A matrix is a rectangular array of numbers.

\[M=\left[ \begin{matrix}   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} & \cdots  & {{a}_{1n}}  \\   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} & \cdots  & {{a}_{2n}}  \\   \vdots  & \vdots  & \vdots  & {} & \vdots   \\   {{a}_{m1}} & {{a}_{m1}} & {{a}_{m3}} & \cdots  & {{a}_{mn}}  \\\end{matrix} \right]\text{An}\ m\times n\ \text{matrix}\]

A matrix with m = n is a sqaure matrix. For square matrices, we can define the corresponding determinants.

In the discussion that follows, an m × n matrix A will be denoted as \(A=\left[ {{a}_{ij}} \right]\) . The index i will vary from 1 to m, while j will vary from 1 to n. The element \({{a}_{ij}}\) is the element in the ith row and the jth column.

(a) Sum/Difference of matrices : If we have two matrices A and B of the same order m × n, their sum/ difference is given by adding/subtracting the corre sponding elements:

\[\begin{align}   A\pm B&=\left[ {{a}_{ij}} \right]\ \pm \ \left[ {{b}_{ij}} \right] \\  & =\left[ {{a}_{ij}}\pm {{b}_{ij}} \right] \\ \end{align}\]

For example,

\[\left[ \ \begin{matrix}   1 & -2  \\   3 & \ \ 4  \\   5 & -3  \\\end{matrix}\  \right]\ \ +\ \ \left[ \begin{matrix}   \ \ 2 & \ 1  \\   -1 & \ 3  \\   \ \ 4 & \ 2  \\\end{matrix}\  \right]\ \ =\ \left[ \ \begin{matrix}   3 & \ -1  \\   2 & \ \ \ 7  \\   9 & \ -1  \\\end{matrix}\  \right]\ \]

(b) Scalar multiplication : If we multiply a matrix \(A=[{{a}_{ij}}]\ \text{by}\ \lambda ,\) we get the matrix \(\lambda A=[\lambda {{a}_{ij}}].\ \) 

Example:

\[4\left[ \ \begin{matrix}   1 & 3 & \ \ 2  \\   2 & 1 & -1  \\\end{matrix}\  \right]\ \ =\ \ \left[ \ \begin{matrix}   4 & 12 & \ \ 8  \\   8 & 4 & -4  \\\end{matrix}\  \right]\]

(c) Multiplication of matrices: To  understand how matrices are multiplied, let us first consider a row vector

\[R=\left[ {{r}_{1}}\ {{r}_{2}}...{{r}_{n}} \right]\ \ \ \ \ \ \left\{ \begin{align}  & \text{A row vector is matrix} \\ 
 & \text{with just one row} \\ \end{align} \right\}\]

and a column vector

\[C=\left[ \begin{align}  \; \ {{c}_{1}}  \;\\  \; \ {{c}_{2}} \; \\  \; \ \ \vdots \; \  \\  \; \ {{c}_{n}} \;\  \\ \end{align} \right]\ \ \ \qquad \left\{ \begin{gathered}   \text{A column vector is a matrix } \\ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\text{with just one column} \\ \end{gathered} \right\}\]

which are both of order n. The product of R and C can be defined as

\[RC=\left[ {{r}_{1}}\ \ {{r}_{2}}\ \ ...\ {{r}_{n}} \right]\ \left[ \begin{align}  & \ {{c}_{1}} \\  & \ {{c}_{2}} \\  & \ \ \vdots \  \\  & \ {{c}_{n}}\  \\ \end{align} \right]\ ={{r}_{1}}{{c}_{1}}+{{r}_{2}}{{c}_{2}}+...+{{r}_{n}}{{c}_{n}}\]

Therefore, RC is a scalar quantity. For example,

\[\left[ 1\ \ 3\ \ 2 \right]\ \ \left[ \begin{align}  & \ \ 2 \\  & -1 \\  & \ \ 4 \\ \end{align} \right]=7\]

Now, we will discuss matrix multiplication. It will soon become evident that to multiply two matrices A and B to find AB, the number of columns in A should equal the number of rows in B.

Let A be of order m × n and B be of order n × p. The matrix AB will be of order m × p and will be obtained by multiplying each row vector of A successively with column vectors in B. Let us understand this using a concrete example:

\[A=\left[ \begin{matrix}   {{a}_{1}} & {{a}_{2}} & {{a}_{3}}  \\   {{b}_{1}} & {{b}_{2}} & {{b}_{3}}  \\   {{c}_{1}} & {{c}_{2}} & {{c}_{3}}  \\\end{matrix} \right]B=\left[ \begin{matrix}   {{\alpha }_{1}} & {{\beta }_{1}}  \\   {{\alpha }_{1}} & {{\beta }_{2}}  \\   {{\alpha }_{3}} & {{\beta }_{3}} \\\end{matrix} \right]\]

To obtain the element a11 of AB, we multiply R1 of A with C1 of B :

To obtain the element  \({{a}_{12}}\ \text{of}\ AB,\) we multiply R1 of A with C2 of B:

To obtain the element \({{a}_{21}}\ \text{of}\ AB,\) we multiply R2 of A with C1 of B:

Proceeding this way, we obtain all the elements of AB.

Let us generalize this: if A is or order m × n, and B of order n × p, then to obtain the element aij in AB, we multiply Ri in A with Cj in B:

Example - 20

If \(A=\left[ \begin{matrix}   2 & 1 & 4  \\   3 & 2 & 1  \\\end{matrix} \right]\ \text{and}\ B=\left[ \begin{matrix}   \ \ 2 & 0  \\   -1 & 1  \\   \ 4 & 3  \\\end{matrix} \right],\ \text{find}\ AB\ \text{and}\ BA\)

Solution: Since A is of order 2 × 3, and B is of order 3 × 2, AB is defined and will be of order 2 × 2. BA is also defined, and will be of order 3 × 3.

\[\begin{align} & \mathbf{AB} \\ \\ && \left[ \begin{matrix}  2 & 1 & 4 \\  3 & 2 & 1 \\\end{matrix} \right]\ \ \left[ \begin{matrix}  \ \ 2 & 0 \\  -1 & 1 \\  \ \ 4 & 3 \\\end{matrix} \right]\ \ =\ \ \left[ \begin{matrix}  19 & 13 \\  8 & 5 \\\end{matrix} \right] \\ \\ \\ & \mathbf{BA} \\ \\ && \left[ \begin{matrix}  \ \ 2 & 0 \\  -1 & 1 \\  \ \ 4 & 3 \\\end{matrix} \right]\ \ \left[ \ \begin{matrix}  2 & 1 & 4 \\  3 & 2 & 1 \\\end{matrix}\ \right]\ \ \ \ =\ \ \left[ \ \begin{matrix}  4 & 2 & \ \ 8 \\  1 & 1 & -3 \\  17 & 10 & 19 \\\end{matrix}\ \right] \\ \end{align}\]

This calculation should have made it clear to you that AB is in general not equal to BA, that is, multiplication of matrices is non-commutative operation

Another fact that is extremely important to note is that for three matrices A, B, C,

\[\boxed {AB=AC\ \text{does not imply }B=C}\]

As a simple example:

\[\begin{align}  & \left[ \ \begin{matrix}   3 & 1 & 6  \\   1 & 3 & 2  \\   2 & 5 & 4  \\\end{matrix}\  \right]\ \ \left[ \ \begin{matrix}   2  \\   0  \\   0  \\\end{matrix}\  \right]=\left[ \ \begin{matrix}   6  \\   2  \\   4  \\\end{matrix}\  \right] \\ \\ 
 &\qquad \qquad \qquad \text{and} \\  \\ 
 & \left[ \ \begin{matrix}   3 & 1 & 6  \\   1 & 3 & 2  \\   2 & 5 & 4  \\\end{matrix}\  \right]\ \ \left[ \ \begin{matrix}   0  \\   0  \\   1  \\\end{matrix}\  \right]=\left[ \ \begin{matrix}   6  \\   2  \\   4  \\\end{matrix}\  \right] \\ \end{align}\]

Let us now write matrix multiplication in more fomral notation, something that will help us in proving more advanced properties pertaining to matrix multiplication.

Let \(\begin{align}A={{[{{a}_{ij}}]}_{m\times n}},B={{[{{b}_{ij}}]}_{n\times p}}\ \text{and}\ C={{[{{c}_{ij}}]}_{m\times p}} \end{align}\) be such that AB = C. As we have seen earlier, \({{C}_{ij~}}\) is obtained by multiplying \({{R}_{i}}\) of A with \({{C}_{j~}}\) of B :

\[\begin{align}   {{C}_{ij}}&=\left( {{R}_{1}}\ \text{of}\ A \right)\times \left( {{C}_{j}}\ \text{of}\ B \right) \\\\  & =\left[ {{a}_{i1}},{{a}_{i2}}...{{a}_{in}} \right]\ \ \left[ \begin{array} \; {{b}_{ij}}\;  \\ \;  {{b}_{ij}}\;  \\   \ \vdots  \\  \;  {{b}_{nj}} \; \\ \end{array} \right] \\ \\ & ={{a}_{i1}}{{b}_{ij}}+{{a}_{i2}}{{b}_{ij}}+...+{{a}_{in}}{{b}_{nj}} \\ \\ & =\sum\limits_{k=1}^{n}{{{a}_{ik}}{{b}_{kj}}} \\ \end{align}\]

Hence, we can write matrix C as

\[C=\left[ {{c}_{ij}} \right]=\left[ \sum\limits_{k=1}^{n}{{{a}_{ik}}{{b}_{kj}}} \right]\]

Now, consider three matrices

\[A={{\left[ {{a}_{ij}} \right]}_{\ m\times n}},B={{\left[ {{b}_{ij}} \right]}_{\ n\times p}},C={{\left[ {{c}_{ij}} \right]}_{\ p\times q}}\]

Let us define the product (AB) C :

\[\left( AB \right)C=\left[ \sum\limits_{k=1}^{n}{{{a}_{ik}}{{b}_{kj}}} \right]\times C\]

Note that in \(\begin{align}AB,\sum\limits_{k=1}^{n}{{{a}_{ik}}{{b}_{kj}}}\end{align}\) is just one of the elements, precisely the element at the position \(\left( i,j \right)\). Since AB is of order \(m\times p\) and C of order \(\left( m,q \right)\).

Let us denote the positive in \(\left( AB \right)C\) using \(\left( x,y \right)\) to avoid confusion. The element \(\left( x,y \right)\) at will be obtained by multiplying the \({{x}^{\text{th}}}\) row in AB with the \({{y}^{\text{th}}}\) row in C.

The elements of  \({{x}^{\text{th}}}\) the row in AB are:

\[\begin{align}\sum\limits_{k=1}^{n}{{{a}_{xk}}{{b}_{k1}}},\ \sum\limits_{k=1}^{n}{{{a}_{xk}}{{b}_{k2}}},\sum\limits_{k=1}^{n}{{{a}_{xk}}{{b}_{k3}}},....\sum\limits_{k=1}^{n}{{{a}_{xk}}{{b}_{kp}}} \end{align}\]

The elements of \({{y}^{\text{th}}}\) the column in C are :

\[\left[ \begin{align}  & {{C}_{1y}} \\  & {{C}_{2y}} \\  & \ \ \vdots  \\  & {{C}_{py}} \\ \end{align} \right]\]

Therefore. the term at \(\left( x,y \right)\text{in}\left( AB \right)C\) is

\[\begin{align}  & {{c}_{iy}}\left( \sum\limits_{k=1}^{n}{{{a}_{xk}}{{b}_{k1}}} \right)+{{c}_{2y}}\left( \sum\limits_{k=1}^{n}{{{a}_{xk}}{{b}_{k2}}} \right)+...+{{c}_{py}}\left( \sum\limits_{k=1}^{n}{{{a}_{xk}}{{b}_{kp}}} \right) \\\\  & =\sum\limits_{k=1}^{n}{\left( {{a}_{xk}}{{b}_{k1}}{{c}_{1y}}+{{a}_{xk}}{{b}_{k2}}{{c}_{2y}}+...+{{a}_{xk}}{{b}_{kp}}{{c}_{py}} \right)} \\\\  & =\sum\limits_{l=1}^{p}{\sum\limits_{k=1}^{n}{\left( {{a}_{xk}}{{b}_{kl}}{{c}_{ly}} \right)}} \\ \end{align}\]

If \(D=\left( AB \right)C,\) then

\[D={{\left[ {{d}_{xy}} \right]}_{m\times q}}=\left[ \sum\limits_{l=1}^{p}{\sum\limits_{k=1}^{n}{\left( {{a}_{xk}}{{b}_{kl}}{{c}_{ly}} \right)}} \right]\]

Note (very) carefully that we can interchanging the order of summation above, by first summing over the index l, and then the index k.

\[\begin{align}   D&=\left[ \sum\limits_{k=1}^{n}{\sum\limits_{l=1}^{p}{\left( {{a}_{xk}}{{b}_{kl}}{{c}_{ly}} \right)}} \right] \\ \\  & =\left[ \sum\limits_{k=1}^{n}{\left( {{a}_{xk}} \right)}\left( \sum\limits_{l=1}^{p}{{{b}_{kl}}{{c}_{ly}}} \right) \right] \\ \\  & =A\left( BC \right) \\ \end{align}\]

This is an important result : we have shown that matrix multiplication is associative :

\[\boxed {D=\left( AB \right)C=A\left( BC \right)}\]

It will be a good exercise to verify this fact using a numerical example.

Example - 21

Let

\[A=\left[ \begin{matrix}   1 & 2 & 3  \\   3 & 1 & 0  \\\end{matrix} \right],\ B=\left[ \begin{matrix}   1 & \ \ 4  \\   3 & -1  \\   2 & -3  \\\end{matrix} \right],\ C=\left[ \begin{matrix}   2 & 1 & 4 & 3  \\   -1 & 0 & 3 & 1  \\\end{matrix} \right]\]

Evaluate (AB) C and A (BC) and verify that they are the same.

Solution: The orders of A, B, C are 2 × 3, 3 × 2 and 2 × 4.

\[\begin{align}AB&=\left[ \begin{matrix}   1 & 2 & 3  \\   3 & 1 & 0  \\\end{matrix} \right]\ \ \left[ \begin{matrix}   1 & \ \ 4  \\   3 & -1  \\   2 & -3  \\\end{matrix} \right]=\left[ \begin{matrix}   13 & -7  \\   6 & 11  \\\end{matrix} \right]\\\\
\left( AB \right)C&=\left[ \begin{matrix}   13 & -7  \\   6 & 11  \\\end{matrix} \right]\ \ \left[ \begin{matrix}
   2 & 1 & 4 & 3  \\   -1 & 0 & 3 & 1  \\\end{matrix} \right]=\left[ \begin{matrix}   33 & 13 & 31 & 32  \\
   1 & 6 & 57 & 29  \\\end{matrix} \right]\\\\
BC&=\left[ \begin{matrix}  1 & 4  \\   3 & -1  \\   2 & -3  \\\end{matrix} \right]\ \ \left[ \begin{matrix}
   2 & 1 & 4 & 3  \\   -1 & 0 & 3 & 1  \\\end{matrix} \right]=\left[ \begin{matrix}   -2 & 1 & 16 & 7  \\
   \begin{gathered}   7 \\   7 \\ \end{gathered} & \begin{gathered}   3 \\   2 \\ \end{gathered} & \begin{gathered}   \ \ 9 \\   -1 \\ \end{gathered} & \begin{gathered}   8 \\   3 \\ \end{gathered}  \\\end{matrix} \right]\\\\
A\left( BC \right)&=\left[ \begin{matrix}   1 & 2 & 3  \\   3 & 1 & 0  \\\end{matrix} \right]\ \ \left[ \begin{matrix}   -2 & 1 & 16 & 7  \\   7 & 3 & 9 & 8  \\   7 & 2 & -1 & 3  \\\end{matrix} \right]=\left[ \begin{matrix}   33 & 13 & 31 & 32  \\   1 & 6 & 57 & 29  \\\end{matrix} \right] \end{align} \]

As expected, (AB) C and A (BC) are the same.

Example - 22

Let \(I\)  be a sqaure matrix of order n × n :

\[I=\left[ \begin{matrix}   1 & 0 & 0 & \cdots  & \cdots   \\   0 & 1 & 0 & \cdots  & \cdots   \\   0 & 0 & 1 & \cdots  & \cdots   \\   \vdots  & \vdots  & \vdots  & \ddots  & \vdots   \\   \vdots  & \vdots  & \vdots  & {} & \vdots   \\\end{matrix} \right] \qquad \left\{ \begin{align}  & \text{All elements are zero except the } \\  & \text{diagonal elements which are 1} \\ \end{align} \right\}\]

Let A be a sqaure matrix of order \(n\times n.\) Find

\[\left( \text{a} \right) \;AI \qquad \qquad \left( \text{b}\right) \;IA\]

Solution: (a) Consider an elemnet  \({{a}_{ij}}\) in A. Let us consider what happens to this element when A is multiplied by \(I\) . To evaluate the element at  \(\left( i,j \right)\) in \(AI\) , we multiply \({{R}_{i}}\) in A by Cj in \(I\)  : 

Therefore, \({{a}_{ij}}\) is unaltered! This happens for each element in A, and thus \(AI=A\)

(b) Similarly, \(IA=A\)

Since \(AI=IA=A\) , the matrix \(I\) is termed the identity matrix (or order n × n) . Multiplying \(I\) with any matrix (of the same order) leaves that matrix unaltered.

Lets say A is of order m × n. We can still define \(AI\), but not \(IA\), when \(I\)  is of order n × n. Note that even now.\(AI\) is the same as A.

The determinant of  \(I\)  is always 1 verfify. Verify this.