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If matrix A=\left[\begin{array}{ccc}{5} & {6} & {7} \\ {0} & {1} & {-1} \\ {0} & {0} & {8}\end{array}\right] than A^T is

Option: 1

A^T=\left[\begin{array}{ccc}{8} & {0} & {0} \\ {6} & {1} & {0} \\ {7} & {-1} & {5}\end{array}\right]


Option: 2

A^T=\left[\begin{array}{ccc}{-5} & {0} & {0} \\ {6} & {-1} & {0} \\ {7} & {-1} & {-8}\end{array}\right]


Option: 3

A^T=\left[\begin{array}{ccc}{5} & {0} & {0} \\ {6} & {1} & {0} \\ {7} & {-1} & {8}\end{array}\right]


Option: 4

A^T=\left[\begin{array}{ccc}{5} & {0} & {0} \\ {6} & {1} & {0} \\ {8} & {-1} & {7}\end{array}\right]


Answers (1)

best_answer

 

 

Transpose of a matrix -

Transpose of a matrix 

In simple language transpose of a matrix is changing its row into columns or columns into rows. Let \\\mathrm{A=\left [ a_{ij} \right ]_{m\times n} }  be a matrix, then matrix obtained by changing rows into columns or vice-versa will give transpose of A which is denoted as A’ or At or AT. Hence \\\mathrm{A'=\left [ a_{ji} \right ]_{n\times m} } 

E.g 

\\\mathrm{A=\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix} \Rightarrow A' = \begin{bmatrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32}\\ a_{13} & a_{23} & a_{33} \end{bmatrix}} \\\\\\\mathrm{If, \;A=\begin{bmatrix} 2 &6 \\ 3& 7\\ 5& 8 \end{bmatrix}\;\;then,\;\;A'=\begin{bmatrix} 2 &3 &5 \\6 &7 & 8 \end{bmatrix}}

 

-

 

 

\\A=\left[\begin{array}{ccc}{5} & {6} & {7} \\ {0} & {1} & {-1} \\ {0} & {0} & {8}\end{array}\right]\\\\\\ A^T=\left[\begin{array}{ccc}{5} & {0} & {0} \\ {6} & {1} & {0} \\ {7} & {-1} & {8}\end{array}\right]

Hence option (C) is correct

Posted by

Gaurav

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