Algebra of Matrices

The algebra of matrices includes

  1. Addition of Matrices
  2. Subtraction of Matrices
  3. Multiplication of a Matrix by Scalar
  4. Multiplication of Matrices

Addition of Matrices:
Two matrices A and B can be added only if order of matrix A is equal to order of matrix B.
Then addition (A + B) of matrices A and B can be obtained by adding the corresponding elements A and B. The order of (A + B) is the same as the order of A the order of B.
Suppose

A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \end{array}} \right],\,\,\,B  = \left[ {\begin{array}{*{20}{c}} {{b_{11}}}&{{b_{12}}}&{{b_{13}}} \\ {{b_{21}}}&{{b_{22}}}&{{b_{23}}} \end{array}} \right]


Then

\begin{gathered} A + B = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \end{array}} \right] +  \left[ {\begin{array}{*{20}{c}} {{b_{11}}}&{{b_{12}}}&{{b_{13}}} \\ {{b_{21}}}&{{b_{22}}}&{{b_{23}}} \end{array}} \right] \\ A + B = \left[ {\begin{array}{*{20}{c}} {{a_{11}} + {b_{11}}}&{{a_{12}} +  {b_{12}}}&{{a_{13}} + {b_{13}}} \\ {{a_{21}} + {b_{21}}}&{{a_{22}} +  {b_{22}}}&{{a_{23}} + {b_{23}}} \end{array}} \right] \\ \end{gathered}

Example:
Let

A = \left[ {\begin{array}{*{20}{c}} 2&0&0 \\ 3&1&1 \\ 1&4&3 \end{array}}  \right],\,\,\,\,B = \left[ {\begin{array}{*{20}{c}} { - 4}&1&4 \\ 3&0&5 \\ 6&{ - 9}&1 \end{array}} \right]

Then

\begin{gathered} A + B = \left[ {\begin{array}{*{20}{c}} 2&0&0 \\ 3&1&1 \\ 1&4&3 \end{array}} \right] +  \left[ {\begin{array}{*{20}{c}} { - 4}&1&4 \\ 3&0&5 \\ 6&{ - 9}&1 \end{array}} \right] \\ A + B = \left[ {\begin{array}{*{20}{c}} {2 + ( - 4)}&{0 + 1}&{0 + 4} \\ {3 + 3}&{1 + 0}&{1 + 5} \\ {1 + 6}&{4 + ( - 9)}&{3 + 1} \end{array}} \right] =  \left[ {\begin{array}{*{20}{c}} { - 2}&1&4 \\ 6&1&6 \\ 7&{ - 5}&4 \end{array}} \right] \\ \end{gathered}

Subtraction of Matrices:
Subtraction of two matrices is the similar to the addition of two matrices. Two matrices A and B are said to be conformable for subtraction A - B, if both A and B have the same order.
Subtraction can be obtained by taking differences of the corresponding elements of matrices A and B. The order of A - B is the same as the order of A and order of B.

Suppose

A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \end{array}} \right],\,\,\,B  = \left[ {\begin{array}{*{20}{c}} {{b_{11}}}&{{b_{12}}}&{{b_{13}}} \\ {{b_{21}}}&{{b_{22}}}&{{b_{23}}} \end{array}} \right]


Then

\begin{gathered} A - B = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \end{array}} \right] -  \left[ {\begin{array}{*{20}{c}} {{b_{11}}}&{{b_{12}}}&{{b_{13}}} \\ {{b_{21}}}&{{b_{22}}}&{{b_{23}}} \end{array}} \right] \\ A - B = \left[ {\begin{array}{*{20}{c}} {{a_{11}} - {b_{11}}}&{{a_{12}} -  {b_{12}}}&{{a_{13}} - {b_{13}}} \\ {{a_{21}} - {b_{21}}}&{{a_{22}} -  {b_{22}}}&{{a_{23}} - {b_{23}}} \end{array}} \right] \\ \end{gathered}

Example:
Let

A = \left[ {\begin{array}{*{20}{c}} 2&0&0 \\ 3&1&1 \\ 1&4&3 \end{array}} \right],\,\,\,B  = \left[ {\begin{array}{*{20}{c}} { - 4}&1&4 \\ 3&0&5 \\ 6&{ - 9}&1 \end{array}} \right]

Then

\begin{gathered} A - B = \left[ {\begin{array}{*{20}{c}} 2&0&0 \\ 3&1&1 \\ 1&4&3 \end{array}} \right] -  \left[ {\begin{array}{*{20}{c}} { - 4}&1&4 \\ 3&0&5 \\ 6&{ - 9}&1 \end{array}} \right] \\ A - B = \left[ {\begin{array}{*{20}{c}} {2 - ( - 4)}&{0 - 1}&{0 - 4} \\ {3 - 3}&{1 - 0}&{1 - 5} \\ {1 - 6}&{4 - ( - 9)}&{3 - 1} \end{array}} \right] =  \left[ {\begin{array}{*{20}{c}} 6&{ - 1}&{ - 4} \\ 0&1&{ - 4} \\ { - 5}&{13}&2 \end{array}} \right] \\ \end{gathered}

Multiplication of a Matrix by Scalar:
Let A be any given matrix and let k be any real number (scalar), then multiplication kA, of the matrix A with the real number k is obtained by multiplying each element of the matrix A by k.
Suppose

A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \end{array}} \right]


Then

kA = \left[ {\begin{array}{*{20}{c}} {k{a_{11}}}&{k{a_{12}}}&{k{a_{13}}} \\ {k{a_{21}}}&{k{a_{22}}}&{k{a_{23}}} \end{array}} \right]


Example:
Let

A = \left[ {\begin{array}{*{20}{c}} 4&1&2 \\ 3&0&5 \\ 6&4&{ - 1} \end{array}} \right]\,and\,k  = 2


Then

2A = \left[ {\begin{array}{*{20}{c}} {2(4)}&{2(1)}&{2(2)} \\ {2(3)}&{2(0)}&{2(5)} \\ {2(6)}&{2(4)}&{2( - 1)} \end{array}} \right] =  \left[ {\begin{array}{*{20}{c}} 8&2&4 \\ 6&0&{10} \\ {12}&8&{ - 2} \end{array}} \right]

Multiplication of Matrices:
Let A and B be any two given matrices, then the multiplication AB can be possible only if number of columns of matrix A is equal to number of rows of matrix B. Then multiplication AB can be obtained by the following method.
The element (1,1) position of AB is obtained by adding the products of corresponding elements of 1st row of A and 1st column of B. Similarly, the element (1,2) position of AB is obtained by adding the products of corresponding elements of 2nd row of A and 2nd column of B and so on.
Suppose

A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right],\,\,\,B  = \left[ {\begin{array}{*{20}{c}} {{b_{11}}}&{{b_{12}}} \\ {{b_{21}}}&{{b_{22}}} \\ {{b_{31}}}&{{b_{32}}} \end{array}} \right]

AB = \left[ {\begin{array}{*{20}{c}} {{a_{11}}&{b_{11}} + {a_{12}}&{b_{21}} + {a_{13}}&{b_{31}}}&{{a_{11}}&{b_{12}} + {a_{12}}&{b_{22}} + {a_{13}}&{b_{32}}}\\<br />
{{a_{21}}&{b_{11}} + {a_{22}}&{b_{21}} + {a_{ & 23}}&{b_{31}}}&{{a_{21}}&{b_{12}} + {a_{22}}{b_{22}} + {a_{23}}&{b_{32}}} \\ {{a_{31}}&{b_{11}} + {a_{32}}&{b_{21}} + {a_{33}}&{b_{31}}}&{{a_{31}}&{b_{12}} + {a_{32}}&{b_{22}} + {a_{33}}&{b_{32}}} \end{array}} \right]

Example:
Let

A = \left[ {\begin{array}{*{20}{c}} 1&2&3 \\ 4&5&6 \end{array}} \right],\,\,\,B  = \left[ {\begin{array}{*{20}{c}} 7&1 \\ 8&2 \\ 9&3 \end{array}} \right]


Since number of columns of A is 3 and number of rows of B is also 3, So AB can be found. i.e.

\begin{gathered} AB = \left[ {\begin{array}{*{20}{c}} 1&2&3 \\ 4&5&6 \end{array}} \right]\left[  {\begin{array}{*{20}{c}} 7&1 \\ 8&2 \\ 9&3 \end{array}} \right] \\ AB = \left[ {\begin{array}{*{20}{c}} {(1)(7) + (2)(8) + (3)(9)}&{(1)(1) +  (2)(2) + (3)(3)} \\ {(4)(7) + (5)(8) + (6)(9)}&{(4)(1) + (5)(2) + (6)(3)} \end{array}} \right] \\ AB = \left[ {\begin{array}{*{20}{c}} {7 + 16 + 27}&{1 + 4 + 9} \\ {28 + 40 + 54}&{4 + 10 + 18} \end{array}} \right] =  \left[ {\begin{array}{*{20}{c}} {50}&{14} \\ {122}&{32} \end{array}} \right] \\ \end{gathered}

Comments

comments