If the system of equations
$2x + \lambda y + 3z = 5$
$3x+2y-z=7$
$4x + 5y + \mu z = 9$
has infinitely many solutions, then $(\lambda^2+ \mu^2)$ is equal to :
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 $Similar Questions
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