If the system of equations (λ−1)x+(λ−4)y+λz=5(\lambda - 1)x + (\lambda - 4)y + \lambda z = 5, λx+(λ−1)y+(λ−4)z=7\lambda x + (\lambda - 1)y + (\lambda - 4)z = 7, (λ+1)x+(λ+2)y−(λ+2)z=9(\lambda + 1)x + (\lambda + 2)y - (\lambda + 2)z = 9 has infinitely many solutions, then λ2+λ\lambda^2 + \lambda is equal to

Answer: 12

Step-by-step solution

Step 1: Form the coefficient matrix

For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero. First, we write down the coefficient matrix AA from the given system of equations.

Step 2: Simplify the determinant using row operations

To simplify the calculation of the determinant, we perform row operations. Subtracting the first row from the second row (R2→R2−R1R_2 \to R_2 - R_1) and subtracting the first row from the third row (R3→R3−R1R_3 \to R_3 - R_1) helps create simpler entries in the matrix, which does not change the determinant's value.

Step 3: Factor out common terms

We can factor out a 2 from the third row of the simplified matrix. This makes the third row entries smaller and easier to work with. The determinant of a matrix changes by a factor of kk if a row is multiplied by kk.

Step 4: Further simplify the determinant

Performing another row operation, R3→R3−R2R_3 \to R_3 - R_2, creates two zeros in the third row. This is a strategic move because the determinant of a matrix with a row containing mostly zeros is much easier to compute. Specifically, we can expand along this row.

Step 5: Calculate the determinant and solve for λ\lambda

Now we calculate the determinant by expanding along the third row. Since the determinant must be zero for infinitely many solutions, we set the expression equal to zero and solve for λ\lambda. This gives us two possible values for λ\lambda.

Step 6: Check for consistency with augmented matrix

For infinitely many solutions, not only must the determinant of the coefficient matrix be zero, but the rank of the coefficient matrix must also be equal to the rank of the augmented matrix, and this rank must be less than the number of variables. We test λ=3\lambda = 3 by substituting it into the augmented matrix and performing row operations. The last row becoming all zeros indicates consistency and infinitely many solutions.

Step 7: Calculate λ2+λ\lambda^2 + \lambda

Since λ=3\lambda = 3 is the value for which the system has infinitely many solutions, we substitute this value into the expression λ2+λ\lambda^2 + \lambda to find the final answer.

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