Question

If \[ A= \begin{bmatrix} 1 & 1 & a+1 1 & a+1 & 1 a+1 & 1 & 1 \end{bmatrix} \] is not an invertible matrix, then the sum of all the values of \(a\) is:

• A. \(-3\)
• B. \(-1\)
• C. \(1\)
• D. \(0\)

Hint:

A square matrix is invertible if and only if its determinant is non-zero. For parameter-based matrix questions, first compute the determinant and then solve the resulting equation.

Answer 1

The Correct Option is A

Step 1: State the singularity condition.
A square matrix fails to be invertible precisely when its determinant equals zero. Set x = a + 1 so that A = [[1, 1, x], [1, x, 1], [x, 1, 1]].

Step 2: Evaluate the determinant.
Expanding along the first row: det(A) = 1·det[[x,1],[1,1]] - 1·det[[1,1],[x,1]] + x·det[[1,x],[x,1]] = (x - 1) - (1 - x) + x(1 - x²) = 2(x - 1) + x - x³ = -x³ + 3x - 2 = -(x³ - 3x + 2).

Step 3: Factor the cubic polynomial.
Setting det(A) = 0 gives x³ - 3x + 2 = 0. Testing x = 1 yields 1 - 3 + 2 = 0, so (x - 1) is a factor. Division produces (x - 1)(x² + x - 2) = (x - 1)(x - 1)(x + 2) = (x - 1)²(x + 2) = 0. Hence x = 1 or x = -2.

Step 4: Recover the values of a.
Using x = a + 1: when x = 1, a = 0; when x = -2, a = -3. The possible values are a = 0, -3.

Step 5: Final conclusion.
Their sum is 0 + (-3) = -3.