The following system of equations has : x+2y+3z=1 x-y+4z=0 2x+y+7z=1

The correct answer is Option (3) → infinitely many solutions

In matrix form the system is AX = B with

    \( A = \begin{bmatrix}1 & 2 & 3\\ 1 & -1 & 4\\ 2 & 1 & 7\end{bmatrix},\quad B = \begin{bmatrix}1\\ 0\\ 1\end{bmatrix} \)

Expanding along the first row:

    \(|A| = 1(-7-4) - 2(7-8) + 3(1+2)\)
    \(= -11 + 2 + 9 = 0\)

Since |A| = 0, there is no unique solution, and the next check is (adj A)B. The cofactors give

    \( \mathrm{adj}\,A = \begin{bmatrix}-11 & -11 & 11\\ 1 & 1 & -1\\ 3 & 3 & -3\end{bmatrix} \)

    \( (\mathrm{adj}\,A)B = \begin{bmatrix}-11 + 0 + 11\\ 1 + 0 - 1\\ 3 + 0 - 3\end{bmatrix} = \begin{bmatrix}0\\ 0\\ 0\end{bmatrix} \)

When |A| = 0 and (adj A)B = O, the system has either infinitely many solutions or none. Adding the first two equations:

    \((x+2y+3z) + (x-y+4z) = 1 + 0\)
    \(2x+y+7z = 1\)

This is exactly the third equation, so the system is consistent and only two equations are independent. Putting z = k in the first two:

    \(x + 2y = 1 - 3k,\quad x - y = -4k\)
    \(3y = 1 + k\)

    \( \displaystyle y = \frac{1+k}{3},\quad x = \frac{1-11k}{3},\quad z = k \)

Every value of k gives a solution, so the system has infinitely many solutions.