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Exercise 3.3 · Q4

Q.If A′=[−2312]A' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix} and B=[−1012]B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}, then find (A+2B)′(A + 2B)'

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Concept understanding — Matrix Transpose

Matrix Transpose

The transpose is one of the simplest yet most useful operations on a matrix: you flip the matrix across its main diagonal, so that its rows become columns and its columns become rows.

The intuition

Picture writing a table of marks with students down the rows and subjects across the columns. If instead you want subjects down the rows and students across the columns, you don't recollect the data — you just turn the table on its side. That turn is the transpose.

The precise definition

If A=[aij]A = [a_{ij}] is a matrix of order m×nm \times n, its transpose, written A′A' (or ATA^{T}), is the n×mn \times m matrix obtained by interchanging rows and columns:

A′=[aji],so the (i,j) entry of A′ is the (j,i) entry of A.A' = [a_{ji}], \qquad \text{so the } (i,j) \text{ entry of } A' \text{ is the } (j,i) \text{ entry of } A.

The entry in row ii, column jj of AA moves to row jj, column ii of A′A'.

A worked look

A=[251034]2×3⟹A′=[205314]3×2.A = \begin{bmatrix} 2 & 5 & 1 \\ 0 & 3 & 4 \end{bmatrix}_{2\times 3} \qquad\Longrightarrow\qquad A' = \begin{bmatrix} 2 & 0 \\ 5 & 3 \\ 1 & 4 \end{bmatrix}_{3\times 2}.

The first row (2,5,1)(2,5,1) of AA has become the first column of A′A'.

Properties you must know

For matrices A,BA,B of suitable orders and a scalar kk:

  • (A′)′=A(A')' = A — transposing twice returns the original.
  • (kA)′=k A′(kA)' = k\,A' — a scalar comes straight through.
  • (A+B)′=A′+B′(A+B)' = A' + B' — transpose distributes over addition.
  • (AB)′=B′A′(AB)' = B'A' — the reversal law: the transpose of a product reverses the order of the factors. …

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