The following system of linear equations 7x + | vedclass

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(A) The given system of equations is homogeneous: $7x + 6y - 2z = 0 \dots (1)$ $3x + 4y + 2z = 0 \dots (2)$ $x - 2y - 6z = 0 \dots (3)$ First,we calcu

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infinitely many solutions,$(x, y, z)$ satisfying $x = 2z$

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(A) The given system of equations is homogeneous: $7x + 6y - 2z = 0 \dots (1)$ $3x + 4y + 2z = 0 \dots (2)$ $x - 2y - 6z = 0 \dots (3)$ First,we calculate the determinant of the coefficient matrix $\Delta$: $\Delta = \begin{vmatrix} 7 & 6 & -2 \ 3 & 4 & 2 \ 1 & -2 & -6 \end{vmatrix}$ $= 7(4(-6) - 2(-2)) - 6(3(-6) - 2(1)) - 2(3(-2) - 4(1))$ $= 7(-24 + 4) - 6(-18 - 2) - 2(-6 - 4)$ $= 7(-20) - 6(-20) - 2(-10)$ $= -140 + 120 + 20 = 0$ Since $\Delta = 0$,the system has infinitely many solutions. Adding equations $(1)$ and $(2)$: $(7x + 6y - 2z) + (3x + 4y + 2z) = 0 + 0$ $10x + 10y = 0 \Rightarrow y = -x$ Substituting $y = -x$ into equation $(1)$: $7x + 6(-x) - 2z = 0$ $7x - 6x - 2z = 0$ $x - 2z = 0 \Rightarrow x = 2z$ Thus,the solutions satisfy $x = 2z$.

The following system of linear equations $7x + 6y - 2z = 0$; $3x + 4y + 2z = 0$; $x - 2y - 6z = 0$ has:

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