System of Equations: Solving for λ² + μ²

We are given a system of three linear equations:

• $2x + \lambda y + 3z = 5$
• $3x + 2y - z = 7$
• $4x + 5y + \mu z = 9$

For this system to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the system must be consistent.

Determinant Condition for Infinite Solutions

The coefficient matrix is:

$ A = \begin{pmatrix} 2 & \lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \mu \end{pmatrix} $

The determinant, det(A), must be zero:

$ \det(A) = 2(2\mu - (-1)(5)) - \lambda(3\mu - (-1)(4)) + 3(3(5) - 2(4)) $ $ \det(A) = 2(2\mu + 5) - \lambda(3\mu + 4) + 3(15 - 8) $ $ \det(A) = 4\mu + 10 - 3\lambda\mu - 4\lambda + 3(7) $ $ \det(A) = -3\lambda\mu - 4\lambda + 4\mu + 31 $

Setting det(A) = 0 gives:

$ -3\lambda\mu - 4\lambda + 4\mu + 31 = 0 \quad (*)$

Consistency Conditions using Determinants

When the main determinant is zero, for the system to have infinitely many solutions (i.e., be consistent), the determinants formed by replacing each column of A with the constant terms vector B must also be zero.

Let $D_y$ be the determinant when the second column is replaced by the constants (5, 7, 9):

$ D_y = \det \begin{pmatrix} 2 & 5 & 3 \\ 3 & 7 & -1 \\ 4 & 9 & \mu \end{pmatrix} $ $ D_y = 2(7\mu - (-1)(9)) - 5(3\mu - (-1)(4)) + 3(3(9) - 7(4)) $ $ D_y = 2(7\mu + 9) - 5(3\mu + 4) + 3(27 - 28) $ $ D_y = 14\mu + 18 - 15\mu - 20 + 3(-1) $ $ D_y = -\mu - 2 - 3 = -\mu - 5 $

Setting $D_y = 0$ gives:

$ -\mu - 5 = 0 \implies \mu = -5 $

Let $D_z$ be the determinant when the third column is replaced by the constants (5, 7, 9):

$ D_z = \det \begin{pmatrix} 2 & \lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9 \end{pmatrix} $ $ D_z = 2(2(9) - 7(5)) - \lambda(3(9) - 7(4)) + 5(3(5) - 2(4)) $ $ D_z = 2(18 - 35) - \lambda(27 - 28) + 5(15 - 8) $ $ D_z = 2(-17) - \lambda(-1) + 5(7) $ $ D_z = -34 + \lambda + 35 = \lambda + 1 $

Setting $D_z = 0$ gives:

$ \lambda + 1 = 0 \implies \lambda = -1 $

Final Calculation

We found $\lambda = -1$ and $\mu = -5$. Let's verify these values in the first condition (*):

$ -3(-1)(-5) - 4(-1) + 4(-5) + 31 = -15 + 4 - 20 + 31 = -11 - 20 + 31 = -31 + 31 = 0 $

The values satisfy the condition.

We need to calculate $(\lambda^2 + \mu^2)$:

$ (\lambda^2 + \mu^2) = (-1)^2 + (-5)^2 $ $ (\lambda^2 + \mu^2) = 1 + 25 $ $ (\lambda^2 + \mu^2) = 26 $