Correct Answer - Option 1 : 1
Concept:
Transpose of a matrix
The transpose of a matrix is obtained by the interchanging of rows and columns of a particular matrix
Let the 2x2 be:
\(A= \left[ {\begin{array}{*{20}{c}} a_1&a_2\\ b_1&b_2 \end{array}\;} \right]\)
The transpose matrix is:
\(A^T = \left[ {\begin{array}{*{20}{c}} a_1&b_1\\ a_2&b_2 \end{array}\;} \right]\)
Echelon matrix
\(A= \left[ {\begin{array}{*{20}{c}} k&-&-&-&-&-&- \\ 0&k&-&-&-&-&- \\ 0&0&k&-&-&-&- \\0&0&0&k&-&-&- \\ 0&0&0&0&k&-&- \end{array}\;} \right]\)
- The number of zeroes should be increased before the non-zero element from one row to another.
- perform the row and column operations to get the above form
- Rank of a matrix = R(A) = Number of non-zero rows in echelon form
Calculation:
Given row vectors are v = (1, 0) and w = (2, 0)
The transpose for the given matrices are:
\(v^T= \left[ {\begin{array}{*{20}{c}} 1 \\ 0 \end{array}\;} \right]\) and \(w^T= \left[ {\begin{array}{*{20}{c}} 2 \\ 0 \end{array}\;} \right]\)
\(v^Tv =\left[ {\begin{array}{*{20}{c}} 1 \\ 0 \end{array}\;} \right] \left[ {\begin{array}{*{20}{c}} 1 & 0 \end{array}\;} \right]\)
\(v^Tv= \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&0 \end{array}\;} \right]\)
\(2v^Tv= \left[ {\begin{array}{*{20}{c}} 2&0 \\ 0&0 \end{array}\;} \right]\)
\(w^Tw =\left[ {\begin{array}{*{20}{c}} 2 \\ 0 \end{array}\;} \right] \left[ {\begin{array}{*{20}{c}} 2 & 0 \end{array}\;} \right]\)
\(w^Tw= \left[ {\begin{array}{*{20}{c}} 4&0 \\ 0&0 \end{array}\;} \right]\)
\(3w^Tw= \left[ {\begin{array}{*{20}{c}} 12&0 \\ 0&0 \end{array}\;} \right]\)
\(2v^Tv + 3w^Tw = \left[ {\begin{array}{*{20}{c}} 14&0 \\ 0&0 \end{array}\;} \right]\)
Rank is defined as the number of non-zero rows in a matrix after performing the row and column operations to convert into an echelon matrix.
The number of non-zero rows in the obtained matrix is 1.
The rank of a matrix is 1