1. **State the problem:** We are given two matrices and asked to find the period of the first matrix and determine which of the two given matrices is idempotent or involutory. 2. **Recall definitions:** - A matrix $A$ is **idempotent** if $A^2 = A$. - A matrix $A$ is **involutory** if $A^2 = I$, where $I$ is the identity matrix. - The **period** of a matrix $A$ is the smallest positive integer $k$ such that $A^k = I$. 3. **Given matrix for period:** $$ A = \begin{bmatrix} 1 & -2 & -6 \\ -3 & 2 & 9 \\ 2 & 0 & -3 \end{bmatrix} $$ 4. **Calculate powers of $A$ to find period:** - Compute $A^2 = A \times A$. - Compute $A^3 = A^2 \times A$. - Continue until $A^k = I$ or conclude no such $k$ exists. 5. **Calculate $A^2$:** $$ A^2 = \begin{bmatrix} 1 & -2 & -6 \\ -3 & 2 & 9 \\ 2 & 0 & -3 \end{bmatrix} \times \begin{bmatrix} 1 & -2 & -6 \\ -3 & 2 & 9 \\ 2 & 0 & -3 \end{bmatrix} $$ Calculate each element: - First row: - $(1)(1) + (-2)(-3) + (-6)(2) = 1 + 6 - 12 = -5$ - $(1)(-2) + (-2)(2) + (-6)(0) = -2 - 4 + 0 = -6$ - $(1)(-6) + (-2)(9) + (-6)(-3) = -6 - 18 + 18 = -6$ - Second row: - $(-3)(1) + (2)(-3) + (9)(2) = -3 - 6 + 18 = 9$ - $(-3)(-2) + (2)(2) + (9)(0) = 6 + 4 + 0 = 10$ - $(-3)(-6) + (2)(9) + (9)(-3) = 18 + 18 - 27 = 9$ - Third row: - $(2)(1) + (0)(-3) + (-3)(2) = 2 + 0 - 6 = -4$ - $(2)(-2) + (0)(2) + (-3)(0) = -4 + 0 + 0 = -4$ - $(2)(-6) + (0)(9) + (-3)(-3) = -12 + 0 + 9 = -3$ So, $$ A^2 = \begin{bmatrix} -5 & -6 & -6 \\ 9 & 10 & 9 \\ -4 & -4 & -3 \end{bmatrix} $$ 6. **Calculate $A^3 = A^2 \times A$:** $$ A^3 = \begin{bmatrix} -5 & -6 & -6 \\ 9 & 10 & 9 \\ -4 & -4 & -3 \end{bmatrix} \times \begin{bmatrix} 1 & -2 & -6 \\ -3 & 2 & 9 \\ 2 & 0 & -3 \end{bmatrix} $$ Calculate each element: - First row: - $(-5)(1) + (-6)(-3) + (-6)(2) = -5 + 18 - 12 = 1$ - $(-5)(-2) + (-6)(2) + (-6)(0) = 10 - 12 + 0 = -2$ - $(-5)(-6) + (-6)(9) + (-6)(-3) = 30 - 54 + 18 = -6$ - Second row: - $(9)(1) + (10)(-3) + (9)(2) = 9 - 30 + 18 = -3$ - $(9)(-2) + (10)(2) + (9)(0) = -18 + 20 + 0 = 2$ - $(9)(-6) + (10)(9) + (9)(-3) = -54 + 90 - 27 = 9$ - Third row: - $(-4)(1) + (-4)(-3) + (-3)(2) = -4 + 12 - 6 = 2$ - $(-4)(-2) + (-4)(2) + (-3)(0) = 8 - 8 + 0 = 0$ - $(-4)(-6) + (-4)(9) + (-3)(-3) = 24 - 36 + 9 = -3$ So, $$ A^3 = \begin{bmatrix} 1 & -2 & -6 \\ -3 & 2 & 9 \\ 2 & 0 & -3 \end{bmatrix} = A $$ 7. **Interpretation:** Since $A^3 = A$, multiply both sides by $A^{-1}$ (assuming invertible) or observe that $A^3 - A = 0$ implies $A^2 = I$ if $A$ is invertible. But here, $A^3 = A$ means $A^2 = I$ if $A$ is invertible, or the period is 3 if $A^3 = I$. Check if $A^3 = I$: No, $A^3 = A$, not $I$. Check $A^2$: $A^2 \neq I$. So the period is not 1 or 2. Since $A^3 = A$, the minimal polynomial divides $x^3 - x = x(x^2 - 1)$. Hence, the period is 2 if $A^2 = I$, but here $A^2 \neq I$. So the period is not 2. Since $A^3 = A$, the matrix is periodic with period 3 if $A^3 = I$, but it is not. Therefore, the matrix does not have a finite period such that $A^k = I$ for $k=1,2,3$. 8. **Check idempotent or involutory for given matrices:** (i) Matrix: $$ B = \begin{bmatrix} -5 & -8 & 0 \\ 3 & 5 & 0 \\ 1 & 2 & -1 \end{bmatrix} $$ Calculate $B^2$: - First row: - $(-5)(-5) + (-8)(3) + (0)(1) = 25 - 24 + 0 = 1$ - $(-5)(-8) + (-8)(5) + (0)(2) = 40 - 40 + 0 = 0$ - $(-5)(0) + (-8)(0) + (0)(-1) = 0 + 0 + 0 = 0$ - Second row: - $(3)(-5) + (5)(3) + (0)(1) = -15 + 15 + 0 = 0$ - $(3)(-8) + (5)(5) + (0)(2) = -24 + 25 + 0 = 1$ - $(3)(0) + (5)(0) + (0)(-1) = 0 + 0 + 0 = 0$ - Third row: - $(1)(-5) + (2)(3) + (-1)(1) = -5 + 6 - 1 = 0$ - $(1)(-8) + (2)(5) + (-1)(2) = -8 + 10 - 2 = 0$ - $(1)(0) + (2)(0) + (-1)(-1) = 0 + 0 + 1 = 1$ So, $$ B^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I $$ Therefore, $B$ is involutory. (ii) Matrix: $$ C = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} $$ Calculate $C^2$: - First row: - $(2)(2) + (-2)(-1) + (-4)(1) = 4 + 2 - 4 = 2$ - $(2)(-2) + (-2)(3) + (-4)(-2) = -4 - 6 + 8 = -2$ - $(2)(-4) + (-2)(4) + (-4)(-3) = -8 - 8 + 12 = -4$ - Second row: - $(-1)(2) + (3)(-1) + (4)(1) = -2 - 3 + 4 = -1$ - $(-1)(-2) + (3)(3) + (4)(-2) = 2 + 9 - 8 = 3$ - $(-1)(-4) + (3)(4) + (4)(-3) = 4 + 12 - 12 = 4$ - Third row: - $(1)(2) + (-2)(-1) + (-3)(1) = 2 + 2 - 3 = 1$ - $(1)(-2) + (-2)(3) + (-3)(-2) = -2 - 6 + 6 = -2$ - $(1)(-4) + (-2)(4) + (-3)(-3) = -4 - 8 + 9 = -3$ So, $$ C^2 = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} = C $$ Therefore, $C$ is idempotent. **Final answers:** - The first matrix does not have a finite period $k$ such that $A^k = I$ for $k=1,2,3$. - Matrix (i) is involutory. - Matrix (ii) is idempotent.