The following system of equations 2x-y+3z = 5, 3x+2y-z = 7, 4x+5y-λ z…

The correct answer is Option (3) → \(\lambda = 5\) and \(\mu = 9\)

Let \(\Delta\) be the determinant of the coefficient matrix. Expanding along the first row:

    \(\displaystyle \Delta = \begin{vmatrix}2&-1&3\\3&2&-1\\4&5&-\lambda\end{vmatrix}\)

    \(= 2(-2\lambda + 5) + 1(-3\lambda + 4) + 3(15 - 8)\)
    \(= -7\lambda + 35\)

If \(\lambda \ne 5\), then \(\Delta \ne 0\) and the system has a unique solution for every value of \(\mu\), so it is consistent anyway. The question is therefore about the case \(\Delta = 0\), that is \(\lambda = 5\), where consistency depends on \(\mu\), and asks which option guarantees a consistent system.

With \(\lambda = 5\), the left side of the third equation is a combination of the first two:

    \(-(2x - y + 3z) + 2(3x + 2y - z) = 4x + 5y - 5z\)

The same combination of the right sides gives

    \(-5 + 2(7) = 9\)

So for \(\lambda = 5\) the third equation agrees with the first two only when \(\mu = 9\), and then the system has infinitely many solutions. For \(\lambda = 5,\ \mu \ne 9\) it is inconsistent.

Option (1) \(\lambda = 5\) and \(\mu \ne 9\) — Incorrect. With \(\lambda = 5\) and \(\mu \ne 9\) the system has no solution.

Option (2) \(\lambda \ne 9\) and \(\mu = 5\) — Incorrect. It allows \(\lambda = 5\) (since \(5 \ne 9\)) with \(\mu = 5\), and then the system has no solution.

Option (3) \(\lambda = 5\) and \(\mu = 9\) — Correct. The third equation is then \(-1\) times the first plus 2 times the second, so the system is consistent.

Option (4) \(\lambda = 5\) and \(\mu = 5\) — Incorrect. With \(\lambda = 5\) the system needs \(\mu = 9\); for \(\mu = 5\) it has no solution.