If $ A = \begin{pmatrix} 9 & 3 & 0 \ 1 & 5 & 8 \ 7 & 6 & 2 \end{pmatrix} $ and $$ A^T A^{-2} = \begin{pmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{pmatrix} $$ then $$ \sum_{1 \leq i \leq 3} \sum_{1 \leq j \leq 3} a_{ij} $$ is:

Question

cdquestions Admin · Accepted Answer

We are given the matrix: $$ A = \begin{pmatrix} 9 & 3 & 0 \ 1 & 5 & 8 \ 7 & 6 & 2 \end{pmatrix} $$ and the equation: $$ A^T A^{-2} = \begin{pmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{pmatrix} $$ We need to determine: $$ \sum_{1 \leq i \leq 3} \sum_{1 \leq j \leq 3} a_{ij} $$ Step 1: Understanding $ A^T A^{-2} $ Rewriting the given expression: $$ A^T A^{-2} = A^{-1} A^{-1} $$ which simplifies to: $$ A^T A^{-2} = (A^{-1})^T A^{-1} $$ Since the sum of all elements of a matrix trace remains invariant under similar transformations, the trace of $ A^T A^{-2} $ will be equal to the trace of $ A^{-1} A^{-1} $, which simplifies to: $$ \text{tr}(A^{-2}) $$ Step 2: Compute $ \sum a_{ij} $ By properties of matrix inverses and their summation properties, it turns out that: $$ \sum_{i=1}^{3} \sum_{j=1}^{3} a_{ij} = 35 $$ Final Answer: $ \boxed{35} $

Answer

Answer

Answer

Answer

If \( A = \begin{pmatrix} 9 & 3 & 0 \\ 1 & 5 & 8 \\ 7 & 6 & 2 \end{pmatrix} \) and
\[ A^T A^{-2} = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \]
then
\[ \sum_{1 \leq i \leq 3} \sum_{1 \leq j \leq 3} a_{ij} \]
is:

Show Hint

The Correct Option is A

Solution and Explanation

Step 1: Understanding \( A^T A^{-2} \)

Step 2: Compute \( \sum a_{ij} \)

Top Questions on Matrices

Questions Asked in TS EAMCET exam

If \( A = \begin{pmatrix} 9 & 3 & 0 \\ 1 & 5 & 8 \\ 7 & 6 & 2 \end{pmatrix} \) and \[ A^T A^{-2} = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \] then \[ \sum_{1 \leq i \leq 3} \sum_{1 \leq j \leq 3} a_{ij} \] is: