Question

Is matrix multiplication associative ?

Answer 1

Hint: In general, the associative law of multiplication states that the order of variables will not make any difference in the result. In the given question we are going to show that matrix multiplication is associative. It means for any three matrices \[A,B\] and \[C\], associative property states: \[(AB)C = A(BC)\]. To prove this, let's consider three matrices in the same order.

Complete step by step answer:
We need to prove that \[(AB)C = A(BC)\], that is L.H.S. \[ = \] R.H.S.
Let \[A = \left( {\begin{array}{*{20}{c}}
  4&{ - 2} \\
  6&{ - 3}
\end{array}} \right)\], \[B = \left( {\begin{array}{*{20}{c}}
  0&2 \\
  1&{ - 1}
\end{array}} \right)\] and \[C = \left( {\begin{array}{*{20}{c}}
  { - 2}&3 \\
  1&{ - 3}
\end{array}} \right)\].
L.H.S. \[ = (AB)C\]
Here first multiply \[AB\] and then multiply \[C\] with its result.
\[AB = \left( {\begin{array}{*{20}{c}}
  4&{ - 2} \\
  6&{ - 3}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  0&2 \\
  1&{ - 1}
\end{array}} \right)\]
As per matrix multiplication, multiply each row of the first matrix with each column of the second matrix,
\[AB = \left( {\begin{array}{*{20}{c}}
  {(4 \times 0) + ( - 2 \times 1)}&{(4 \times 2) + ( - 2 \times - 1)} \\
  {(6 \times 0) + ( - 3 \times 1)}&{(6 \times 2) + ( - 3 \times - 1)}
\end{array}} \right)\]
Multiply the values inside the bracket,
\[ AB = \left( {\begin{array}{*{20}{c}}
  {0 - 2}&{8 + 2} \\
  {0 - 3}&{12 + 3}
\end{array}} \right)\]
Perform the action specified in each elements,
\[AB = \left( {\begin{array}{*{20}{c}}
  { - 2}&{10} \\
  { - 3}&{15}
\end{array}} \right)\]
Now multiplying \[C\] with the result of \[AB\],
\[(AB)C = \left( {\begin{array}{*{20}{c}}
  { - 2}&{10} \\
  { - 3}&{15}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  { - 2}&3 \\
  1&{ - 3}
\end{array}} \right)\]
By multiplying each row of the first matrix with each column of the second matrix,

\[(AB)C = \left( {\begin{array}{*{20}{c}}
  {( - 2 \times - 2) + (10 \times 1)}&{( - 2 \times 3) + (10 \times - 3)} \\
  {( - 3 \times - 2) + (15 \times 1)}&{( - 3 \times 3) + (15 \times - 3)}
\end{array}} \right)\]
Perform multiplication inside the bracket,
\[(AB)C = \left( {\begin{array}{*{20}{c}}
  {4 + 10}&{ - 6 - 30} \\
  {6 + 15}&{ - 9 - 45}
\end{array}} \right)\]
By performing addition, we will get,
\[(AB)C = \left( {\begin{array}{*{20}{c}}
  {14}&{ - 36} \\
  {21}&{ - 54}
\end{array}} \right)\]
L.H.S. \[ = (AB)C = \left( {\begin{array}{*{20}{c}}
  {14}&{ - 36} \\
  {21}&{ - 54}
\end{array}} \right)\]
R.H.S. \[ = A(BC)\]
Whereas here, multiply \[BC\] first and then multiply its result with \[A\].
\[BC = \left( {\begin{array}{*{20}{c}}
  0&2 \\
  1&{ - 1}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  { - 2}&3 \\
  1&{ - 3}
\end{array}} \right)\]
By performing matrix multiplication,
\[BC = \left( {\begin{array}{*{20}{c}}
  {(0 \times - 2) + (2 \times 1)}&{(0 \times 3) + (2 \times - 3)} \\
  {(1 \times - 2) + ( - 1 \times 1)}&{(1 \times 3) + ( - 1 \times - 3)}
\end{array}} \right)\]
\[ BC= \left( {\begin{array}{*{20}{c}}
  {0 + 2}&{0 - 6} \\
  { - 2 - 1}&{3 + 3}
\end{array}} \right)\]
Performing addition,
\[BC = \left( {\begin{array}{*{20}{c}}
  2&{ - 6} \\
  { - 3}&6
\end{array}} \right)\]
Multiplying \[BC\] with \[A\],
\[A(BC) = \left( {\begin{array}{*{20}{c}}
  4&{ - 2} \\
  6&{ - 3}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  2&{ - 6} \\
  { - 3}&6
\end{array}} \right)\]
Again by performing matrix multiplication,
\[A(BC) = \left( {\begin{array}{*{20}{c}}
  {(4 \times 2) + ( - 2 \times - 3)}&{(4 \times - 6) + ( - 2 \times 6)} \\
  {(6 \times 2) + ( - 3 \times - 3)}&{(6 \times - 6) + ( - 3 \times 6)}
\end{array}} \right)\]
\[A(BC) = \left( {\begin{array}{*{20}{c}}
  {8 + 6}&{ - 24 - 12} \\
  {12 + 9}&{ - 36 - 18}
\end{array}} \right)\]
By adding the values,
\[A(BC) = \left( {\begin{array}{*{20}{c}}
  {14}&{ - 36} \\
  {21}&{ - 54}
\end{array}} \right)\]
R.H.S. \[ = A(BC) = \left( {\begin{array}{*{20}{c}}
  {14}&{ - 36} \\
  {21}&{ - 54}
\end{array}} \right)\].
The result of both \[(AB)C\] and \[A(BC)\] are equal.Hence we have proved that L.H.S. \[ = \] R.H.S., that is \[(AB)C = A(BC)\]. Therefore matrix multiplication is associative.

Note: A matrix is a rectangular arrangement of number enclosed by a pair of brackets \[\left( {\begin{array}{*{20}{c}}
  {\text{ }}&{\text{ }} \\
  {\text{ }}&{\text{ }}
\end{array}} \right)\], in the form of horizontal and vertical lines. The numbers inside the brackets are called elements of the matrix. Matrices are usually denoted by capital letters \[{\text{A,B,C,D,}}\] etc. Horizontal lines are called rows and vertical lines are called columns. If a matrix has \[{\text{m}}\] rows and \[{\text{n}}\] columns, then the order of the matrix is said to be \[{\text{m}} \times {\text{n}}\] (read as \[{\text{m}}\] by \[{\text{n}}\]) and it has \[{\text{mn}}\] elements.

The multiplication of two matrices \[A\] and \[B\] is only possible when the number of columns in \[A\] is equal to the number of rows in \[B\]. If \[A\] be an \[{\text{m}} \times {\text{n}}\] order and \[B\] be an \[n \times p\] order matrix, then their product \[AB\] has \[m \times p\] order is defined as \[AB = {[{C_{ik}}]_{m \times p}}\] where \[{C_{ik}} = \] sum of the elements of the \[i\] th row of \[A\] with corresponding elements of the \[{k^{th}}\] column of \[B\].

Associative law for addition of any three matrices: \[A = [{a_{ij}}]\], \[B = [{b_{ij}}]\] and \[C = [{c_{ij}}]\] of the same order, say \[{\text{m}} \times {\text{n}}\], \[(A + B) + C = A + (B + C)\].
The associative law for multiplication of any three matrices: \[A,B\] and \[C\] is \[(AB)C = A(BC)\], whenever both sides of the equality are defined.