Question

Matrix Inverse Sum Calculation

Given the matrix:

    A =    | 1  2  2 |    | 3  2  3 |    | 1  1  2 |    

The inverse matrix is represented as:

    A^-1 =    | a₁₁  a₁₂  a₁₃ |    | a₂₁  a₂₂  a₂₃ |    | a₃₁  a₃₂  a₃₃ |    

The sum of all elements in A^-1 is:

∑ a_ij = 3

• A. \(\tfrac{2}{3}\)
• B. \(\tfrac{1}{3}\)
• C. \(1\)
• D. \(17\)

Hint:

The sum of all entries of an inverse matrix can be found via \(\mathbf{1}^T A^{-1} \mathbf{1}\), i.e., solve \(A x = \mathbf{1}\) and sum the components of \(x\).

Answer 1

The Correct Option is B

Step 1: Interpret the sum of elements of \(A^{-1}\). Define \(A^{-1} = [a_{ij}]\). The sum of all entries in \(A^{-1}\) is

\[\sum_{i=1}^{3}\,\sum_{j=1}^{3} a_{ij}.\] One efficient way to find this is to observe that

\[\sum_{i,j} a_{ij} \;=\; \mathbf{1}^T\,A^{-1}\,\mathbf{1},\]

where \(\mathbf{1} = (1,\,1,\,1)^T\).

Step 2: Solve \(A\,x = \mathbf{1}\) to locate row-sums of \(A^{-1}\). If \(x = A^{-1}\mathbf{1}\), then each component of \(x\) is precisely the sum of one row of \(A^{-1}\). Hence

\[ A\,x \;=\; \mathbf{1}. \] Let \(x = (x_1,\,x_2,\,x_3)^T\). Then:

\[\begin{pmatrix} 1 & 2 & 2 \\ 3 & 2 & 3 \\ 1 & 1 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}.\]

Step 3: Solve for \((x_1,\,x_2,\,x_3)\).

Subtracting the first equation from the third gives \(x_2 = 0\).

From the second minus thrice the first: \(4x_2 + 3x_3 = 2\). With \(x_2=0\), we get \(3x_3 = 2\implies x_3 = \tfrac{2}{3}\).

Back-substituting into the first:

\(x_1 + 2\cdot 0 + 2\cdot \tfrac{2}{3} = 1 \implies x_1 + \tfrac{4}{3} = 1 \implies x_1 = -\tfrac{1}{3}.\)

Thus \[ x = (x_1,\,x_2,\,x_3) = \Bigl(-\tfrac{1}{3},\,0,\,\tfrac{2}{3}\Bigr). \]

Step 4: Sum of all elements of \(A^{-1}\).

That sum is \[ \mathbf{1}^T(A^{-1}\mathbf{1}) = \mathbf{1}^T\,x = x_1 + x_2 + x_3 = -\tfrac{1}{3} + 0 + \tfrac{2}{3} = \tfrac{1}{3}. \] Hence \(\boxed{\tfrac{1}{3}}\) is the value of \(\sum_{i,j} a_{ij}\).