-1
$\begingroup$

I just want to know if this is the right answer
Given $$A = \begin{pmatrix} 2&0&0\\ 1&3&0\\ 1&1&1 \end{pmatrix} , b = \begin{pmatrix} b_1\\ b_2\\ b_3 \end{pmatrix}$$

A)Write the linear system corresponding to $Ax = b$: \begin{align*} x_1 &= b_1\\ x_2 &= b_2+b_1\\ x_3 &= b_3+b_2+b_1 \end{align*} B)Solve the linear equation \begin{align*} b_1 &= 2\\ x_2 &= 4\\ x_3 &= 3 \end{align*}

C)Find the inverse of matrix A $$ \begin{pmatrix} 2&0&0\\ 3&1&0\\ 1&1&1 \end{pmatrix}$$

$\endgroup$

closed as off-topic by Davide Giraudo, Henrik, iadvd, Shailesh, Claude Leibovici Sep 6 '16 at 5:35

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, Henrik, iadvd, Shailesh, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Scientifica Sep 5 '16 at 18:18
  • 2
    $\begingroup$ No. They are not correct. $\endgroup$ – Siong Thye Goh Sep 5 '16 at 18:22
  • $\begingroup$ So is it suppose to be x1 = b1,x2=b1+b2 and x3 = b1+b2+b3 $\endgroup$ – user263146 Sep 5 '16 at 18:32
  • $\begingroup$ Which parts are wrong? $\endgroup$ – user263146 Sep 5 '16 at 18:33
  • $\begingroup$ Hmm... all the parts are wrong. DonAntonio has helped you with part A. In part b, try to express $x_i$ in terms of linear combination of $b_j$. $\endgroup$ – Siong Thye Goh Sep 5 '16 at 18:50
1
$\begingroup$

You seem to have (if I understood correctly your symbols):

$$\begin{pmatrix}2&0&0\\1&3&0\\1&1&1\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}\iff \begin{cases}&2x_1&=b_1\\&x_1+3x_2&=b_2\\&x_1+x_2+x_3&=b_3\end{cases}$$

Try now to calculate the inverse of $\;A\;$, and observe that $\;\det A=6\;$

Added on request: Observe that we have

$$\text{First line:}\;\;2x_1=b_1\implies x_1=\frac{b_1}2$$

and now

$$\text{Second line:}\;\;\frac{b_1}2+3x_2=b_2\implies x_2=\frac{b_2-\frac{b_1}2}3$$

and etc.

$\endgroup$
  • $\begingroup$ Still don't understand part b. I thought that I just add the results from part a and that should be my answer to part b. $\endgroup$ – user263146 Sep 5 '16 at 19:18
  • $\begingroup$ @user263146 HAve you studied matrix reduction, Gauss method and etc.? Iin this case it is even easier as your coefficients matrix is upper triangular, so you can begin solving directy $\endgroup$ – DonAntonio Sep 5 '16 at 19:21
  • 1
    $\begingroup$ @user263146 Check what I added to my answer. $\endgroup$ – DonAntonio Sep 5 '16 at 19:26

Not the answer you're looking for? Browse other questions tagged or ask your own question.