To solve for [tex]\(\left|A^2 - B^2\right|\)[/tex], we need to follow several steps that involve matrix operations. Let's break it down step-by-step.
### Step 1: Calculate [tex]\(A^2\)[/tex]
Given the matrix [tex]\(A = \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}\)[/tex],
[tex]\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix} \][/tex]
Perform the matrix multiplication:
[tex]\[
A^2 = \begin{pmatrix}
(1 \cdot 1 + 2 \cdot 2) & (1 \cdot 2 + 2 \cdot -1) \\
(2 \cdot 1 + -1 \cdot 2) & (2 \cdot 2 + -1 \cdot -1)
\end{pmatrix}
= \begin{pmatrix}
1 + 4 & 2 - 2 \\
2 - 2 & 4 + 1
\end{pmatrix}
= \begin{pmatrix}
5 & 0 \\
0 & 5
\end{pmatrix}
\][/tex]
### Step 2: Calculate [tex]\(B^2\)[/tex]
Given the matrix [tex]\(B = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}\)[/tex],
[tex]\[ B^2 = B \cdot B = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \][/tex]
Perform the matrix multiplication:
[tex]\[
B^2 = \begin{pmatrix}
(1 \cdot 1 + 0 \cdot 2) & (1 \cdot 0 + 0 \cdot 1) \\
(2 \cdot 1 + 1 \cdot 2) & (2 \cdot 0 + 1 \cdot 1)
\end{pmatrix}
= \begin{pmatrix}
1 + 0 & 0 \\
2 + 2 & 1
\end{pmatrix}
= \begin{pmatrix}
1 & 0 \\
4 & 1
\end{pmatrix}
\][/tex]
### Step 3: Calculate [tex]\(A^2 - B^2\)[/tex]
[tex]\[
A^2 - B^2 = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 4 & 1 \end{pmatrix}
= \begin{pmatrix} 5 - 1 & 0 - 0 \\ 0 - 4 & 5 - 1 \end{pmatrix}
= \begin{pmatrix} 4 & 0 \\ -4 & 4 \end{pmatrix}
\][/tex]
### Step 4: Calculate the determinant of [tex]\(\begin{pmatrix} 4 & 0 \\ -4 & 4 \end{pmatrix}\)[/tex]
The determinant of a 2x2 matrix [tex]\(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\)[/tex] is given by [tex]\(ad - bc\)[/tex].
[tex]\[
\left|\begin{pmatrix} 4 & 0 \\ -4 & 4 \end{pmatrix}\right| = (4 \cdot 4) - (0 \cdot -4) = 16 - 0 = 16
\][/tex]
Therefore, the value of [tex]\(\left|A^2 - B^2\right|\)[/tex] is [tex]\(16\)[/tex].
