If the system of equations x+2y+3z=3,4x+3y-4z | vedclass

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(C) The given system of equations is:$x+2y+3z=3$ ... $(i)$$4x+3y-4z=4$ ... (ii)$8x+4y-\lambda z=9+\mu$ ... (iii)For the system to have infinitely many

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$\left(\frac{72}{5}, \frac{-21}{5}\right)$

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(C) The given system of equations is:$x+2y+3z=3$ ... $(i)$$4x+3y-4z=4$ ... (ii)$8x+4y-\lambda z=9+\mu$ ... (iii)For the system to have infinitely many solutions,the determinant of the coefficient matrix must be zero,and the augmented matrix must have a rank less than $3$.First,calculate the determinant of the coefficient matrix $D$:$D = \begin{vmatrix} 1 & 2 & 3 \ 4 & 3 & -4 \ 8 & 4 & -\lambda \end{vmatrix} = 1(-3\lambda + 16) - 2(-4\lambda + 32) + 3(16 - 24) = -3\lambda + 16 + 8\lambda - 64 - 24 = 5\lambda - 72$.For infinite solutions,$D = 0 \Rightarrow 5\lambda - 72 = 0 \Rightarrow \lambda = \frac{72}{5}$.Now,consider the augmented matrix $[A|B]$:$\begin{bmatrix} 1 & 2 & 3 & | & 3 \ 4 & 3 & -4 & | & 4 \ 8 & 4 & -\frac{72}{5} & | & 9+\mu \end{bmatrix}$.Perform row operations: $R_2 	o R_2 - 4R_1$ and $R_3 	o R_3 - 8R_1$:$\begin{bmatrix} 1 & 2 & 3 & | & 3 \ 0 & -5 & -16 & | & -8 \ 0 & -12 & -\frac{72}{5}-24 & | & 9+\mu-24 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 & | & 3 \ 0 & -5 & -16 & | & -8 \ 0 & -12 & -\frac{192}{5} & | & \mu-15 \end{bmatrix}$.For infinite solutions,the third row must be a multiple of the second row. The ratio of coefficients is $\frac{-12}{-5} = 2.4$.Thus,$\mu - 15 = 2.4 	imes (-8) = -19.2 \Rightarrow \mu = 15 - 19.2 = -4.2 = -\frac{21}{5}$.Therefore,$(\lambda, \mu) = \left(\frac{72}{5}, -\frac{21}{5}ight)$.

If the system of equations $x+2y+3z=3$,$4x+3y-4z=4$,and $8x+4y-\lambda z=9+\mu$ has infinitely many solutions,then the ordered pair $(\lambda, \mu)$ is equal to

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