Subtraction of Matrices

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

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

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


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 = \(\begin{bmatrix} 1 & 3\\ 2 & 4 \end{bmatrix}\) and B =  \(\begin{bmatrix} 2 & 1\\ 3 & 0 \end{bmatrix}\) are m × n matrices and 4 is scalar. Then 

\[4\left (\begin{bmatrix} 1 & 3\\ 2 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 1\\ 3 & 0 \end{bmatrix}\right ) = 4\begin{bmatrix} 1 & 3\\ 2 & 4 \end{bmatrix} - 4 \begin{bmatrix} 2 & 1\\ 3 & 0 \end{bmatrix}\]


II. (c - d)A = cA - dA.

For Example:

If A = \(\begin{bmatrix} 2 & 0\\ -1 & 5 \end{bmatrix}\) be m × n matrix and 4, 2 are scalars. Then 

\[\left (4 - 2\right )\begin{bmatrix} 2 & 0\\ -1 & 5 \end{bmatrix} = 4\begin{bmatrix} 2 & 0\\ -1 & 5 \end{bmatrix} - 2\begin{bmatrix} 2 & 0\\ -1 & 5 \end{bmatrix}\]





10th Grade Math

From Subtraction of Matrices to HOME


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