Let A = begin{bmatrix} 1 & 2 & 3 2 & 0 & 5 | vedclass

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(A) To determine the nature of the solutions for the system $Ax = b$,we first calculate the determinant of matrix $A$,denoted as $|A|$.$|A| = 1(0 	im

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$Ax = b$ has a unique solution.

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(A) To determine the nature of the solutions for the system $Ax = b$,we first calculate the determinant of matrix $A$,denoted as $|A|$.$|A| = 1(0 	imes 1 - 5 	imes 2) - 2(2 	imes 1 - 5 	imes 0) + 3(2 	imes 2 - 0 	imes 0)$$|A| = 1(0 - 10) - 2(2 - 0) + 3(4 - 0)$$|A| = -10 - 4 + 12 = -2$Since $|A| = -2$,which is not equal to $0$,the matrix $A$ is non-singular (invertible).$A$ system of linear equations $Ax = b$ has a unique solution if and only if the matrix $A$ is invertible $(|A| 
eq 0)$.Therefore,the system $Ax = b$ has a unique solution.

Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 0 & 5 \\ 0 & 2 & 1 \end{bmatrix}$ and $b = \begin{bmatrix} 0 \\ -3 \\ 1 \end{bmatrix}$. Which of the following is true?

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