# From the given set of Linear equations. Find all solutions and the Rank of the matrix

Question :

Write the following system of equations :

$$\begin{bmatrix} x_1+x_2+2x_3=4 \\ 3x_1+2x_2+x_3=7 \end{bmatrix}$$

in the form $$A\vec x = \vec b$$. Find all the solutions. Determine the rank of A.

$$\,\,\,\, \underbrace{\begin{bmatrix} 1 & 1 & 2 \\ 3 & 2 & 1 \end{bmatrix} }_{\text { 2 by 3 Matrix}}~~ \underbrace{\begin{bmatrix} x_1\\ x_2 \\ x_3 \end{bmatrix}}_{\text{3 by 1 Matrix}} = \underbrace{\begin{bmatrix} y_1 \\ y_2 \end{bmatrix}}_{\text{2 by 1 Matrix}}$$ $$A\cdot \vec x= \vec b$$

Solving second part of the Question :

$$\tag{1}R_2-3R_1 \rightarrow R_1$$

$$\begin{bmatrix} x_1+x_2+2x_3=4 \\ 0-x_2-5x_3=-5 \end{bmatrix}$$

$$\tag{2}R_1+R_2 \rightarrow R_1$$

$$\begin{bmatrix} x_1+0-3x_3=-1 \\ 0-x_2-5x_3=-5 \end{bmatrix}$$

$$\tag{3} R_2 \div (-1) \rightarrow R_2$$

$$\begin{bmatrix} x_1-3x_3=-1 \\ x_2+5x_3=5 \end{bmatrix}$$

$$\begin{bmatrix} x_1\\ x_2\\x_3 \end{bmatrix} = \begin{bmatrix} -1+3t\\ 5-5t\\t \end{bmatrix} \, ,t \in \Bbb R$$

Last part of the Question :

Rank of a matrix, I have learned this way - firstly, convert the matrix into a upper triangular Matrix and then count the number of non-zero rows. The number of non-zero rows give the Rank$$(A)$$.

I think a $$2$$ by $$3$$ Upper triangular matrix would look like :

$$\begin{bmatrix} 1& a& b \\ 0& 1 & c \end{bmatrix}$$

Making $$A$$ an upper triangular Matrix :

$$\begin{bmatrix} 1 & 1 & 2 \\ 3 & 2 & 1 \end{bmatrix}$$

$$R_2-3R_1 \rightarrow R_2$$

$$\begin{bmatrix} 1 & 1 & 2 \\ 0 & -1 & -5 \end{bmatrix} = \underbrace{\begin{bmatrix} 1 & 1 & 2 \\ 0 & 1& 5 \end{bmatrix}}_{\text{ Both rows are non-zero rows}}$$

$$Rank(A) = 2$$

Yes, it is correct for that reason.

Another way of justifying that is as follows: the rank of $$A$$ is the dimension of the range of the map $$f_A\colon\Bbb R^3\longrightarrow\Bbb R^2$$ whose matrix with respect to the standard bases is $$A$$. But $$f_A(1,0,0)=(1,3)$$ and $$f_A(0,1,0)=(1,2)$$. Since $$\{(1,3),(1,2)\}$$ spans $$\Bbb R^2$$, the dimension of the range is $$2$$.

• Thank you very much for the response. Could you please suggest a reason for not choosing $f_A(0,0,1) = (2,1)$ Jul 9 at 19:37
• Since the two other vectors that I got already span $\Bbb R^2$, there was no reason for me to take into account a third vector. Jul 9 at 20:47
• Alright, Thank you very much for your time and response Jul 9 at 21:12
• I'm glad I could help. Jul 9 at 21:23