Matrix Operations

Matrix operations mainly involve three operations which are addition of matrices, subtraction of matrices, and multiplication of matrices. All these operations on matrices are covered in this lesson along with the properties of matrix operation and solved examples.

All Contents in Matrices

Before starting with the operation, it is important to know about the elementary operation of a matrix in detail which is given in the linked article. Also, the concept of equality of matrices is given below.

Important Operation on Matrix

Addition of Matrices

If A[aij]mxn and B[bij]mxn are two matrices of the same order then their sum A + B is a matrix, and each element of that matrix is the sum of the corresponding elements. i.e. A + B = [aij + bij]mxn

Properties of Matrix Addition: If a, B and C are matrices of same order, then

(a) A + B = B + A(Commutative law),

(b) (A + B) + C = A + (B + C) (Associative law),

(c) A + O + O + A = A, where O is zero matrix which is additive identity of the matrix,

(d) A + (-A) = 0 = (-A) + A, where (-A) is obtained by changing the sign of every element of A which is additive inverse of the matrix,

(e) \(\left. \begin{matrix} A+B=A+C \\ B+A=C+A \\ \end{matrix} \right\}\Rightarrow B=C\)

(f) \(tr\left( A\pm B \right)=tr\left( A \right)\pm tr\left( B \right)\)

(g) Additive Inverse: If A + B = 0 = B + A, then B is called additive inverse of A and also A is called the additive inverse of A.

(h) Existence of Additive Identity: Let \(A=\left[ {{a}_{ij}} \right]\) be anmatrix and O be an mxn zero matrix, then A+O=O+A=A. In other words, O is the additive identity for matrix addition.

Subtraction of Matrices

If A and B are two matrices of the same order, then we define \(A-B=A+\left( -B \right).\)

Scalar Multiplication of Matrices

If \(A={{\left[ {{a}_{ij}} \right]}_{m\times n}}\) is a matrix and k any number, then the matrix which is obtained by multiplying the elements of A by k is called the scalar multiplication of A by k and it is denoted by k A thus if \(A={{\left[ {{a}_{ij}} \right]}_{m\times n}}\) Then \(k{{A}_{m\,\times n}}={{A}_{m\,\times \,n}}k=\left[ k{{a}_{i\times j}} \right]\)

Properties of Scalar Multiplication: If A, B are matrices of the same order and are any two scalars then;

(a) \(\lambda \left( A+B \right)=\lambda A+\lambda B\) (b) \(\left( \lambda +\mu \right)A=\lambda A+\mu A\)

(c) \(\lambda \left( \mu A \right)=\left( \lambda \,\mu A \right)=\mu \left( \lambda A \right)\) (d) \(\left( -\lambda A \right)=-\left( \lambda A \right)=\lambda \left( -A \right)\)

(e) \(tr\left( kA \right)=k\,\,tr\,\,\left( A \right)\)

Multiplication of Matrices

If A and B be any two matrices, then their product AB will be defined only when the number of columns in A is equal to the number of rows in B.

If \(A{{\left[ {{a}_{ij}} \right]}_{m\,\times n}}. and B{{\left[ {{b}_{ij}} \right]}_{n\,\times p}} then\; their\; product AB=C=\left[ {{C}_{ij}} \right],\) will be a matrix of order mxp where \({{\left( AB \right)}_{ij}}={{C}_{ij}}=\sum\limits_{r=1}^{n}{{{a}_{ir}}{{b}_{rj}}}\)

Properties of matrix multiplication

(a) Matrix multiplication is not commutative in general, i.e. in general \(AB\ne BA.\)

(b) Matrix multiplication is associative, i.e. (AB)C = A(BC).

(c) Matrix multiplication is distributive over matrix addition, i.e. A.(B + C) = A.B + A.C and (A + B)C = AC + BC.

(d) If A is an m × n matrix, then \({{I}_{m}}A=A=A{{I}_{n}}.\)

(e) The product of two matrices can be a null matrix while neither of them is null, i.e. if AB = 0, it is not necessary that either A = 0 or B = 0.

(f) If A is an m × n matrix and O is a null matrix then \({{A}_{m\,\times n}}.{{O}_{n\,\times p}}={{O}_{m\,\times p}}.\) i.e. the product of the matrix with a null matrix is always a null matrix.

(g) If AB = 0 (It does not mean that A = 0 or B = 0, again the product of two non-zero matrices may be a zero matrix).

(h) If AB = AC B C (Cancellation Law is not applicable).

(i) \(tr\left( AB \right)=tr\left( BA \right).\)

Problems on Matrix Operations

Illustration 7: If \(A=\left[ \begin{matrix} 2 & 1 & 3 \\ 3 & -2 & 1 \\ -1 & 0 & 1 \\ \end{matrix} \right] and B=\left[ \begin{matrix} 1 \\ 2 \\ 4 \\ \end{matrix}\,\,\,\,\begin{matrix} -2 \\ 1 \\ -2 \\ \end{matrix} \right]\)

find AB and BA if possible.

Solution:

Using matrix multiplication. Here, A is a 3 × 3 matrix and B is a 3 × 2 matrix, therefore, A and B are conformable for the product AB and it is of the order 3 × 2 such that

(First row of A) (First column of B) \(=\left[ 2\,1\,3 \right]\left[ \begin{matrix} 1 \\ 2 \\ 4 \\ \end{matrix} \right]=2\times 1+1\times 2+3\times 4=16\).

(First row of A) (Second column of B) \(=\left[ 2\,\,1\,\,3 \right]\left[ \begin{matrix} -2 \\ 1 \\ -3 \\ \end{matrix} \right]=2\times \left( -2 \right)+1\times 1+3\times \left( -3 \right)=-12\)

(Second row of A) (First column of B) \(=\left[ 3\,\,-2\,\,1 \right]\left[ \begin{matrix} 1 \\ 2 \\ 4 \\ \end{matrix} \right]=3\times 1+\left( -2 \right)\times 2+1\times 4=3\)

Similarly \({{\left( AB \right)}_{22}}=-11,{{\left( AB \right)}_{31}}=3 \;and \;{{\left( AB \right)}_{32}}=-1\)

∴ AB = \(\left[ \begin{matrix} 16 \\ 3 \\ 3 \\ \end{matrix}\,\,\,\begin{matrix} -12 \\ -11 \\ -1 \\ \end{matrix} \right]\)

BA is not possible since number of columns of \(B\ne\) number of rows of A.

 

Illustration 8: Find the value of x and y if \(2\left[ \begin{matrix} 1 & 3 \\ 0 & x \\ \end{matrix} \right]+\left[ \begin{matrix} y & 0 \\ 1 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 5 & 6 \\ 1 & 8 \\ \end{matrix} \right]\)

Solution:

Using the method of multiplication and addition of matrices, then equating the corresponding elements of L.H.S. and R.H.S., we can easily get the required values of x and y.

We have, \(2\left[ \begin{matrix} 1 & 3 \\ 0 & x \\ \end{matrix} \right]+\left[ \begin{matrix} y & 0 \\ 1 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 5 & 6 \\ 1 & 8 \\ \end{matrix} \right]\Rightarrow \left[ \begin{matrix} 2 & 6 \\ 0 & 2x \\ \end{matrix} \right]+\left[ \begin{matrix} y & 0 \\ 1 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 5 & 6 \\ 1 & 8 \\ \end{matrix} \right]\Rightarrow \left[ \begin{matrix} 2+y & 6+0 \\ 0+1 & 2x+2 \\ \end{matrix} \right]=\left[ \begin{matrix} 5 & 6 \\ 1 & 8 \\ \end{matrix} \right]\)

Equating the corresponding elements, a11 and a22 we get

\(2+y=5\Rightarrow \,\,\,y=3;\,\,\,2x+2=8\,\,\,\Rightarrow 2x=6\,\,\,\Rightarrow x=3;\)

Hence x = 3 and y = 3 .

 

Illustration 9: Find the value of a, b, c and d, if \(\left[ \begin{matrix} a-b & 2a+c \\ 2a-b & 3c+d \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & 5 \\ 0 & 13 \\ \end{matrix} \right]\)

Solution:

As the two matrices are equal, their corresponding elements are equal. Therefore, by equating the corresponding elements of given matrices we will obtain the value of a, b, c and d.

\(\left[ \begin{matrix} a-b & 2a+c \\ 2a-b & 3c+d \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & 5 \\ 0 & 13 \\ \end{matrix} \right]\)

(given)

a – b = 1 … (i)

2a + c = 5 … (ii)

2a-b=0 … (iii)

3c + d = 13 … (iv)

Subtracting equation (i) from (iii), we have a = 1;

Putting the value of a in equation (i), we have\(1-b=-1 \Rightarrow b = 2\);

Putting the value of a in equation (ii), we have \(2+c=5\Rightarrow c=3;\)

Putting the value of c in equation (iv), we find \(9+x=13\Rightarrow d=\)

Hence a=1,b=2,c=3,d=4.

 

Illustration 10: find x and y, if \(2x+3y=\left[ \begin{matrix} 2 & 3 \\ 4 & 0 \\ \end{matrix} \right] and 3x+2y=\left[ \begin{matrix} 2 & -2 \\ -1 & 5 \\ \end{matrix} \right]\)

Solution:

Solving the given equations simultaneously, we will obtain the values of x and y.

We have \(2x+3y=\left[ \begin{matrix} 2 & 3 \\ 4 & 0 \\ \end{matrix} \right]\) … (i)

\(3x+2y=\left[ \begin{matrix} 2 & -2 \\ -1 & 5 \\ \end{matrix} \right]\) … (ii)

Multiplying (i) by 3 and (ii) by 2, we get\(6x+9y=\left[ \begin{matrix} 6 & 9 \\ 12 & 0 \\ \end{matrix} \right]\) … (iii)

\(6x+9y=\left[ \begin{matrix} 6 & 9 \\ 12 & 0 \\ \end{matrix} \right]\) … (iv)

Subtracting (iv) from (iii), we get \(5y=\left[ \begin{matrix} 6-4 & 9+4 \\ 12+2 & 0-10 \\ \end{matrix} \right]=\left[ \begin{matrix} 2 & 13 \\ 14 & -10 \\ \end{matrix} \right]\) \(\Rightarrow y=\left[ \begin{matrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & \frac{-10}{5} \\ \end{matrix} \right]\Rightarrow y=\left[ \begin{matrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \\ \end{matrix} \right]\)

Putting the value of y in (iii), we get \(2x+3\left[ \begin{matrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \\ \end{matrix} \right]=\left[ \begin{matrix} 2 & 3 \\ 4 & 0 \\ \end{matrix} \right]\) \(\Rightarrow 2x=\left[ \begin{matrix} 2 & 3 \\ 4 & 0 \\ \end{matrix} \right]-\left[ \begin{matrix} \frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6 \\ \end{matrix} \right]=\left[ \begin{matrix} 2-\frac{6}{5} & 3-\frac{39}{5} \\ 4-\frac{42}{5} & 0+6 \\ \end{matrix} \right]=\left[ \begin{matrix} \frac{4}{5} & -\frac{24}{5} \\ -\frac{22}{5} & 6 \\ \end{matrix} \right]\Rightarrow x=\left[ \begin{matrix} \frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3 \\ \end{matrix} \right]\)

Hence \(x=\left[ \begin{matrix} \frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3 \\ \end{matrix} \right] \;and\; y=\left[ \begin{matrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \\ \end{matrix} \right]\)

 

Illustration 11: If \(\left[ \begin{matrix} x+3 & z+4 & 2y-7 \\ -6 & a-1 & 0 \\ b-3 & -21 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 6 & 3y-2 \\ -6 & -3 & 2c+z \\ 2b+4 & -21 & 0 \\ \end{matrix} \right]\)

then find the values of a, b, c, x, y and z.

Solution:

As the two matrices are equal, their corresponding elements are also equal. Therefore, by equating the corresponding elements of the given matrices, we will obtain the values of a, b, c, x, y and z.

\(\left[ \begin{matrix} x+3 & z+4 & 2y-7 \\ -6 & a-1 & 0 \\ b-3 & -21 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 6 & 3y-2 \\ -6 & -3 & 2c+z \\ 2b+4 & -21 & 0 \\ \end{matrix} \right]\)

Comparing both sides, we get [x + 3 = 0] ⇒ x = – 3 . . . . (i)

And \(z+4=6 \;\;\;\;\;\; \Rightarrow z=6-4 \;\;\;\;\;\; \Rightarrow z=2\) … (ii)

and \(2y-7=3y-2\;\;\;\;\;\;\;\;\;\; \Rightarrow 2y-3y=-2+7\;\;\;\;\;\;\;\;\;\; \Rightarrow -y=5\;\;\;\;\;\;\;\;\;\; \Rightarrow y=-5\) … (iii)

and \(a-1=-3 \;\;\;\;\;\;\;\;\;\; \Rightarrow a=-3+1 \;\;\;\;\;\;\;\;\;\; \Rightarrow a=-2\) … (iv)

and \(b-3=2b+4\;\;\;\;\;\;\; \Rightarrow b-2b=4+3 \;\;\;\;\;\;\; \Rightarrow -b=7 \;\;\;\;\;\;\; \Rightarrow b=-7\) … (v)

and \(2c+2=0 \;\;\;\;\;\; \Rightarrow 2c+2=0 \;\;\;\;\;\; \Rightarrow 2c=-2 \Rightarrow c=\frac{-2}{2}\;\;\;\;\;\;\Rightarrow c=-1\) … (vi)

[from (2)] Thus; \(a=-2,b=-7,c=-1,x=-3,y=-5 and z=2\)

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