Subtraction of Matrices | Examples on Difference of Two Matrices

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Two matrices A and B are said to be conformable for subtraction
if they have the same order (i.e. same number of rows and columns) and their
difference A – B is defined to be the addition of A and (-B).

i.e., A – B = A + (-B)

For example:

(begin{bmatrix} a_{11} & a_{12} & a_{13}\ a_{21}
& a_{22} & a_{23}\ a_{31} & a_{32} & a_{33} end{bmatrix}) – (begin{bmatrix}
b_{11} & b_{12} & b_{13}\ b_{21} & b_{22} & b_{23}\ b_{31}
& b_{32} & b_{33} end{bmatrix})

= (begin{bmatrix} a_{11} & a_{12} & a_{13}\
a_{21} & a_{22} & a_{23}\ a_{31} & a_{32} & a_{33}
end{bmatrix}) + (begin{bmatrix} – b_{11} & – b_{12} & – b_{13}\ – b_{21}
& – b_{22} & – b_{23}\ -b_{31} & – b_{32} & – b_{33}
end{bmatrix})

= (begin{bmatrix} a_{11} – b_{11} & a_{12} – b_{12}
& a_{13} – b_{13}\ a_{21} – b_{21} & a_{22} – b_{22} & a_{23} –
b_{23}\ a_{31} – b_{31} & a_{32} – b_{32} & a_{33} – b_{33}
end{bmatrix})


Again, if A  = (aij)m,
n
and B = (bij)m, n then their difference A – B is the matrix C = (cij)m,n
where cij = aij – bij, i = 1, 2, 3, …… , m, j = 1, 2, 3, …., n.

For example:

If A = (begin{bmatrix} a_{11} & a_{12} & a_{13}\
a_{21} & a_{22} & a_{23}\ a_{31} & a_{32} & a_{33}
end{bmatrix}) and B = (begin{bmatrix} b_{11} & b_{12} & b_{13}\ b_{21}
& b_{22} & b_{23}\ b_{31} & b_{32} & b_{33} end{bmatrix}),
then

A – B = (begin{bmatrix} a_{11} – b_{11} & a_{12} –
b_{12} & a_{13} – b_{13}\ a_{21} – b_{21} & a_{22} – b_{22} &
a_{23} – b_{23}\ a_{31} – b_{31} & a_{32} – b_{32} & a_{33} – b_{33}
end{bmatrix}) = C

Note: If A and B be matrices of different orders, then A – B
is not defined.

Example on Subtraction of Matrices:

1. If A = (begin{bmatrix} 1 & 2\ 3 & 1 end{bmatrix}) and B = (begin{bmatrix} 2 & 4\ 1 & 3 end{bmatrix}), then 

A – B = (begin{bmatrix} 1 & 2\ 3 & 1 end{bmatrix}) – (begin{bmatrix} 2 & 4\ 1 & 3 end{bmatrix})

         = (begin{bmatrix} 1 – 2 & 2 – 4\ 3 – 1 & 1 – 3end{bmatrix})

         = (begin{bmatrix} -1 & -2\ 2 & -2 end{bmatrix})

2. If A = (begin{bmatrix} 0 & 1 & 2\ 2 & -3 & 1\ 1 & -2 & 0 end{bmatrix}), B = (begin{bmatrix} -1 & 0 & 2\ 3 & 2 & 1\ -2 & -1 & 0 end{bmatrix}) and M = (begin{bmatrix} 4 & 2\ 1 & 3 end{bmatrix}), then 

A – B = (begin{bmatrix} 0 & 1 & 2\ 2 & -3 & 1\ 1 & -2 & 0 end{bmatrix}) – (begin{bmatrix} -1 & 0 & 2\ 3 & 2 & 1\ -2 & -1 & 0 end{bmatrix}) 

        = (begin{bmatrix} 0 – 1 & 1 – 0 & 2 – 2\ 2 – 3 & -3 – 2 & 1 – 1\ 1 – (-2) & -2 – (-1) & 0 – 0 end{bmatrix})

       = (begin{bmatrix} -1 & 1 & 0\ -1 & -5 & 0\ 3 & -1 & 0 end{bmatrix})

A – M is not defined since the order of matrix M is not equal to the order of matrix A.

B – M is also not defined since the order of matrix M is not equal to the order of matrix B.

Note: Let A and B are m × n matrices and c, d are scalars. Then the following results are obvious. 

I. c(A – B) = cA – cB,

For Example:

If A = (begin{bmatrix} 2 & 5\ 3 & 1 end{bmatrix}) and B =  (begin{bmatrix} 3 & 2\ 0 & 7 end{bmatrix}) are m × n matrices and 5 is scalar. Then 

[5left (begin{bmatrix} 2 & 5\ 3 & 1 end{bmatrix} + begin{bmatrix} 3 & 2\ 0 & 7 end{bmatrix} right ) = 5begin{bmatrix} 2 & 5\ 3 & 1 end{bmatrix} + 5 begin{bmatrix} 3 & 2\ 0 & 7 end{bmatrix}]

II. (c – d)A = cA – dA.

For Example:

If A = (begin{bmatrix} 2 & 5\ 3 & 1 end{bmatrix}) be m × n matrix and 5, 3 are scalars. Then 

[left (5 + 3right )begin{bmatrix} 2 & 5\ 3 & 1 end{bmatrix} = 5begin{bmatrix} 2 & 5\ 3 & 1 end{bmatrix} + 3begin{bmatrix} 2 & 5\ 3 & 1 end{bmatrix}]

`

10th Grade Math

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