Let the system of equations: 2x + 3y + 5z = 9 | vedclass

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(B) For the system to have infinitely many solutions,the determinant of the coefficient matrix must be zero: $\left|\begin{array}{ccc} 2 & 3 & 5 \ 7

Vedclass · Accepted Answer

$\frac{7}{5}$

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(B) For the system to have infinitely many solutions,the determinant of the coefficient matrix must be zero: $\left|\begin{array}{ccc} 2 & 3 & 5 \ 7 & 3 & -2 \ 12 & 3 & -(4+\lambda) \end{array}ight| = 0$.Expanding along the first row: $2(-3(4+\lambda) + 6) - 3(-7(4+\lambda) + 24) + 5(21 - 36) = 0$.$2(-12 - 3\lambda + 6) - 3(-28 - 7\lambda + 24) + 5(-15) = 0$.$2(-6 - 3\lambda) - 3(-4 - 7\lambda) - 75 = 0$.$-12 - 6\lambda + 12 + 21\lambda - 75 = 0 \Rightarrow 15\lambda = 75 \Rightarrow \lambda = 5$.For infinitely many solutions,the augmented matrix determinant must also be zero: $\left|\begin{array}{ccc} 9 & 3 & 5 \ 8 & 3 & -2 \ 16-\mu & 3 & -9 \end{array}ight| = 0$.Using row operations $R_1 	o R_1 - R_2$: $\left|\begin{array}{ccc} 1 & 0 & 7 \ 8 & 3 & -2 \ 16-\mu & 3 & -9 \end{array}ight| = 0$.$1(-27 + 6) - 0 + 7(24 - 3(16-\mu)) = 0$.$-21 + 7(24 - 48 + 3\mu) = 0 \Rightarrow -21 + 7(3\mu - 24) = 0$.$-3 + 3\mu - 24 = 0 \Rightarrow 3\mu = 27 \Rightarrow \mu = 9$.The center of the circle is $(5, 9)$. The line is $4x - 3y = 0$.The radius is the perpendicular distance from $(5, 9)$ to $4x - 3y = 0$: $r = \frac{|4(5) - 3(9)|}{\sqrt{4^2 + (-3)^2}} = \frac{|20 - 27|}{5} = \frac{7}{5}$.

Let the system of equations: $2x + 3y + 5z = 9$,$7x + 3y - 2z = 8$,$12x + 3y - (4 + \lambda)z = 16 - \mu$ have infinitely many solutions. Then the radius of the circle centred at $(\lambda, \mu)$ and touching the line $4x = 3y$ is

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