The equations x + 2y + 3z = 1, 2x + y + 3z = | vedclass

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(A) The given system of equations is:$x + 2y + 3z = 1$$2x + y + 3z = 2$$5x + 5y + 9z = 4$We can represent this as $AX = B$,where $A = \begin{bmatrix}

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(A) The given system of equations is:$x + 2y + 3z = 1$$2x + y + 3z = 2$$5x + 5y + 9z = 4$We can represent this as $AX = B$,where $A = \begin{bmatrix} 1 & 2 & 3 \ 2 & 1 & 3 \ 5 & 5 & 9 \end{bmatrix}$.Now,calculate the determinant $|A|$:$|A| = 1(1 	imes 9 - 3 	imes 5) - 2(2 	imes 9 - 3 	imes 5) + 3(2 	imes 5 - 1 	imes 5)$$|A| = 1(9 - 15) - 2(18 - 15) + 3(10 - 5)$$|A| = 1(-6) - 2(3) + 3(5)$$|A| = -6 - 6 + 15 = 3$Since $|A| = 3 
eq 0$,the system of equations has a unique solution.

The equations $x + 2y + 3z = 1,$ $2x + y + 3z = 2,$ and $5x + 5y + 9z = 4$ have:

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