Let's consider a [tex]\(2 \times 2\)[/tex] matrix [tex]\(A\)[/tex] and its inverse [tex]\(A^{-1}\)[/tex]. The inverse of a matrix [tex]\(A\)[/tex], denoted as [tex]\(A^{-1}\)[/tex], is a matrix such that when it is multiplied with [tex]\(A\)[/tex], it yields the identity matrix of the same dimension.
Mathematically, this can be expressed as:
[tex]\[ A \cdot A^{-1} = I \][/tex]
where [tex]\(I\)[/tex] is the identity matrix.
For a [tex]\(2 \times 2\)[/tex] matrix, the identity matrix is:
[tex]\[ I = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix} \][/tex]
1. Given that [tex]\(A\)[/tex] is a [tex]\(2 \times 2\)[/tex] matrix, let's denote it as:
[tex]\[ A = \begin{pmatrix}
a & b \\
c & d
\end{pmatrix} \][/tex]
2. Furthermore, suppose [tex]\(A^{-1}\)[/tex] is the inverse of [tex]\(A\)[/tex] and can be represented as:
[tex]\[ A^{-1} = \begin{pmatrix}
e & f \\
g & h
\end{pmatrix} \][/tex]
3. Upon multiplying [tex]\(A\)[/tex] with [tex]\(A^{-1}\)[/tex], we should get:
[tex]\[ A \cdot A^{-1} = \begin{pmatrix}
a & b \\
c & d
\end{pmatrix} \cdot \begin{pmatrix}
e & f \\
g & h
\end{pmatrix} = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix} \][/tex]
4. Performing the matrix multiplication, we get:
[tex]\[ \begin{pmatrix}
(a \cdot e + b \cdot g) & (a \cdot f + b \cdot h) \\
(c \cdot e + d \cdot g) & (c \cdot f + d \cdot h)
\end{pmatrix} = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix} \][/tex]
This shows that each product of corresponding elements and their sums must yield the elements of the identity matrix, confirming that:
[tex]\[ a \cdot e + b \cdot g = 1 \][/tex]
[tex]\[ a \cdot f + b \cdot h = 0 \][/tex]
[tex]\[ c \cdot e + d \cdot g = 0 \][/tex]
[tex]\[ c \cdot f + d \cdot h = 1 \][/tex]
Therefore, the product of matrix [tex]\(A\)[/tex] and its inverse [tex]\(A^{-1}\)[/tex] results in the identity matrix:
[tex]\[ A \cdot A^{-1} = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix} \][/tex]
So, the answer is:
[tex]\[ A \cdot A^{-1} = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix} \][/tex]
