Question

If $A = \begin{bmatrix} 2 & -1 \\ 4 & -2 \end{bmatrix}$, then the value of $det(I+2A+3A^2+4A^3+5A^4+\dots \text{upto} \infty)$ is

Answer 1

1

Answer 2

The given series is $S = I+2A+3A^2+4A^3+5A^4+\dots \text{upto} \infty$. This is a matrix power series. We can relate it to a known scalar series. Consider the scalar geometric series: $G(x) = 1 + x + x^2 + x^3 + \dots = \frac{1}{1-x}$ for $|x|<1$. Differentiating $G(x)$ with respect to $x$: $\frac{d}{dx} G(x) = 0 + 1 + 2x + 3x^2 + \dots = \sum_{n=1}^{\infty} nx^{n-1}$. Also, $\frac{d}{dx} (1-x)^{-1} = -1(1-x)^{-2}(-1) = (1-x)^{-2}$. Thus, the sum of the scalar series $S(x) = 1+2x+3x^2+4x^3+\dots$ is $(1-x)^{-2}$ for $|x|<1$.

Extending this to matrices, the sum of the matrix series $S = I+2A+3A^2+4A^3+\dots$ is $(I-A)^{-2}$, provided the series converges. The convergence condition for such a series is that the spectral radius of matrix $A$, denoted $\rho(A)$, must be less than 1. The spectral radius is the maximum magnitude of the eigenvalues of $A$.

Step 1: Find the eigenvalues of matrix A to check for convergence. Given $A = \begin{bmatrix} 2 & -1 \\ 4 & -2 \end{bmatrix}$. The characteristic equation is $det(A - \lambda I) = 0$. $det \begin{bmatrix} 2-\lambda & -1 \\ 4 & -2-\lambda \end{bmatrix} = 0$ $(2-\lambda)(-2-\lambda) - (-1)(4) = 0$ $-(4-\lambda^2) + 4 = 0$ $-4 + \lambda^2 + 4 = 0$ $\lambda^2 = 0$ The eigenvalues are $\lambda_1 = 0$ and $\lambda_2 = 0$. The spectral radius $\rho(A) = \max(|\lambda_1|, |\lambda_2|) = \max(|0|, |0|) = 0$. Since $\rho(A) = 0 < 1$, the series converges, and its sum is $S = (I-A)^{-2}$.

Step 2: Calculate $I-A$. $I-A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 1-2 & 0-(-1) \\ 0-4 & 1-(-2) \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ -4 & 3 \end{bmatrix}$.

Step 3: Calculate the determinant of $I-A$. $det(I-A) = (-1)(3) - (1)(-4) = -3 - (-4) = -3 + 4 = 1$.

Step 4: Calculate the determinant of the series sum $S = (I-A)^{-2}$. Using the properties of determinants: $det(M^k) = (det(M))^k$ and $det(M^{-1}) = \frac{1}{det(M)}$. Therefore, $det(S) = det((I-A)^{-2}) = (det((I-A)^{-1}))^2 = \left(\frac{1}{det(I-A)}\right)^2$. Substitute the value of $det(I-A)$: $det(S) = \left(\frac{1}{1}\right)^2 = 1^2 = 1$.

The final answer is 1.

Explanation of the solution:

1. Recognize the given infinite series as $S = \sum_{n=1}^{\infty} n A^{n-1}$.
2. Recall that the scalar series $1+2x+3x^2+\dots$ sums to $(1-x)^{-2}$ for $|x|<1$.
3. Extend this to matrices: $S = (I-A)^{-2}$, provided the spectral radius $\rho(A) < 1$.
4. Calculate the eigenvalues of $A$: $\lambda^2=0 \implies \lambda=0$. So $\rho(A)=0 < 1$, confirming convergence.
5. Calculate $I-A = \begin{bmatrix} -1 & 1 \\ -4 & 3 \end{bmatrix}$.
6. Calculate $det(I-A) = (-1)(3) - (1)(-4) = 1$.
7. Use the determinant property $det((I-A)^{-2}) = (det(I-A)^{-1})^2 = \left(\frac{1}{det(I-A)}\right)^2$.
8. Substitute $det(I-A)=1$ to get $det(S) = \left(\frac{1}{1}\right)^2 = 1$.