The following system of equations $2x - y + 3z = 5, 3x + 2y - z = 7, 4x + 5y - \lambda z = \mu$ is consistent. Then

A system of equations is consistent when it has at least one solution. If the equations contradict each other, the system is inconsistent (no solution exists).

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We have three equations:

$2x - y + 3z = 5$ ... (Equation 1)

$3x + 2y - z = 7$ ... (Equation 2)

$4x + 5y - \lambda z = \mu$ ... (Equation 3)

For the system to be consistent, Equation 3 must not contradict Equations 1 and 2. If we can create Equation 3 by combining Equations 1 and 2, then it's just repeating what we already know, which guarantees consistency.

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Assume: Equation 3 $= a \times$ Equation 1 $+ b \times$ Equation 2

$a(2x - y + 3z) + b(3x + 2y - z) = 4x + 5y - \lambda z$

Expanding:

$(2a + 3b)x + (-a + 2b)y + (3a - b)z = 4x + 5y - \lambda z$

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Comparing coefficients:

For $x$: $2a + 3b = 4$ ... (i)

For $y$: $-a + 2b = 5$ ... (ii)

For $z$: $3a - b = -\lambda$ ... (iii)

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From equation (ii):

$a = 2b - 5$

Substitute into equation (i):

$2(2b - 5) + 3b = 4$

$4b - 10 + 3b = 4$

$7b = 14$

$b = 2$

Therefore:

$a = 2(2) - 5 = -1$

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Using equation (iii):

$3a - b = -\lambda$

$3(-1) - 2 = -\lambda$

$-5 = -\lambda$

$\lambda = 5$

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The right-hand sides must satisfy the same relationship:

$a \times 5 + b \times 7 = \mu$

$(-1) \times 5 + 2 \times 7 = \mu$

$-5 + 14 = \mu$

$\mu = 9$

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Therefore, $\lambda = 5$ and $\mu = 9$.