Skip to main content

Problem 10: Solve the following systems of linear equations by Gaussian elimination method

 Problem 10: Solve the following systems of linear equations by Gaussian elimination method:

\begin{align*} 2x-2y+3z=2\\ x+2y-z=3\\ 3x-y+2z=1 \end{align*}

Solution:

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

\[\left[\begin{array}{c|c} A & B \end{array} \right] = \begin{align*} \left[\begin{array}{rrr|r} 2 & -2 & 3& 2 \\ 1 & 2& -1 & 3 \\ 3 & -1 & 2 & 1 \\ \end{array}\right] \xrightarrow{R_1\leftrightarrow R_2} \left[\begin{array}{rrr|r} 1 & 2& -1 & 3 \\ 2 & -2 & 3& 2 \\ 3 & -1 & 2 & 1 \\ \end{array}\right] \end{align*}\] \[\xrightarrow{R2\rightarrow R_2-2R_1,R3\rightarrow R_3-3R_1} \left[\begin{array}{rrr|r} 1 & 2& -1 & 3 \\ 0 & -6 & 5& -4 \\ 0 & -7 & 5 & -8 \\ \end{array}\right] \xrightarrow{R3\rightarrow R_3-R_2} \left[\begin{array}{rrr|r} 1 & 2& -1 & 3 \\ 0 & -6 & 5& -4 \\ 0 & -1 & 0 & -4 \\ \end{array}\right]\] \[\xrightarrow{R_3\rightarrow -R_3} \left[\begin{array}{rrr|r} 1 & 2& -1 & 3 \\ 0 & -6 & 5& -4 \\ 0 & 1 & 0 & 4 \\ \end{array}\right]\]

For Gaussian elimination we write down the equations corresponding to REF. Then we get

\begin{align*} x+2y+-z=3\\ -6y+5z=-4\\ y=4 \end{align*}

By solving we get

Solution set S is \begin{align*} y=4\\ z=4\\ x=-1 \end{align*}

Labels:

Comments

Post a Comment

Popular posts from this blog

Problem 85: A manufacturer of external hard drives claims that only 10% of his drives require repairs within the warranty peroid of 12 months.

Problem 85: A manufacturer of external hard drives claims that only 10% of his drives require repairs within the warranty peroid of 12 months. a. If 5 of 20 of his drives required repairs within the first year, does this tend to support or refuse the claim? b. If 3 of 20 of his drives required repairs within the first year, does this tend to support or refuse the claim? Solution: a. If 5 of 20 of his drives required repairs within the first year Given that $n=20,$ and $p=0.3$ $\sum _{x=5}^{20}b(x;n,p)=\sum _{x=5}^{20}b(x;20,0.1)=1-\sum _{x=0}^{4}b(x;20,0.1)$ $ = 1 - [b(0;20,0.1)+b(1;20,0.1)+b(2;20,0.1)+b(3;20,0.1)+b(4;20,0.1)]$ $b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$ $b(0;20,0.1) = \left(\begin{array}{c}20\\ 0\end{array}\right){0.1}^{0}(1-0.1)^{20-0} = 0.121577$ $b(1;20,0.1) = \left(\begin{array}{c}20\\ 1\end{array}\right){0.1}^{1}(1-0.1)^{20-1} = 0.27017$ $b(2;20,0.1) = \left(\begin{array}{c}20\\ 2\end{array}\right){0.1}^{...
Post a Comment
Read more

Problem 56: Let $S = \{x \in \R^3 : x = (r-2s,3r+s,s), r,s \in \R \}$. Establish that $S$ is a subspace of $\R^3$.

Problem 56: Let $S = \{x \in \R^3 : x = (r-2s,3r+s,s), r,s \in \R \}$. Establish that $S$ is a subspace of $\R^3$. Solution: $S = \{x \in \R^3 : x = (r-2s,3r+s,s), r,s \in \R \}$. S is certainly nonempty. Let $x, y \in S$. Then for some $r, s, u,v \in \R$, $x = (r-2s,3r+s,s)$ and $y = (u-2v,3u+v,v)$. Hence, $x + y = (r-2s,3r+s,s) + (u-2v,3u+v,v)$ $= (r-2s+u-2v,3r+s+3u+v,s+v) = ((r+u)-2(s+v),3(r+u)+(s+v),(s+v))$ $= (k-2l,3k+l,l)$, where $k = r + u, l=s+v$. Consequently, $S$ is closed under addition. Further, if $c \in \R$, then $cx = c(r-2s,3r+s,s) = (c(r-2s),c(3r+s),cs) $ $= (cr-2cs,3cr+cs,cs) = (a-2b,3a+b,b)$, where $a = cr,b=cs$. Therefore $S$ is also closed under scalar multiplication. It follows that $S$ is a subspace of $\R^3$.
Post a Comment
Read more

Problem 107: In a given city, 6% of all drivers get at least one parking ticket per year. Use the Poisson approximation to the binomial distribution to determine the probabilities that among 80 drivers randomly chosen in this given city:
a. 4 will get at least one parking ticket in any given year,
b. At least 3 will get at least at least one parking ticket in any given year,
c. Anywhere from 3 to 6, inclusive will get at least one parking ticket in any given year.

Problem 107: In a given city, 6% of all drivers get at least one parking ticket per year. Use the Poisson approximation to the binomial distribution to determine the probabilities that among 80 drivers randomly chosen in this given city: a. 4 will get at least one parking ticket in any given year, b. At least 3 will get at least at least one parking ticket in any given year, c. Anywhere from 3 to 6, inclusive will get at least one parking ticket in any given year. Solution: Poisson distribution, $f(x;\lambda )=\frac{{\lambda }^{x}{e}^{-\lambda }}{x!}$ a. 4 will get at least one parking ticket in any given year Given that $x=4, n=80, p=0.06, np = 4.8$ $f(4;4.8)=\frac{{4.8}^{4}{e}^{-4.8}}{4!} = 0.182029$ b. At least 3 will get at least at least one parking ticket in any given year Given that $x\ge 3, n=80, p=0.06, np = 4.8$ $P(x\ge 3) = 1-P(x $P(x=0) = f(0;4.8)=\frac{{4.8}^{0}{e}^{-4.8 }}{0!} = 0.00823$ $P(x=1) = f(1;4.8)=\frac{{4.8}^{1}{e}^{-4.8 }}{1!}...
Post a Comment
Read more