Question

A matrix $M$ is said to be idempotent if $M^2 = M$. Now, consider a matrix $A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$ and $A^n = A^{-1}, n \in N$. The minimum value of $n$ is $n_1$.

Answer 1

3

Answer 2

To find the minimum value of $n$ such that $A^n = A^{-1}$, we first need to calculate $A^{-1}$ and then compare it with powers of $A$.

Given matrix $A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$.

Step 1: Calculate the determinant of A. $\det(A) = 3 \begin{vmatrix} -3 & 4 \\ -1 & 1 \end{vmatrix} - (-3) \begin{vmatrix} 2 & 4 \\ 0 & 1 \end{vmatrix} + 4 \begin{vmatrix} 2 & -3 \\ 0 & -1 \end{vmatrix}$ $\det(A) = 3((-3)(1) - (4)(-1)) + 3((2)(1) - (4)(0)) + 4((2)(-1) - (-3)(0))$ $\det(A) = 3(-3 + 4) + 3(2 - 0) + 4(-2 - 0)$ $\det(A) = 3(1) + 3(2) + 4(-2)$ $\det(A) = 3 + 6 - 8 = 1$

Since $\det(A) = 1 \neq 0$, the inverse exists.

Step 2: Calculate the adjoint of A. First, find the cofactor matrix $C$: $C_{11} = \begin{vmatrix} -3 & 4 \\ -1 & 1 \end{vmatrix} = -3 + 4 = 1$ $C_{12} = -\begin{vmatrix} 2 & 4 \\ 0 & 1 \end{vmatrix} = -(2 - 0) = -2$ $C_{13} = \begin{vmatrix} 2 & -3 \\ 0 & -1 \end{vmatrix} = -2 - 0 = -2$

$C_{21} = -\begin{vmatrix} -3 & 4 \\ -1 & 1 \end{vmatrix} = -(-3 + 4) = -1$ $C_{22} = \begin{vmatrix} 3 & 4 \\ 0 & 1 \end{vmatrix} = 3 - 0 = 3$ $C_{23} = -\begin{vmatrix} 3 & -3 \\ 0 & -1 \end{vmatrix} = -(-3 - 0) = 3$

$C_{31} = \begin{vmatrix} -3 & 4 \\ -3 & 4 \end{vmatrix} = -12 + 12 = 0$ $C_{32} = -\begin{vmatrix} 3 & 4 \\ 2 & 4 \end{vmatrix} = -(12 - 8) = -4$ $C_{33} = \begin{vmatrix} 3 & -3 \\ 2 & -3 \end{vmatrix} = -9 + 6 = -3$

The cofactor matrix is $C = \begin{bmatrix} 1 & -2 & -2 \\ -1 & 3 & 3 \\ 0 & -4 & -3 \end{bmatrix}$.

The adjoint of A is the transpose of the cofactor matrix: $\text{adj}(A) = C^T = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$.

Step 3: Calculate $A^{-1}$. $A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \frac{1}{1} \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$.

Step 4: Calculate powers of A ($A^1, A^2, A^3, \dots$) and compare with $A^{-1}$. $A^1 = A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \neq A^{-1}$.

Now, calculate $A^2$: $A^2 = A \cdot A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$ $A^2 = \begin{bmatrix} (3)(3)+(-3)(2)+(4)(0) & (3)(-3)+(-3)(-3)+(4)(-1) & (3)(4)+(-3)(4)+(4)(1) \\ (2)(3)+(-3)(2)+(4)(0) & (2)(-3)+(-3)(-3)+(4)(-1) & (2)(4)+(-3)(4)+(4)(1) \\ (0)(3)+(-1)(2)+(1)(0) & (0)(-3)+(-1)(-3)+(1)(-1) & (0)(4)+(-1)(4)+(1)(1) \end{bmatrix}$ $A^2 = \begin{bmatrix} 9-6+0 & -9+9-4 & 12-12+4 \\ 6-6+0 & -6+9-4 & 8-12+4 \\ 0-2+0 & 0+3-1 & 0-4+1 \end{bmatrix} = \begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix}$.

$A^2 \neq A^{-1}$.

Now, calculate $A^3$: $A^3 = A^2 \cdot A = \begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$ $A^3 = \begin{bmatrix} (3)(3)+(-4)(2)+(4)(0) & (3)(-3)+(-4)(-3)+(4)(-1) & (3)(4)+(-4)(4)+(4)(1) \\ (0)(3)+(-1)(2)+(0)(0) & (0)(-3)+(-1)(-3)+(0)(-1) & (0)(4)+(-1)(4)+(0)(1) \\ (-2)(3)+(2)(2)+(-3)(0) & (-2)(-3)+(2)(-3)+(-3)(-1) & (-2)(4)+(2)(4)+(-3)(1) \end{bmatrix}$ $A^3 = \begin{bmatrix} 9-8+0 & -9+12-4 & 12-16+4 \\ 0-2+0 & 0+3+0 & 0-4+0 \\ -6+4+0 & 6-6+3 & -8+8-3 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$.

Comparing $A^3$ with $A^{-1}$: $A^3 = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$. Thus, $A^3 = A^{-1}$.

Since $A^1 \neq A^{-1}$ and $A^2 \neq A^{-1}$, the minimum value of $n$ for which $A^n = A^{-1}$ is $n=3$. So, $n_1 = 3$.

The definition of an idempotent matrix ($M^2=M$) is not directly used in this problem, as matrix A is not idempotent.