Question

Test whether the system of linear equations is consistent or inconsistent using rank method. Find the solution if exists.
1) x + 2y + 3z = 3
2) 2x - y + z = 6
3) 3x + y - z = 4 ​

Answer 1

Answer:

Step-by-step explanation:

(i) Here the number of unknowns = 3. The matrix form of the system is AX = B where (i.e) AX = B  The augmented matrix (A, B) is Applying Gaussian elimination method on [A,B] we get The above matrix is in echelon form also ρ(A, B)  = ρ(A) = 3 = number of unknowns The system of equations is consistent with a unique solution. To find the solution. Now writing the equivalent equations we get x – y + 2z = 2  3y = 3 ⇒ y = 1  7z = -7 ⇒ z = 1  Substituting z = y = 1 in (1) we get x – 1 + 2 = 2 ⇒ x = 1  ⇒ x = y = z = 1 (ii) Here the number of unknowns is 3.  The matrix form of the given system of equations is: AX = B (i.e) Now the augmented matrix [A, B] is The above matrix is in echelon form also  ρ(A, B) = ρ(A) = 2 < number of unknowns  The system of equations is consistent with the infinite number of solutions.  To find the solution:  Now writing the equivalent equations we get (iii) Here the number of unknowns is 3.  The matrix form of the given equation is AX = B The above matrix is in echelon form.  Here ρ(A, B) = 3; ρ(A) = 2  So ρ(A, B) ≠ ρ(A)  The system of equations is inconsistent with no solution.  (iv) Here the number of unknowns is 3.  The matrix form of the given equation is AX = B The augmented matrix [A, B] is The above matrix is in echelon form also ρ(A, B)  = ρ(A) = 1 < number of unknowns  The system of equations is consistent with the infinite number of solutions.  To find the Solution  Now writing the equivalent equations we get Read more on Sarthaks.com - https://www.sarthaks.com/864871/test-for-consistency-and-possible-solve-the-following-systems-equations-rank-method-2z-4z

Answer 2

$\large\underline{\sf{Solution-}}$

Given system of equations are

$\rm :\longmapsto\:x + 2y + 3z = 3$

$\rm :\longmapsto\:2x - y + z = 6$

$\rm :\longmapsto\:3x + y - z = 4$

So, its Augmented matrix can be written as

$\rm :\longmapsto\: \rho \: [A : B]$

$\rm \:  =  \: \begin{gathered}\sf \left[\begin{array}{cccc}1&2&3&3\\2& - 1&1&6\\ 3&1& - 1&4\end{array}\right]\end{gathered}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ OP \: R_2 \: \to \: R_2 - 2R_1 \: }}}$

and

$\purple{\rm :\longmapsto\:\boxed{\tt{ OP \: R_3 \: \to \: R_3 - 3R_1 \: }}}$

$\rm \:  =  \: \begin{gathered}\sf \left[\begin{array}{cccc}1&2&3&3\\0& - 5& - 5&0\\ 0& - 5& - 10& - 5\end{array}\right]\end{gathered}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ OP \: R_2 \: \to \: \frac{R_2}{ - 5} \: }}}$

and

$\purple{\rm :\longmapsto\:\boxed{\tt{ OP \: R_3 \: \to \: \frac{R_3}{ - 5} \: }}}$

$\rm \:  =  \: \begin{gathered}\sf \left[\begin{array}{cccc}1&2&3&3\\0&1&1&0\\ 0&1&2& 1\end{array}\right]\end{gathered}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ OP \: R_3 \: \to \: R_3 - R_2 \: }}}$

$\rm \:  =  \: \begin{gathered}\sf \left[\begin{array}{cccc}1&2&3&3\\0&1&1&0\\ 0&0&1& 1\end{array}\right]\end{gathered}$

$\rm\implies \: \rho \: [A : B] = \: \rho \: [A] = 3$

$\rm\implies \:System \: of \: equation \: is \: consistent \: having \: unique \: solution$

So, we have

$\purple{\rm :\longmapsto\:z = 1}$

$\purple{\rm :\longmapsto\:y + z = 0\rm\implies \:y = - 1}$

$\purple{\rm :\longmapsto\:x + 2y + 3z = 3}$

$\purple{\rm :\longmapsto\:x - 2+ 3= 3}$

$\purple{\rm :\longmapsto\:x + 1= 3}$

$\purple{\rm :\longmapsto\:x= 3 - 1}$

$\purple{\rm :\longmapsto\:x = 2}$

Hence,

Solution of given system of equations is

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{x = 2} \\ \\ &\sf{y = - 1} \\ \\ &\sf{z = 1} \end{cases}\end{gathered}\end{gathered}$