Skip to main content

Problem 21: Discuss for all values of $\lambda$, the system of equations

Problem 21: Discuss for all values of $\lambda$, the system of equations

\begin{align*} x+y+4z=6,\\ x+2y-2z=6,\\ \lambda x+y+z = 6, \end{align*} as regards existence and nature of solutions.

Solution:

Consider the augmented matrix of this system and apply row operations.

\[\left[\begin{array}{c|c} A & B \end{array} \right] = \begin{align*} \left[\begin{array}{rrr|r} 1 & 1 & 4 & 6 \\ 1 & 2 & -2& 6 \\ \lambda & 1 & 1 & 6 \\ \end{array}\right] \end{align*}\] \[\xrightarrow{R2\rightarrow R_2-R_1,R3\rightarrow R_3-\lambda R_1} \left[\begin{array}{rrr|r} 1 & 1 & 4 & 6 \\ 0 & 1 & -6& 0 \\ 0 & 1-\lambda & 1-4\lambda & 6-6\lambda \\ \end{array}\right]\]

Therefore the coefficient matrix will be non-singular if and only if, \begin{align*} 1-4\lambda + 6-6\lambda \ne 0,\\ \lambda \ne \frac{7}{10}. \end{align*}

In Case $\lambda = \frac{7}{10}$, \begin{align*} \left[\begin{array}{rrr|r} 1 & 1 & 4 & 6 \\ 0 & 1 & -6& 0 \\ 0 & \frac{3}{10} & -\frac{18}{10} & \frac{18}{10} \\ \end{array}\right] \end{align*} \[\xrightarrow{R3\rightarrow R_3-\frac{3}{10}R_2} \left[\begin{array}{rrr|r} 1 & 1 & 4 & 6 \\ 0 & 1 & -6& 0 \\ 0 & 0 & 0 & \frac{18}{10} \\ \end{array}\right]\] showing that the equations are not consistent in this case and no solution is possible.

Labels:

Comments

Post a Comment

Popular posts from this blog

Problem 98: If 6 of 18 new buildings in a city violate the building code, what is the probability that a building inspector, who randomly selects 4 of the new buildings for inspection, will catch
a. none of the buildings that violate the building code?
b. 1 of the new buildings that violate the building code?
c. 2 of the new buildings that violate the building code?
d. at least 3 of the new buildings that violate the building code?

Problem 98: If 6 of 18 new buildings in a city violate the building code, what is the probability that a building inspector, who randomly selects 4 of the new buildings for inspection, will catch a. none of the buildings that violate the building code? b. 1 of the new buildings that violate the building code? c. 2 of the new buildings that violate the building code? d. at least 3 of the new buildings that violate the building code? Solution: a. none of the buildings that violate the building code? Given that $x=0,n=4,a=6, N=18$ Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$ $P(x=0) = h(0;4,6,18)=\frac{\left(\begin{array}{c}6\\ 0\end{array}\right)\left(\begin{array}{c}18-6\\ 4-0\end{array}\right)}{\left(\begin{array}{c}18\\ 4\end{array}\right)} = 0.16176.$ b. 1 of the new buildings that violate the building code? Given ...
Post a Comment
Read more

Problem 105: If a bank receives on the average α = 6 bad checks per day, what are the probabilities that it will receive (a) 4 bad checks on any given day?
(b) 10 bad checks over any 2 consecutive days?

Problem 105: If a bank receives on the average α = 6 bad checks per day, what are the probabilities that it will receive (a) 4 bad checks on any given day? (b) 10 bad checks over any 2 consecutive days? Solution: Poisson distribution, $f(x;\lambda )=\frac{{\lambda }^{x}{e}^{-\lambda }}{x!}$ a. 4 bad checks on any given day Given that $x=4, \lambda = \alphaT = 6\dot 1 = 6$ $f(4;6 )=\frac{{6 }^{4}{e}^{-6 }}{4!} = 0.134$ b. 10 bad checks over any 2 consecutive days Given that $x=10, \lambda = \alphaT = 6\dot 2 = 12$ $f(10;12 )=\frac{{12 }^{10}{e}^{-12 }}{10!} = 0.104837$
Post a Comment
Read more

Problem 57: Determine whether it is a subspace of the given vector space $\V$
$\V = \R^2$, and $S$ is the set of all vectors $(x, y)$ in $\V$ satisfying $3x + 2y = 0$.

Problem 57: Determine whether it is a subspace of the given vector space $\V$ $\V = \R^2$, and $S$ is the set of all vectors $(x, y)$ in $\V$ satisfying $3x + 2y = 0$. Solution: $S = \{x \in \R^2 : 3x+2y=0\}$. Let $(x_1,x_2),(y_1,y_2) \in S$. Then $3x_1+2x_2 = 0$ and $3y_1+2y_2 = 0$ Hence, $x + y = (x_1,x_2)+(y_1,y_2) =3x_1+2x_2 + 3y_1+2y_2 = 3(x_1+y_1) + 2(x_2+y_2)$ which implies that $(x_1+y_1,x_2+y_2) \in S$ Consequently, $S$ is closed under addition. Let $a \in \R$ and $(x_1,x_2) \in S$, then $ax = a(x_1,x_2) = a(3x_1+2x_2) = a0 = 0$ $=3(ax_1) + 2(ax_2) =0 $, which implies that $(ax_1,ax_2) \in S$ Therefore $S$ is also closed under scalar multiplication. It follows that $S$ is a subspace of $\R^2$.
Post a Comment
Read more