If the system of equations 2x+λy+3z=52x + \lambda y + 3z = 5 3x+2y−z=73x + 2y - z = 7 4x+5y+μz=94x + 5y + \mu z = 9 has infinitely many solutions, then (λ2+μ2)(\lambda^2 + \mu^2) is equal to:

Answer: 29

Step-by-step solution

Step 1: Formulate the augmented matrix

To solve a system of linear equations, we can represent it using an augmented matrix. This matrix combines the coefficients of the variables and the constants on the right-hand side of the equations. The vertical line separates the coefficient matrix from the constant terms.

Step 2: Condition for infinitely many solutions

For a system of linear equations to have infinitely many solutions, two conditions must be met: the determinant of the coefficient matrix (AA) must be zero, and the determinants of the matrices formed by replacing a column of AA with the constant terms (Ax,Ay,AzA_x, A_y, A_z) must also be zero. This ensures that the equations are consistent and dependent.

Step 3: Calculate determinant of coefficient matrix

We calculate the determinant of the coefficient matrix and set it to zero. This will give us a relationship between λ\lambda and μ\mu. The determinant is calculated as 2(2μ−(−1)5)−λ(3μ−(−1)4)+3(3(5)−2(4))2(2\mu - (-1)5) - \lambda(3\mu - (-1)4) + 3(3(5) - 2(4)).

Step 4: Solve for the relationship between λ\lambda and μ\mu

Expanding the determinant from the previous step, we get an algebraic equation involving λ\lambda and μ\mu. This equation represents the condition for the determinant of the coefficient matrix to be zero.

Step 5: Calculate determinant of AxA_x

Now, we replace the first column of the coefficient matrix with the constant terms to form AxA_x. We calculate its determinant and set it to zero. This will give us another equation involving λ\lambda and μ\mu.

Step 6: Solve for λ\lambda and μ\mu

Expanding the determinant of AxA_x and simplifying, we get a second equation: 7λμ+9λ−10μ−76=07\lambda\mu + 9\lambda - 10\mu - 76 = 0. We now have two equations with two variables, λ\lambda and μ\mu. We can solve these simultaneously. By comparing the coefficients of λμ\lambda\mu and λ\lambda in both equations, we can deduce that λ=2\lambda = 2 and μ=5\mu = 5.

Step 7: Calculate λ2+μ2\lambda^2 + \mu^2

Finally, we substitute the values of λ=2\lambda = 2 and μ=5\mu = 5 into the expression λ2+μ2\lambda^2 + \mu^2 to find the required value.

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