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The system of equations

                        x + y + z = 6

                        x + 2y + 3z = 14

                        x + 4y + 7z = 30 has

Option: 1

no solution


Option: 2

unique solution


Option: 3

infinite solutions


Option: 4

none of these


Answers (1)

best_answer

We have

                        x + y + z = 6

                        x + 2y + 3z = 14

                        z + 4y + 7z = 30

The given system of equations in the matrix form are written as below:

                        \begin{bmatrix} 1 &1 &1 \\ 1& 2 & 3\\ 1& 4 &7 \end{bmatrix}\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 6\\ 14\\ 30 \end{bmatrix}

    AX = B      ......(1)

Where A =  \begin{bmatrix} 1 &1 &1 \\ 1& 2 & 3\\ 1& 4 &7 \end{bmatrix},X=\begin{bmatrix} x\\ y\\ z \end{bmatrix} and\; \; B=\begin{bmatrix} 6\\ 14\\ 30 \end{bmatrix}

      |A| = 1 (14 – 12) – 1 (7 – 3) + 1 (4 – 2)

            = 2 – 4 + 2 = 0

\therefore The equation either has no solution or an infinite number of solutions. To decide about this, we need to find adj(A).B

 

 

 

            On comparing

                        x + y + z = 6, y + 2z = 8

Taking z = k \epsilon R

            \therefore        y = 8 – 2k

            and       x = k – 2

            Since k is arbitrary, hence the number of solutions is infinite.

 

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Gunjita

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