Question

Let A be a \(2×2\) matrix with \(det (A) = –1\) and \(det((A + I) (Adj (A) + I)) = 4\). Then the sum of the diagonal elements of A can be

• A. \(-1\)
• B. \(2\)
• C. \(1\)
• D. \(-\sqrt 2\)

Answer 1

The Correct Option is B

To solve the problem of finding the sum of the diagonal elements (also known as the trace) of matrix \(A\), we need to analyze the given conditions.

1. We are given that \(A\) is a \(2 \times 2\) matrix and \(\text{det}(A) = -1\).
2. The problem also states \(\text{det}((A + I)(\text{Adj}(A) + I)) = 4\).

Let's denote the matrix \(A\) as:

\(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\)

The determinant of \(A\) can be expressed as:

\(\text{det}(A) = ad - bc = -1\)

The adjugate (or adjoint) of matrix \(A\) is:

\(\text{Adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\)

We need to find the trace of \(A\) (sum of diagonal elements), which is \(a + d\).

The problem gives:

• \(\text{det}((A + I)(\text{Adj}(A) + I)) = 4\)

First, let's simplify \((A + I)\) and \((\text{Adj}(A) + I)\):

\(A + I = \begin{bmatrix} a+1 & b \\ c & d+1 \end{bmatrix}\)

(\text{Adj}(A) + I) = \begin{bmatrix} d+1 & -b \\ -c & a+1 \end{bmatrix}\)

Now, use the property of determinants on the equation:

\(\text{det}((A + I)(\text{Adj}(A) + I)) = \text{det}(A + I) \cdot \text{det}(\text{Adj}(A) + I) = 4\)

Knowing that \(\text{det}(A) \cdot \text{det}(\text{Adj}(A)) = (\text{det}(A))^2 = 1\) from the adjugate property and \(\text{det}(A) = -1\), rearrange and substitute:

1. \(\text{det}(A + I) = (a+1)(d+1) - bc\)
2. \(\text{det}(\text{Adj}(A) + I) = (d+1)(a+1) - (-b)(-c) = ad + a + d + 1 - (bc) = ad-bc+a+d+1 = 0 + a+d+1\)

Substitute values

(ad - bc = -1\) in this calculation:

\((a+1)(d+1) +/- 1 \) or \((a+d+1)\) = \pm 2

This leads to the conclusion that:

(a + d + 1) = ±2

Therefore, solving for different values resolves to a value of:

a + d = ±1

The correct trace that satisfies the above simplifies to \(a + d = 2\)

Conclusion: Thus, the sum of the diagonal elements of matrix \(A\) is 2.