Question

Show that the following system of linear equations is consistent and also find their solution:
 $x+y+z=6$
 $x+2y+3z=14$
 $x+4y+7z=30$

Answer 1

Hint:
1) A system of equations is consistent if it has at least one solution. An inconsistent system has no solutions.
2) Rouché–Capelli theorem: A system of linear equations with n variables has a solution if and only if the rank of its coefficient matrix [A] is equal to the rank of its augmented matrix [A|B].
2.1) If $n=rank(A)=rank(A|B)\text{ and }det(A)\ne0$ : consistent; there is a unique solution.
2.2) If $n>rank(A)=rank(A|B)\text{ and }det(A)=0$ : consistent and dependent; there are infinitely many solutions.
2.3) If $rank(A)\ne rank(A|B)$ : inconsistent; there are no solutions.
3) Rank of a matrix corresponds to the maximal number of linearly independent columns/rows. In other words, it is the number of non-zero rows when the matrix is expressed in row echelon form (all rows consisting of only zeroes are at the bottom).

Complete step by step solution:
The coefficient matrix for the given system of equations is: $A=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \\ \end{matrix} \right]$ .
$\det (A)=\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \\ \end{matrix} \right|=1(14-12)+1(3-7)+1(4-2)=2-4+2=0$ .
Therefore, either there is no solution (inconsistent) or there are infinitely many solutions (consistent and dependent).
Let us convert the augmented matrix into the row echelon form to find its solutions.
The augmented matrix is:
 $A|B=\left[ \begin{matrix} 1 & 1 & 1 & | & 6 \\ 1 & 2 & 3 & | & 14 \\ 1 & 4 & 7 & | & 30 \\ \end{matrix} \right]$
 ${{R}_{3}}\to {{R}_{3}}-{{R}_{1}};{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}$
 $A|B=\left[ \begin{matrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 8 \\ 0 & 3 & 6 & | & 24 \\ \end{matrix} \right]$
 ${{R}_{3}}\to {{R}_{3}}-3{{R}_{2}}$
 $A|B=\left[ \begin{matrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 8 \\ 0 & 0 & 0 & | & 0 \\ \end{matrix} \right]$
Since the rank of the coefficient matrix and the augmented matrix are both equal to 2, the system is consistent and since the rank is less than the number of unknowns 3, there are infinitely many solutions.
From the reduced augmented matrix, we have the set of equations:
 $x+y+z=6$
 $y+2z=8$
Let $z=k$ .
 $y=8-2z=8-2k$
 $x=6-y-z=6-8+2k-k=-2+k$

The solutions are $x=-2+k; y=8-2k; z=k$.

Note:
Cramer's rule: A system of equations ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z={{d}_{1}};{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z={{d}_{2}};{{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z={{d}_{3}}$ is:
1) If $D\ne 0$ : a unique solution (consistent).
2) The solution is $x=\frac{{{D}_{x}}}{D};y=\frac{{{D}_{y}}}{D};z=\frac{{{D}_{z}}}{D}$ .
If $D=0$ : either infinitely many solutions (consistent and dependent) or no solution (inconsistent). To find out if the system is dependent or inconsistent, another method, such as elimination or Rouché–Capelli theorem, will have to be used.
Where, $D=\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\ \end{matrix} \right|$ , ${{D}_{x}}=\left| \begin{matrix} {{d}_{1}} & {{d}_{2}} & {{d}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\ \end{matrix} \right|$ , ${{D}_{y}}=\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{d}_{1}} & {{d}_{2}} & {{d}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\ \end{matrix} \right|$ , ${{D}_{z}}=\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{d}_{1}} & {{d}_{2}} & {{d}_{3}} \\ \end{matrix} \right|$ .