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$ 2x + 2y - 2z = 1 $

$ 4x + 4y - z = 2 $

$ 6x + 6y + 2z = 3 $

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So we have the equation given as

$ 2x + 2y - 2z = 1 $

$ 4x + 4y - z = 2 $

$ 6x + 6y + 2z = 3 $

Now we will check the value of $ \vartriangle ,{\vartriangle _1},{\vartriangle _2},{\vartriangle _3} $ and then we will come to know whether the system of linear equations is consistent or not.

So forming the determinant for $ \vartriangle $ , it will be as

$ \Rightarrow \vartriangle = \left| {\begin{array}{*{20}{c}}

2&2&{ - 2} \\

4&4&{ - 1} \\

6&6&2

\end{array}} \right| $

Now on solving the above determinant we will get

$ \Rightarrow 2\left| {\begin{array}{*{20}{c}}

4&{ - 1} \\

6&2

\end{array}} \right| - 2\left| {\begin{array}{*{20}{c}}

4&{ - 1} \\

6&2

\end{array}} \right| - 2\left| {\begin{array}{*{20}{c}}

4&4 \\

6&6

\end{array}} \right| $

Now by doing the cross multiplication, we have

$ \Rightarrow 2\left( {8 + 6} \right) - 2\left( {8 + 6} \right) - 2\left( {24 - 24} \right) $

And on solving it we get

$ \Rightarrow \vartriangle = 0 $

Since the determinant is zero then it means that two of its rows or columns are the same. So by using this property we can directly write that the determinant will be zero.

Now, forming the determinant for $ {\vartriangle _1} $ , it will be as

For this just changing any one row or column with the constant, then the determinant will be

$ \Rightarrow {\vartriangle _1} = \left| {\begin{array}{*{20}{c}}

1&2&{ - 2} \\

2&4&{ - 1} \\

3&6&2

\end{array}} \right| $

And it can also be written as

$ \Rightarrow {\vartriangle _1} = 2\left| {\begin{array}{*{20}{c}}

1&1&{ - 2} \\

2&2&{ - 1} \\

3&3&2

\end{array}} \right| $

Again we can see that two of its rows or columns are the same. So by using this property we can directly write that the determinant will be zero.

$ \Rightarrow {\vartriangle _1} = 0 $

Similarly for $ {\vartriangle _2} = \left| {\begin{array}{*{20}{c}}

2&1&{ - 2} \\

4&2&{ - 1} \\

6&3&2

\end{array}} \right| $

And it can also be written as

$ \Rightarrow {\vartriangle _2} = 2\left| {\begin{array}{*{20}{c}}

1&1&{ - 2} \\

2&2&{ - 1} \\

3&3&2

\end{array}} \right| $

Therefore, \[{\vartriangle _2} = 0\]

Now for $ {\vartriangle _3} $ , it will be

$ \Rightarrow {\vartriangle _3} = \left| {\begin{array}{*{20}{c}}

2&2&1 \\

4&4&2 \\

6&6&3

\end{array}} \right| $

Therefore, \[{\vartriangle _3} = 0\]

Since $ \vartriangle = {\vartriangle _1} = {\vartriangle _2} = {\vartriangle _3} = 0 $ so the given system will be consistent as it has an infinite number of solutions.

Now for finding the solutions, Let us assume $ x $ will be equal to $ k $

Then, $ 6y + 2z = 3 - 6k $

And the other equation formed will be $ 4y - z = 2 - 4k $

Now on solving both of this equation, but for this, we have to keep the constant to one side so we get

$ \Rightarrow 6y + 2z + 6k = 3 $ and we will name it equation $ 1 $

$ \Rightarrow 4y - z + 4k = 2 $ and we will name it equation $ 2 $

Now on solving both of the equation by making one term similar, we will get the values as

$ y = \dfrac{1}{2} - k $ and $ z = 0 $

Hence, we can say that the system of linear equations is consistent having the solution as $ y = \dfrac{1}{2} - k $ and $ z = 0 $ .

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