Question

Let $A$ be a 3 × 3 symmetric invertible matrix with real positive elements. Then the number of zero elements in $A^{-1}$ are less than or equal to:

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

2

Answer 2

Let

$A^{-1}=\begin{pmatrix} x & y & z \\ y & w & v \\ z & v & u \end{pmatrix}$.

Since $A$ has all positive entries and is invertible and symmetric, it must be irreducible. In any symmetric matrix the off‐diagonal zeros occur in symmetric pairs. Hence if one off‐diagonal entry is zero the corresponding symmetric entry must also be zero. In a $3\times3$ matrix the only possible counts for zeros (aside from the diagonal, which are nonzero for invertible matrices) are 0, 2, 4, or 6.

However, if more than 2 off–diagonal entries of $A^{-1}$ were zero (say, if $A^{-1}$ were block–diagonal or even diagonal) then by the formula

$A = \frac{\operatorname{adj}(A)}{\det(A)}$,

the matrix $A$ would inherit zeros in the corresponding off–diagonals. This contradicts the condition that every element of $A$ is positive.

A concrete example is:

$A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}$.

Its inverse is computed as:

$A^{-1} = \begin{pmatrix} 3 & -1 & -1 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{pmatrix}$,

which shows exactly 2 zero entries (namely, the $(2,3)$ and $(3,2)$ elements).

Thus, the number of zero elements in $A^{-1}$ is less than or equal to 2.