Sure! Let's solve the problem step-by-step. We need to display the given data in a matrix [tex]\( A \)[/tex] with columns indicating the years, and then identify the element [tex]\( a_{15} \)[/tex].
Here's the data presented:
[tex]\[
\begin{array}{|l|c|c|c|c|c|c|}
\hline
& 1980 & 1985 & 1990 & 1992 & 1995 & 1997 \\
\hline
\text{ColorTVs (millions)} & 63 & 78 & 90 & 91 & 94 & 97 \\
\hline
\text{VCRs (millions)} & 1 & 18 & 63 & 69 & 77 & 82 \\
\hline
\end{array}
\][/tex]
1. Constructing Matrix [tex]\( A \)[/tex]:
We organize the data into a matrix where the first row represents the number of households with Color TVs, and the second row represents the number of households with VCRs. The columns represent the different years: 1980, 1985, 1990, 1992, 1995, and 1997.
[tex]\[
A = \begin{pmatrix}
63 & 78 & 90 & 91 & 94 & 97 \\
1 & 18 & 63 & 69 & 77 & 82
\end{pmatrix}
\][/tex]
2. Identifying Element [tex]\( a_{15} \)[/tex]:
In the notation [tex]\( a_{ij} \)[/tex], [tex]\( i \)[/tex] represents the row number, and [tex]\( j \)[/tex] represents the column number, both in 1-based indexing.
Therefore, [tex]\( a_{15} \)[/tex] indicates the element in the 1st row and the 5th column. Referring to matrix [tex]\( A \)[/tex], we locate this element:
[tex]\[
A = \begin{pmatrix}
63 & 78 & 90 & 91 & 94 & 97 \\
1 & 18 & 63 & 69 & 77 & 82
\end{pmatrix}
\][/tex]
From the first row (Color TVs), the elements are: [tex]\( 63, 78, 90, 91, 94, 97 \)[/tex].
The 5th column element in the first row is [tex]\( 94 \)[/tex].
Therefore, [tex]\( a_{15} \)[/tex] is [tex]\( 94 \)[/tex].
Conclusion:
The element [tex]\( a_{15} \)[/tex] in matrix [tex]\( A \)[/tex] is [tex]\( 94 \)[/tex].
