Let the matrix A = begin{bmatrix} 10^{30} + 5 | vedclass

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(A) To determine if $A$ is invertible,we check if the determinant $|A| 
eq 0$. We consider the parity of the elements in the matrix by taking the mat

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$A$ is invertible for all $n \in N$

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(A) To determine if $A$ is invertible,we check if the determinant $|A| 
eq 0$. We consider the parity of the elements in the matrix by taking the matrix modulo $2$. Modulo $2$,the matrix $A$ becomes: $A \equiv \begin{bmatrix} 0+1 & 0+0 & 0+0 \ 0+0 & 0+1 & 0+0 \ 0+0 & 0+0 & 0+1 \end{bmatrix} \pmod{2}$ $A \equiv \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \pmod{2}$ The determinant $|A| \pmod{2}$ is the determinant of the identity matrix,which is $1$. Since $|A| \equiv 1 \pmod{2}$,the determinant $|A|$ must be an odd number. An odd number cannot be $0$. Therefore,$|A| 
eq 0$ for all $n \in N$. Thus,the matrix $A$ is invertible for all $n \in N$.

Let the matrix $A = \begin{bmatrix} 10^{30} + 5 & 10^{20} + 4 & 10^{20} + 6 \\ 10^4 + 2 & 10^8 + 7 & 10^{10} + 2n \\ 10^4 + 8 & 10^6 + 4 & 10^{15} + 9 \end{bmatrix}$,where $n \in N$. Then:

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