Question

Examine whether the following system of equations is consistent or inconsistent. If consistent, find the complete solution
x + y + z = 1
2x + y + z = 2
x + 2y + 2z = 1

Answer 1

Answer:

Given linear equations are consistent,but have infinite solution

Step-by-step explanation:

Analysis of equation: Since here  

$A=\left[\begin{array}{ccc}1&1&1\\2&1&1\\1&2&2\end{array}\right] \\\\\\X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right] \\\\\\B=\left[\begin{array}{ccc}1\\2\\1\end{array}\right]$

Augmented matrix:

$A=\left[\begin{array}{cccc}1&1&1&1\\2&1&1&2\\1&2&2&1\end{array}\right]\\\\R_{2} -> R_{2}-2R_{1}\\\\\left[\begin{array}{cccc}1&1&1&1\\0&-1&-1&0\\1&2&2&1\end{array}\right]\\\\R_{3} -> R_{3}-R_{1}\\\\\left[\begin{array}{cccc}1&1&1&1\\0&-1&-1&0\\0&1&1&0\end{array}\right]\\\\R_{3} -> R_{3}+R_{2}\\\\\left[\begin{array}{cccc}1&1&1&1\\0&-1&-1&0\\0&0&0&0\end{array}\right]$

$Rank \left[\begin{array}{cccc}1&1&1&1\\2&1&1&2\\1&2&2&1\end{array}\right]=2$

By the same way

$A=\left[\begin{array}{ccc}1&1&1\\2&1&1\\1&2&2\end{array}\right]\\\\R_{2} -> R_{2}-2R_{1}\\\\\left[\begin{array}{ccc}1&1&1\\0&-1&-1\\1&2&2\end{array}\right]\\\\R_{3} -> R_{3}-R_{1}\\\\\left[\begin{array}{ccc}1&1&1\\0&-1&-1\\0&1&1\end{array}\right]\\\\R_{3} -> R_{3}+R_{2}\\\\\left[\begin{array}{ccc}1&1&1\\0&-1&-1\\0&0&0&\end{array}\right]$

$Rank \left[\begin{array}{ccc}1&1&1\\2&1&1\\1&2&2\end{array}\right]=2$

Both rank are equal thus given linear equations are consistent,but have infinite solution,Because rank is less than number of variables.