If the system of equations (λ−1)x+(λ−4)y+λz=5, λx+(λ−1)y+(λ−4)z=7, (λ+1)x+(λ+2)y−(λ+2)z=9 has infinitely many solutions, then λ2+λ is equal to
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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 A 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−R1) and subtracting the first row from the third row (R3→R3−R1) 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 k if a row is multiplied by k.
Step 4: Further simplify the determinant
Performing another row operation, R3→R3−R2, 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
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. This gives us two possible values for λ.
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 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
Since λ=3 is the value for which the system has infinitely many solutions, we substitute this value into the expression λ2+λ to find the final answer.