Question
Find the transpose of the matrix: $\left[\begin{array}{c} {5} \\ {\frac{1}{2}} \\ {-1} \end{array}\right]$
✓
Answer
We know that transpose of a matrix is obtained by interchanging the elements of the rows and columns. In other words, we can say, if
$\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{m \times \mathrm{n}} \text { then } \mathrm{A}^{\prime}=\left[\mathrm{a}_{\mathrm{ji}}\right]_{\mathrm{n\times m}}$
So, let $\left[\begin{array}{c} {5} \\ {\frac{1}{2}} \\ {-1} \end{array}\right]=A$
Therefore, transpose of the given matrix A is denoted by A’
Hence, A' = $\left[\begin{array}{ccc} {5} & {\frac{1}{2}} & {-1} \end{array}\right]$
The transpose of the given matrix is $\left[\begin{array}{ccc} {5} & {\frac{1}{2}} & {-1} \end{array}\right]$
$\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{m \times \mathrm{n}} \text { then } \mathrm{A}^{\prime}=\left[\mathrm{a}_{\mathrm{ji}}\right]_{\mathrm{n\times m}}$
So, let $\left[\begin{array}{c} {5} \\ {\frac{1}{2}} \\ {-1} \end{array}\right]=A$
Therefore, transpose of the given matrix A is denoted by A’
Hence, A' = $\left[\begin{array}{ccc} {5} & {\frac{1}{2}} & {-1} \end{array}\right]$
The transpose of the given matrix is $\left[\begin{array}{ccc} {5} & {\frac{1}{2}} & {-1} \end{array}\right]$
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