & c \end{bmatrix}\) and \(B = \begin{bmatrix}d & 0 & 0 \\0 & e & 0 \\0 & 0 & f \end{bmatrix}\) respectively, then their product \(AB = \begin{bmatrix}ad & 0 & 0 \\0 & be & 0 \\0 & 0 & cf \end{bmatrix}\). Therefore, the product of two diagonal matrices is a diagonal matrix whose entries are the products of the corresponding entries in the original matrices.

Step 1: Defining the Matrices

Let us first define two 3x3 diagonal matrices \(A\) and \(B\). We state \(A =\begin{bmatrix}a & 0 & 0 \\0 & b & 0 \\0 & 0 & c \end{bmatrix}\)and \(B = \begin{bmatrix}d & 0 & 0 \\0 & e & 0 \\0 & 0 & f \end{bmatrix}\) where \(a\), \(b\), \(c\), \(d\), \(e\), and \(f\) are non-zero entries.

Step 2: Calculating the Product AB

When we multiply the matrices \(A\) and \(B\) we get, \(AB = \begin{bmatrix}a & 0 & 0 \\0 & b & 0 \\0 & 0 & c \end{bmatrix}\begin{bmatrix}d & 0 & 0 \\0 & e & 0 \\0 & 0 & f \end{bmatrix} = \begin{bmatrix}ad & 0 & 0 \\0 & be & 0 \\0 & 0 & cf \end{bmatrix}\)

Step 3: Making a Conjecture

From the multiplication, we can observe that the entries on the diagonal of the resulting matrix are product of the corresponding entries in the original matrices. Remembering that diagonal matrices are zero everywhere except the diagonal, we can conclude with: The product of two diagonal matrices is a diagonal matrix whose entries are the products of the corresponding entries from the original matrices.