Question

If the system of linear equations \[x-4y+7z=g\], \[3y-5z=h\] and \[-2x+5y-9z=k\] is consistent then:
\[\left( A \right)g+h+k=0\]
\[\left( B \right)2g+h+k=0\]
\[\left( C \right)g+h+2k=0\]
\[\left( D \right)g+2h+k=0\]

Answer 1

Hint: We start solving this question by considering all the given three equations. Given the system of linear equations is consistent which means there exist at least one solution which satisfies all the three equations \[x-4y+7z=g\] , \[3y-5z=h\] and \[-2x+5y-9z=k\]. We solve the equations by multiplying first equation with 2 and adding to the third equation. Then we compare the obtained result to the second equation to find the result.

Complete step-by-step answer:
Let us consider the given three equations,
\[x-4y+7z=g................\left( 1 \right)\]
\[3y-5z=h..............\left( 2 \right)\]
\[-2x+5y-9z=k...........\left( 3 \right)\]
Now, the equations (1), (2) and (3) are of the form \[ax+by+cz=n\]
From equation (1), \[a=1\], \[b=-4\] ,\[c=7\]and \[n=g\]
From equation (2), \[a=0\] , \[b=3\] , \[c=-5\] and \[n=h\]
From equation (3), \[a=-2\],\[b=5\] , \[c=-9\]and \[n=k\]
Now we multiply equation (1) with the number 2, just to make both the ‘a’ term of equation (1) and (3) equal and opposite sign so that we can add both the equations and make the ‘a’ term of the resultant equation 0.
\[2\left( x-4y+7z \right)=2g\]
\[2x-8y+14z=2g................\left( 4 \right)\]
Now, we add equation (3) and (4), we get,
\[(2x-8y+14z)+(-2x+5y-9z)=2g+k\]
\[(2x-2x)+(-8y+5y)+(14z-9z)=2g+k\]
\[-3y+5z=2g+k\]
Let us multiply the resultant equation by (-1), we get,
\[3y-5z=-2g-k.............\left( 5 \right)\]

By comparing the equations (2) and (5)
We see that the values of ‘b’ and ‘c’ are same.
We already know if two linear equations in two variables coincides then the ratios of the coefficients is equal to the ratio of the constant if not equal then the two lines are parallel. If the two lines are parallel then it states that there does not exist solution which satisfies all the given three equation in the problem.
So, the lines must coincide.
From equation (2) and (5), we get,
\[\dfrac{h}{-2g-k}=\dfrac{3}{3}=\dfrac{-5}{-5}\]
\[\dfrac{h}{-2g-k}=1\]
\[h=-2g-k\]
\[2g+h+k=0\]

So, the correct answer is “Option B”.

Note: We can also solve this question using matrices as AX=B. So, let us write them in matrix form.
$\left[ \begin{matrix}
   1 & -4 & 7 \\
   0 & 3 & -5 \\
   -2 & 5 & -9 \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   g \\
   h \\
   k \\
\end{matrix} \right]$
Now let us take it in the form
$\left| \begin{matrix}
   1 & -4 & 7 \\
   0 & 3 & -5 \\
   -2 & 5 & -9 \\
\end{matrix}\text{ }\begin{matrix}
   g \\
   h \\
   k \\
\end{matrix} \right|$
Now let us apply row elementary operations to the above matrix.
Multiplying row 1 with 2 and adding two row 3 we get
$\begin{align}
  & \left| \begin{matrix}
   1 & -4 & 7 \\
   0 & 3 & -5 \\
   -2+2 & 5-8 & -9+14 \\
\end{matrix}\text{ }\begin{matrix}
   g \\
   h \\
   k+2g \\
\end{matrix} \right| \\
 & \\
 & \left| \begin{matrix}
   1 & -4 & 7 \\
   0 & 3 & -5 \\
   0 & -3 & 5 \\
\end{matrix}\text{ }\begin{matrix}
   g \\
   h \\
   k+2g \\
\end{matrix} \right| \\
\end{align}$
Now let us add row 2 to row three. Then we get,
$\begin{align}
  & \left| \begin{matrix}
   1 & -4 & 7 \\
   0 & 3 & -5 \\
   0 & -3+3 & 5-5 \\
\end{matrix}\text{ }\begin{matrix}
   g \\
   h \\
   k+2g+h \\
\end{matrix} \right| \\
 & \\
 & \left| \begin{matrix}
   1 & -4 & 7 \\
   0 & 3 & -5 \\
   0 & 0 & 0 \\
\end{matrix}\text{ }\begin{matrix}
   g \\
   h \\
   k+2g+h \\
\end{matrix} \right| \\
\end{align}$
We can see in the third row that all the coefficients are zero. As the equations are consistent, when all coefficients in the equations are zero, their value is zero. So, we need to equate $k+2g+h$ to zero. So, by doing so we get,
$\Rightarrow 2g+h+k=0$
Hence answer is option B.