Matrix multiplication is associative, i.e. for any 3 matrices $A, B$ & $C$ such that $A$ & $B$are compatible for multiplying and B & C are compatible for multiplying, then

$A \cdot(B \cdot C)= (A \cdot B) \cdot C$

Let, $A$ be a $n \times m$ matrix

$B$ be a $m \times r$ matrix

& $C$ be a $r \times s$ matrix

(This is possible because of the given hypothesis)

Let, $A=\left[a_{i j}\right], {1\leqslant i \leqslant n}$, $~$ ${1\leqslant j \leqslant m}$

$B=\left[b_{i j}\right], {1\leqslant i \leqslant m}$, $~$ ${1\leqslant j \leqslant r}$

$C=\left[c_{i j}\right], {1\leqslant i \leqslant r}$, $~$ ${1\leqslant j \leqslant s}$

$A \cdot(B \cdot C) =A \cdot\left(\left[b_{i j}\right] \cdot\left[c_{i j}\right]\right)$

=$A \cdot\left(\left[\sum_{k=1}^r b_{i k} c_{k j}\right]\right)$

$=\left[a_{i j}\right]\left(\left[\sum_{k=1}^r b_{i k} c_{k j}\right]\right)$

$=\left[\sum_{t=1}^m a_{i t} d_{t j}\right]$

$\text { where } d_{t j} \text { is the }(t, j)^{th}$ entry in the matrix $B \cdot C$

$=\left[\sum_{t=1}^m a_{i t} \sum_{k=1}^r b_{t k} c_{k j}\right]$

$=\left[\sum_{t=1}^m \sum_{k=1}^r a_{i t} b_{t k} c_{k j}\right]$

$(A \cdot B) \cdot C=\left(\left[a_{i j}\right]\left[b_{i j}\right]\right) \cdot C$

$=\left(\left[\sum_{k=1}^m a_{i k} b_{k j}\right]\right)$

$=\left(\left[\sum_{k=1}^m a_{i k} b_{k j}\right]\right) \cdot\left[c_{i j}\right]$

$=\left[\left(\sum_{t=1}^r \sum_{k=1}^m a_{i k} b_{k t}\right) \cdot c_{t j}\right]$

$=\left[\sum_{t=1}^r \sum_{k=1}^m a_{i k} b_{k t} c_{t j}\right]$

$=\left[\sum_{k=1}^r \sum_{t=1}^m a_{i t} b_{t k} c_{k j}\right]=A \cdot(B \cdot C)$

Example-1

Let, $A=\left[\begin{array}{ccc}1 & -2 & 3 \\ -4 & 2 & 5\end{array}\right]$ &

$B=\left[\begin{array}{ll}2 &3 \\ 4 & 5 \\ 2 & 1\end{array}\right]$

$A B=\left[\begin{array}{ccc}1 & -2 & 3 \\ -4 & 2 & 5\end{array}\right]$ $\left[\begin{array}{ll}2 & 3 \\ 4 & 5 \\ 2 &1\end{array}\right]$

$=\left[\begin{array}{cc}2-8+6 & 3-10+3 \\ -8+8+10 & -12+10+5\end{array}\right]=\left[\begin{array}{cc}0 & -4 \\ 10 & 3\end{array}\right]$

$\begin{aligned} & B A=\left[\begin{array}{ll} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & -2 & 2 \\ 4 & 2 & 5 \end{array}\right] \\ & =\left[\begin{array}{llll} 2 - 12 & -4+6 & 6+15 \\ 4 - 20 & -8+10 & 12+25 \\ 2 - 4 & -4+2 & 6+5 \end{array}\right] \\ & =\left[\begin{array}{ccc} 10 & 2 & 21 \\ 16 & 2 & 37 \\ 2 & -2 & 11 \end{array}\right] \end{aligned}$

$A ~ B$ is a matrix of order $2 \times 2, ~ B ~ A$ is a matrix of order $3 \times 2$

$\Rightarrow A B \neq B A \text {. }$

Let, $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$ & $B=\left[\begin{array}{ll}5 & 6 \\ 7 & 8\end{array}\right]$

$\begin{aligned} A B & =\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}5 & 6 \\ 7 & 8\end{array}\right] \\ & =\left[\begin{array}{ll}5+14 & 6+16 \\ 15+28 & 18+32\end{array}\right] \\ & =\left[\begin{array}{ll}19 & 22 \\ 43 & 50\end{array}\right]\end{aligned}$

$B A =\left[\begin{array}{ll}5 & 6 \\7 & 8\end{array}\right]\left[\begin{array}{ll}1 & 2 \\3 & 4\end{array}\right]$

$=\left[\begin{array}{ll}5+18 & 10+24 \\7+24 & 14+32\end{array}\right]$

$=\left[\begin{array}{ll}23 & 34 \\31 & 46\end{array}\right]$

Thus, $A B=\left[\begin{array}{ll}19 & 32 \\ 43 & 50\end{array}\right] \neq\left[\begin{array}{ll}23 & 34 \\ 31 & 46\end{array}\right]=B A$

$\therefore A B \neq B A$

If $\alpha$ & $\beta$ are any two scalars such that $\alpha \cdot \beta=0$, then either $\alpha=0$ or $\beta=0$.

$A=\left[\begin{array}{cc} 0 & -1 \\ 0 & 2 \end{array}\right]$ and $B=\left[\begin{array}{ll} 3 & 5 \\ 0 & 0 \end{array}\right]$

$A B=\left[\begin{array}{cc} 0 & -1 \\ 0 & 2 \end{array}\right]$

$\left[\begin{array}{ll} 3 & 5 \\ 0 & 0 \end{array}\right]$$=\left[\begin{array}{cc} 0+0 & 0+0 \\ 0+0 & 0+0 \end{array}\right]$

$=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \text { the zero matrix. }$

Definition :- A matrix is called row reduced echelon if the following properties hold:

i) Every zero row is below every non-zero row.

ii) The Leading coefficient (first non-zero coefficient) of every row is 1 .

iii) A column which contains the leading coefficient of A, now has all the other coefficients equal to zero.

1. $\left(\begin{array}{lll}1 & 0 & 2 \\ 0 & 0 & 0 \\ 0 & 1 & 0\end{array}\right)$
   
   

2. $\left(\begin{array}{lll}\ 1 & 0 & 1 \\ 0 & \ 2 & 0 \\ 0 & 0 & 0\end{array}\right)$

The first non-zero coefficient in the second row is 2 & hence not a row reduced echelon matrix.

3. $\left(\begin{array}{lll}\ 1 & 1 & 2 \\ 0 & \ 1 & 1 \\ 0 & 0 & 0\end{array}\right)$

Hence, this is not a row reduced echelon matrix.

4. $\left(\begin{array}{lll}\ 0 & 1 & 2 \\ 0 & 0 & 3 \\ 0 & 0 & 0\end{array}\right)$ $\quad k_1=2, \quad k_2=1$, $k_2

$\therefore$ This is not row reduced.

5) $\left(\begin{array}{lll}\ 1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 0\end{array}\right)$

The zero row, the third row is below all the non-zero rows.

The leading coefficients in both $1^{\text {st }}$ & 2nd rows are just 1.

All the other elements in a column containing a leading coefficient are zero

$k_1=1 \quad k_2=2 \quad k_1<k_2$

Given a matrix $A$, is there any procedure to convert it into a row reduced echelon matrix?

Row elementary operations.

1) Multiplying the $i^{\text {th }}$ row by a non-zero scalar

say $\lambda . \quad R_i \rightarrow \lambda R_i$.

Example: $\left(\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right) R_1 \rightarrow 2 R_1\left(\begin{array}{lll}2 & 4 & 6 \\ 4 & 5 & 6\end{array}\right)$.

2. Interchanging $i^{\text {th }}$ row & $j^{\text {th }}$ now.

$R_i \longleftrightarrow R_j$

Eg. $\left(\begin{array}{lll}0 & 1 & 2 \\ 1 & 0 & 3 \\ 0 & 0 & 0\end{array}\right) R_1 \longleftrightarrow R_2\left(\begin{array}{lll}1 & 0 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0\end{array}\right)$.

3. Replace the $i^{\text {th }}$ row by the sum of $i^{\text {th }}$ row & $\mu$-multiple of $j^{\text {th }}$ now.

$R_i \longrightarrow R_i+\mu R_j \text {. }$

Example: $\left(\begin{array}{llll}1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 3\end{array}\right) R_1 \rightarrow R_1+2 R_2\left(\begin{array}{llll}1 & 2 & 2 & 6 \\ 0 & 0 & 1 & 3\end{array}\right)$

Step1: Apply interchange of rows to push down the zero rows to the end of the matrix.

Step2: Find the first non-zero column (from left) (let us suppose that it is $k_1$)

Step3: Again apply interchange of rows to push up a row where leading non-zero coefficient occurs in first non- zero colum to the first row.

Divide the first row by the leading non-zero

coefficient, so that the leading non-zero coefficient becomes 1.

Step 4. Next apply $R_i \longmapsto R_i+\mu R_{1}$ for suitable values of $i$ & $\mu$ so that the first non-zero column has non-zero coefficients only in the first row.

Step 5: Repeat steps 2 to 4 for the submatrix obtained by deleting the $1^{\text {st }}$ row & $1^{\text {nd }}$ column until all the non-zero rows are exhausted.

Let, $\left(\begin{array}{lll}1 & 1 & 2 \\ 1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right)$

$R_2 \rightarrow R_2+(-1) R_1\\ R_3 \rightarrow R_3+(-1) R_1 \left(\begin{array}{lll}1 & 1 & 2 \\ 0 & 1 & -1 \\ 0 & 1 & 1\end{array}\right)$

$R_1 \longrightarrow R_1+(-1) R_2\\ R_3 \longrightarrow R_3+(-1) R_2 \left(\begin{array}{ccc}1 & 0 & 3 \\ 0 & 1 & -1 \\ 0 & 0 & 2 \end{array}\right)$

$R_1 \rightarrow R_1+(-3) R_3$

$R_2 \rightarrow R_2+R_3$

$\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$

Thus, the row reduced echelon matrix obtained after applying the procedure to the matrix $\left(\begin{array}{lll}1 & 1 & 2 \\ 1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right)$ is just the identity matrix.