Let A = [aij] be a square matrix of order 3 such that aij = 2j–i, for all i, j = 1, 2, 3. Then, the matrix A2 + A3 + … + A10 is equal to
- \((\frac{3^{10}-3}{2})A\)
- \((\frac{3^{10}-1}{2})A\)
- \((\frac{3^{10}+1}{2})A\)
- \((\frac{3^{10}+3}{2})A\)
The Correct Option is A
Solution and Explanation
The correct option is(A): \((\frac{3^{10}-3}{2})A\)
A3 = A.A2 = A(3A) = 3A2 = 32A
A4 = 33A
Now
A2 + A3 + … + A10
A[31 + 32 + 33 + … + 39]
\((\frac{3^{10}-3}{2})A\)
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Concepts Used:
Matrix Transformation
The numbers or functions that are kept in a matrix are termed the elements or the entries of the matrix.
Transpose Matrix:
The matrix acquired by interchanging the rows and columns of the parent matrix is termed the Transpose matrix. The definition of a transpose matrix goes as follows - “A Matrix which is devised by turning all the rows of a given matrix into columns and vice-versa.”