Reversal law for transpose
\[(AB)^{T} = B^{T}A^{T}\]
Also \((A^T)^T = A\), \((A+B)^T = A^T + B^T\), \((kA)^T = kA^T\).
Matrices
The transpose \(A^T\) (also written \(A\')\) is obtained by turning rows into columns: if \(A = [a_{ij}]_{m\times n}\) then \(A^T = [a_{ji}]_{n\times m}\).
Example: \(\begin{bmatrix} 1 & 4 & -2 \\ 0 & 3 & 5 \end{bmatrix}^{T} = \begin{bmatrix} 1 & 0 \\ 4 & 3 \\ -2 & 5 \end{bmatrix}\).
\[(AB)^{T} = B^{T}A^{T}\]
Also \((A^T)^T = A\), \((A+B)^T = A^T + B^T\), \((kA)^T = kA^T\).
\[(A^T)^T=A,\quad (kA)^T=kA^T,\quad (A+B)^T=A^T+B^T,\quad (AB)^T=B^TA^T\]
Writing \((AB)^T = A^TB^T\).
Correct: \((AB)^T = B^TA^T\). Check with orders: if \(A\) is \(2\times3\) and \(B\) is \(3\times4\), \(A^TB^T\) is not even defined.
A fresh set of questions from this topic. Change answers freely – solutions appear after you submit.
Start Practice\(A^T\): rows ↔ columns. \((A^T)^T = A\), \((A+B)^T = A^T + B^T\), \((kA)^T = kA^T\), \((AB)^T = B^TA^T\).
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