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 $$ \begin{pmatrix} 2&0&0\\ 3&1&0\\ 1&1&1 \end{pmatrix}$$


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  • $\begingroup$ meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Scientifica Sep 5 '16 at 18:18
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    $\begingroup$ 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

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.

  • $\begingroup$ 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
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    $\begingroup$ @user263146 Check what I added to my answer. $\endgroup$ – DonAntonio Sep 5 '16 at 19:26

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