Question

If the characteristic equation of a $3 \times 3$ square matrix A is $ax^3+bx^2+cx+d=0$, $(a \neq 0)$ then the Trace of A + det A =

• A. $\frac{d-b}{a}$
• B. $\frac{b+c+d}{2a}$
• C. $-\frac{(b+d)}{a}$
• D. $-\frac{(a+b)}{d}$

Hint:

• For a characteristic polynomial $a_n \lambda^n + a_{n-1} \lambda^{n-1} + \dots + a_1 \lambda + a_0 = 0$:
  • Trace(A) = Sum of eigenvalues = $-a_{n-1}/a_n$.
  • det(A) = Product of eigenvalues = $(-1)^n a_0/a_n$.
• In this $3 \times 3$ case, $ax^3+bx^2+cx+d=0$. So $n=3$.
  • Trace(A) = $-b/a$.
  • det(A) = $(-1)^3 d/a = -d/a$.
• Then sum Trace(A) + det(A).

Answer 1

The Correct Option is C

Given the characteristic equation of a 3×3 matrix A is:

$$ ax^3 + bx^2 + cx + d = 0, \quad a \neq 0 $$

Step 1: Recall the characteristic polynomial of a 3×3 matrix

The characteristic polynomial of a 3×3 matrix A is generally written as:

$$ \det(xI - A) = x^3 - (\text{trace}(A))x^2 + (\text{sum of principal minors})x - \det(A) = 0 $$

In the standard form, the coefficients correspond to:

• Coefficient of \(x^3\) is 1.
• Coefficient of \(x^2\) is \(-\text{trace}(A)\).
• Coefficient of \(x\) is the sum of principal minors of \(A\).
• Constant term is \(-\det(A)\).

Step 2: Compare with the given equation

Given equation is:

$$ ax^3 + bx^2 + cx + d = 0 $$

Divide both sides by \(a\) (since \(a \neq 0\)) to make the coefficient of \(x^3\) equal to 1:

$$ x^3 + \frac{b}{a}x^2 + \frac{c}{a}x + \frac{d}{a} = 0 $$

Step 3: Relate coefficients to matrix properties

By comparing the standard form with the normalized form, we have:

• Coefficient of \(x^2\): $$ \frac{b}{a} = -\text{trace}(A) $$ so, $$ \text{trace}(A) = -\frac{b}{a} $$
• Constant term: $$ \frac{d}{a} = -\det(A) $$ so, $$ \det(A) = -\frac{d}{a} $$

Step 4: Find \(\text{trace}(A) + \det(A)\)

\[ \text{trace}(A) + \det(A) = -\frac{b}{a} - \frac{d}{a} = -\frac{b + d}{a} \]

Answer:

$$ \boxed{-\frac{b + d}{a}} $$

This corresponds to option 3.