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: Condition for a matrix to be non-invertible.
A square matrix is not invertible if and only if its determinant is zero.
Therefore, \[ \det(A)=0. \] Let \[ x=a+1. \] Then \[ A= \begin{bmatrix} 1 & 1 & x 1 & x & 1 x & 1 & 1 \end{bmatrix}. \]

Step 2: Evaluate the determinant.
Expanding along the first row, \[ \det(A) = 1 \begin{vmatrix} x & 1 1 & 1 \end{vmatrix} -1 \begin{vmatrix} 1 & 1 x & 1 \end{vmatrix} +x \begin{vmatrix} 1 & x x & 1 \end{vmatrix}. \] Simplifying, \[ \det(A) = (x-1)-(1-x)+x(1-x^2). \] \[ \det(A) = 2(x-1)+x-x^3. \] \[ \det(A) = -x^3+3x-2. \] Hence, \[ \det(A)=-(x^3-3x+2). \]

Step 3: Factorize the polynomial.
For non-invertibility, \[ x^3-3x+2=0. \] Testing \(x=1\), \[ 1-3+2=0. \] So \((x-1)\) is a factor. \[ x^3-3x+2 = (x-1)(x^2+x-2). \] Further factorizing, \[ x^2+x-2=(x-1)(x+2). \] Therefore, \[ (x-1)^2(x+2)=0. \] Hence, \[ x=1 \quad \text{or} \quad x=-2. \]

Step 4: Find the corresponding values of \(a\).
Since \[ x=a+1, \] for \(x=1\), \[ a+1=1 \] \[ a=0. \] For \(x=-2\), \[ a+1=-2 \] \[ a=-3. \] Thus the values of \(a\) are \[ a=0,\,-3. \]

Step 5: Final conclusion.
The sum of all possible values of \(a\) is \[ 0+(-3)=-3. \] Therefore, \[ \boxed{-3} \]