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To determine which matrix [tex]\( A \)[/tex] satisfies [tex]\( A^2 = I \)[/tex], where [tex]\( I \)[/tex] is the 2x2 identity matrix, we need to verify each given matrix by squaring it and checking if the result is the identity matrix. The identity matrix [tex]\( I \)[/tex] in 2x2 form is:
[tex]\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \][/tex]
We will examine each option:
Option A:
[tex]\[ A = \begin{pmatrix} -5 & 6 \\ 4 & 5 \end{pmatrix} \][/tex]
Calculating [tex]\( A^2 \)[/tex] involves:
[tex]\[ A^2 = A \cdot A = \begin{pmatrix} -5 & 6 \\ 4 & 5 \end{pmatrix} \cdot \begin{pmatrix} -5 & 6 \\ 4 & 5 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} (-5)(-5) + (6)(4) & (-5)(6) + (6)(5) \\ (4)(-5) + (5)(4) & (4)(6) + (5)(5) \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 25 + 24 & -30 + 30 \\ -20 + 20 & 24 + 25 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 49 & 0 \\ 0 & 49 \end{pmatrix} \][/tex]
Clearly, [tex]\( A^2 \neq I \)[/tex].
Option B:
[tex]\[ A = \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix} \][/tex]
Calculating [tex]\( A^2 \)[/tex]:
[tex]\[ A^2 = A \cdot A = \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} (-1)(-1) + (1)(1) & (-1)(1) + (1)(1) \\ (1)(-1) + (1)(1) & (1)(1) + (1)(1) \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 1 + 1 & -1 + 1 \\ -1 + 1 & 1 + 1 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \][/tex]
Clearly, [tex]\( A^2 \neq I \)[/tex].
Option C:
[tex]\[ A = \begin{pmatrix} 3 & -2 \\ 4 & -3 \end{pmatrix} \][/tex]
Calculating [tex]\( A^2 \)[/tex]:
[tex]\[ A^2 = A \cdot A = \begin{pmatrix} 3 & -2 \\ 4 & -3 \end{pmatrix} \cdot \begin{pmatrix} 3 & -2 \\ 4 & -3 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} (3)(3) + (-2)(4) & (3)(-2) + (-2)(-3) \\ (4)(3) + (-3)(4) & (4)(-2) + (-3)(-3) \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 9 - 8 & -6 + 6 \\ 12 - 12 & -8 + 9 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \][/tex]
So, [tex]\( A^2 = I \)[/tex], which means this matrix does satisfy [tex]\( A^2 = I \)[/tex].
Option D:
[tex]\[ A = \begin{pmatrix} -3 & 5 \\ 2 & 3 \end{pmatrix} \][/tex]
Calculating [tex]\( A^2 \)[/tex]:
[tex]\[ A^2 = A \cdot A = \begin{pmatrix} -3 & 5 \\ 2 & 3 \end{pmatrix} \cdot \begin{pmatrix} -3 & 5 \\ 2 & 3 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} (-3)(-3) + (5)(2) & (-3)(5) + (5)(3) \\ (2)(-3) + (3)(2) & (2)(5) + (3)(3) \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 9 + 10 & -15 + 15 \\ -6 + 6 & 10 + 9 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 19 & 0 \\ 0 & 19 \end{pmatrix} \][/tex]
Clearly, [tex]\( A^2 \neq I \)[/tex].
Therefore, the correct matrix [tex]\( A \)[/tex] that satisfies [tex]\( A^2 = I \)[/tex] is:
[tex]\[ \boxed{C} \][/tex]
[tex]\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \][/tex]
We will examine each option:
Option A:
[tex]\[ A = \begin{pmatrix} -5 & 6 \\ 4 & 5 \end{pmatrix} \][/tex]
Calculating [tex]\( A^2 \)[/tex] involves:
[tex]\[ A^2 = A \cdot A = \begin{pmatrix} -5 & 6 \\ 4 & 5 \end{pmatrix} \cdot \begin{pmatrix} -5 & 6 \\ 4 & 5 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} (-5)(-5) + (6)(4) & (-5)(6) + (6)(5) \\ (4)(-5) + (5)(4) & (4)(6) + (5)(5) \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 25 + 24 & -30 + 30 \\ -20 + 20 & 24 + 25 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 49 & 0 \\ 0 & 49 \end{pmatrix} \][/tex]
Clearly, [tex]\( A^2 \neq I \)[/tex].
Option B:
[tex]\[ A = \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix} \][/tex]
Calculating [tex]\( A^2 \)[/tex]:
[tex]\[ A^2 = A \cdot A = \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} (-1)(-1) + (1)(1) & (-1)(1) + (1)(1) \\ (1)(-1) + (1)(1) & (1)(1) + (1)(1) \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 1 + 1 & -1 + 1 \\ -1 + 1 & 1 + 1 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \][/tex]
Clearly, [tex]\( A^2 \neq I \)[/tex].
Option C:
[tex]\[ A = \begin{pmatrix} 3 & -2 \\ 4 & -3 \end{pmatrix} \][/tex]
Calculating [tex]\( A^2 \)[/tex]:
[tex]\[ A^2 = A \cdot A = \begin{pmatrix} 3 & -2 \\ 4 & -3 \end{pmatrix} \cdot \begin{pmatrix} 3 & -2 \\ 4 & -3 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} (3)(3) + (-2)(4) & (3)(-2) + (-2)(-3) \\ (4)(3) + (-3)(4) & (4)(-2) + (-3)(-3) \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 9 - 8 & -6 + 6 \\ 12 - 12 & -8 + 9 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \][/tex]
So, [tex]\( A^2 = I \)[/tex], which means this matrix does satisfy [tex]\( A^2 = I \)[/tex].
Option D:
[tex]\[ A = \begin{pmatrix} -3 & 5 \\ 2 & 3 \end{pmatrix} \][/tex]
Calculating [tex]\( A^2 \)[/tex]:
[tex]\[ A^2 = A \cdot A = \begin{pmatrix} -3 & 5 \\ 2 & 3 \end{pmatrix} \cdot \begin{pmatrix} -3 & 5 \\ 2 & 3 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} (-3)(-3) + (5)(2) & (-3)(5) + (5)(3) \\ (2)(-3) + (3)(2) & (2)(5) + (3)(3) \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 9 + 10 & -15 + 15 \\ -6 + 6 & 10 + 9 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 19 & 0 \\ 0 & 19 \end{pmatrix} \][/tex]
Clearly, [tex]\( A^2 \neq I \)[/tex].
Therefore, the correct matrix [tex]\( A \)[/tex] that satisfies [tex]\( A^2 = I \)[/tex] is:
[tex]\[ \boxed{C} \][/tex]
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