Let $A = [a_{ij}]$ be a 3 × 3 matrix such that $a_{ij} = i + j$, for all $i, j$. If $A^3 + kA^2 + lA + mI = 0$, then $k + l + m$ is equal to

Looking at the matrix $A$ where $a_{ij} = i + j$ for a 3×3 matrix:

$A = \begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix}$

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By the Cayley-Hamilton theorem, every matrix satisfies its own characteristic equation. The characteristic polynomial is $\det(A - \lambda I) = 0$.

The characteristic polynomial for a 3×3 matrix has the form:

$p(\lambda) = -\lambda^3 + (\text{tr}(A))\lambda^2 - (\text{sum of principal 2×2 minors})\lambda + \det(A)$

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Finding $\text{tr}(A)$:

$\text{tr}(A) = 2 + 4 + 6 = 12$

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Finding $\det(A)$:

$\det(A) = 2(4 \cdot 6 - 5 \cdot 5) - 3(3 \cdot 6 - 5 \cdot 4) + 4(3 \cdot 5 - 4 \cdot 4)$

$= 2(24 - 25) - 3(18 - 20) + 4(15 - 16)$

$= 2(-1) - 3(-2) + 4(-1)$

$= -2 + 6 - 4$

$= 0$

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Finding the sum of principal 2×2 minors:

$\begin{vmatrix} 2 & 3 \\ 3 & 4 \end{vmatrix} + \begin{vmatrix} 2 & 4 \\ 4 & 6 \end{vmatrix} + \begin{vmatrix} 4 & 5 \\ 5 & 6 \end{vmatrix}$

$= (8-9) + (12-16) + (24-25)$

$= -1 - 4 - 1$

$= -6$

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The characteristic polynomial is:

$p(\lambda) = -\lambda^3 + 12\lambda^2 - (-6)\lambda + 0$

$= -\lambda^3 + 12\lambda^2 + 6\lambda$

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By the Cayley-Hamilton theorem:

$-A^3 + 12A^2 + 6A = 0$

Multiplying by $-1$:

$A^3 - 12A^2 - 6A = 0$

This can be written as:

$A^3 + (-12)A^2 + (-6)A + 0 \cdot I = 0$

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Comparing with $A^3 + kA^2 + lA + mI = 0$:

$k = -12$

$l = -6$

$m = 0$

Therefore:

$k + l + m = -12 + (-6) + 0 = -18$

The answer is Option 3: $-18$