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  • Given A =2 4 0 and B =1 4 . Is (AB)’ = B’A’?

Given A=\left[\begin{array}{lll} 2 & 4 & 0 \\ 3 & 9 & 6 \end{array}\right]_{\text {and }} B=\left[\begin{array}{ll} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{array}\right] . Is (AB)' = B'A'?

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We have two given matrices A and B,

A=\left[\begin{array}{lll} 2 & 4 & 0 \\ 3 & 9 & 6 \end{array}\right]_{\text {and }} B=\left[\begin{array}{ll} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{array}\right]

We need to verify whether  (AB)' = B'A'

Let us see what a transpose is.

In linear algebra, the transpose of a matrix is an operator which flips a matrix over its diagonal, that is it switches the row and column indices of the matrix by producing another matrix denoted as A^T.

Take L.H.S = (AB)'

So, let us compute AB.

A B=\left[\begin{array}{lll} 2 & 4 & 0 \\ 3 & 9 & 6 \end{array}\right]\left[\begin{array}{ll} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{array}\right]

Multiply 1st row of matrix A by matching members of 1st column of matrix B, then finally end by summing it up.

\\(2, 4, 0)(1, 2, 1) = (2 \times 1) + (4 \times 2) + (0 \times 1) \\ \Rightarrow (2, 4, 0)(1, 2, 1) = 2 + 8 + 0 \\ \Rightarrow (2, 4, 0)(1, 2, 1) = 10

\left[\begin{array}{lll} 2 & 4 & 0 \\ 3 & 9 & 6 \end{array}\right]\left[\begin{array}{ll} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{array}\right]= \begin{bmatrix} 10 & \\ & \end{bmatrix}

Multiply 1st row of matrix A by matching members of 2nd column of matrix B, then finally end by summing them up.

\\(2, 4, 0)(4, 8, 3) = (2 \times 4) + (4 \times 8) + (0 \times 3) \\ \Rightarrow (2, 4, 0)(4, 8, 3) = 8 + 32 + 0 \\ \Rightarrow (2, 4, 0)(4, 8, 3) = 40

\left[\begin{array}{lll} 2 & 4 & 0 \\ 3 & 9 & 6 \end{array}\right]\left[\begin{array}{ll} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{array}\right]= \begin{bmatrix} 10 &40 \\ & \end{bmatrix}

Similarly, let us fill for the rest of the elements.

\left[\begin{array}{lll} 2 & 4 & 0 \\ 3 & 9 & 6 \end{array}\right]\left[\begin{array}{ll} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{array}\right]= \begin{bmatrix} 10 &40 \\(3\times1)+(9\times2)+(6\times1) & (3\times4)+(9\times8)+(6\times3) \end{bmatrix}

\left[\begin{array}{lll} 2 & 4 & 0 \\ 3 & 9 & 6 \end{array}\right]\left[\begin{array}{ll} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{array}\right]= \begin{bmatrix} 10 &40 \\3+18+6 & 12+72+18 \end{bmatrix}

\left[\begin{array}{lll} 2 & 4 & 0 \\ 3 & 9 & 6 \end{array}\right]\left[\begin{array}{ll} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{array}\right]= \begin{bmatrix} 10 &40 \\27 & 102 \end{bmatrix}

So, AB= \begin{bmatrix} 10 &40 \\27 & 102 \end{bmatrix}

Now, for transpose of AB, rows will become columns.

(AB)'= \begin{bmatrix} 10 &27 \\40 & 102 \end{bmatrix}

Now, take R.H.S = B’A’

If B = \begin{bmatrix} 1 &4 \\2 &8 \\1 &3 \end{bmatrix}

Then, if (1, 4) are the elements of 1st row, it will become elements of 1st column, and so on.

B' = \begin{bmatrix} 1 &2 &1 \\4 &8 &3 \end{bmatrix}

Also,

A = \begin{bmatrix} 2 &4 &0 \\3 &9 &6 \end{bmatrix}

Then, if (2, 4, 0) are the elements of 1st row, it will become elements of 1st column, and so on.

A'= \begin{bmatrix} 2 &3 \\4 &9 \\0 &6 \end{bmatrix}

Now, multiply B’A’.

B' A'= \begin{bmatrix} 1 &2 &1 \\4 &8 &3 \end{bmatrix}\begin{bmatrix} 2 &3 \\4 &9 \\0 &6 \end{bmatrix}

Multiply 1st row of matrix B’ by matching members of 1st column of matrix A’, then finally end by summing them up.

\\(1, 2, 1)(2, 4, 0) = (1 \times 2) + (2 \times 4) + (1 \times 0) \\ \Rightarrow (1, 2, 1)(2, 4, 0) = 2 + 8 + 0 \\ \Rightarrow (1, 2, 1)(2, 4, 0) = 10.


\begin{bmatrix} 1 &2 &1 \\4 &8 &3 \end{bmatrix} \begin{bmatrix} 2 &3 \\4 &9 \\0 &6 \end{bmatrix} = \begin{bmatrix} 10 & & \\ & & \\ & & \end{bmatrix}

  Multiply 1st row of matrix B’ by matching members of 2nd column of matrix A’, then finally end by summing it up.

\\(1, 2, 1)(3, 9, 6) = (1 \times 3) + (2 \times 9) + (1 \times 6) \\ \Rightarrow (1, 2, 1)(3, 9, 6) = 3 + 18 + 6 \\ \Rightarrow (1, 2, 1)(3, 9, 6) = 27

\begin{bmatrix} 1 &2 &1 \\4 &8 &3 \end{bmatrix} \begin{bmatrix} 2 &3 \\4 &9 \\0 &6 \end{bmatrix} = \begin{bmatrix} 10 &27 & \\ & & \\ & & \end{bmatrix}

Filling up the rest of the elements in the similar manner

\begin{bmatrix} 1 &2 &1 \\4 &8 &3 \end{bmatrix} \begin{bmatrix} 2 &3 \\4 &9 \\0 &6 \end{bmatrix} = \begin{bmatrix} 10 &27 \\ 4*2+8*4+3*0 &4*3+8*9+ 3*6 \end{bmatrix}

\begin{bmatrix} 1 &2 &1 \\4 &8 &3 \end{bmatrix} \begin{bmatrix} 2 &3 \\4 &9 \\0 &6 \end{bmatrix} = \begin{bmatrix} 10 &27 \\8+32+0 &12+72+ 18 \end{bmatrix}

\begin{bmatrix} 1 &2 &1 \\4 &8 &3 \end{bmatrix} \begin{bmatrix} 2 &3 \\4 &9 \\0 &6 \end{bmatrix} = \begin{bmatrix} 10 &27 \\40 &102\end{bmatrix}

⇒ L.H.S = R.H.S

Therefore, (AB)' = B'A'

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infoexpert22

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