Let, $A$ be an $n \times m$ matrix i.e., $A=\left[a_{i j}\right]$ & let $\alpha \in \mathbb{R}$ or $\mathbb{C}$. We define the matrix $A$ as follows:

$(\alpha . A)_ij=\alpha . a_ij$.

The $\alpha \cdot A=\left[\alpha a_{i j}\right]$

i) $(\alpha+\beta) \cdot A=\alpha \cdot A+\beta \cdot A$

Let, $A=\left[a_{i j}\right] then,$

$(\alpha+\beta) \cdot A =\left[(\alpha+\beta) \cdot a_{i j}\right]$

$=\left[\alpha \cdot a_{i j}+\beta \cdot a_{i j}\right]$

$=\left[\alpha \cdot a_{i j}\right]+\left[\beta \cdot a_{i j}\right]$

$=\alpha \cdot\left[a_{i j}\right]+\beta \cdot\left[a_{i j}\right]$

$=\alpha \cdot A+\beta \cdot A$

For any scalar $\alpha$ and for any two matrices $A$ and $B$ of same order, $\alpha \cdot(A+B)=\alpha \cdot A+\alpha \cdot B$.

Proof:

Let, $A=\left[a_{i j}\right]$ and $B=\left[b_{i j}\right]$.

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

$=\alpha \cdot\left(\left[a_{i j}+b_{i j}\right]\right)$

$=\left[\alpha \cdot\left(a_{i j}+b_{i j}\right)\right]$

$=\left[\alpha \cdot a_{i j}+\alpha \cdot b_{i j}\right]$

$=\left[\alpha \cdot a_{i j}\right]+\left[\alpha \cdot b_{i j}\right]$

$=\alpha \cdot\left[a_{i j}\right]+\alpha \cdot\left[b_{i j}\right]$

$=\alpha \cdot A+\alpha \cdot B .$

$(\alpha \cdot \beta) \cdot A =\alpha \cdot(\beta \cdot A)=\beta \cdot(\alpha \cdot A)$

$(\alpha \cdot \beta) \cdot A =(\alpha \cdot \beta) \cdot\left[a_{i j}\right]$

$=\left[(\alpha \beta) \cdot a_{i j}\right]$

$=\left[\alpha \cdot\left(\beta \cdot a_{i j}\right)\right]$

$=\alpha\left[\beta \cdot a_{i j}\right]$

$=\alpha \cdot\left(\left[\beta \cdot a_{i j}\right]\right)$

$=\alpha \cdot\left(\beta \cdot\left[a_{i j}\right]\right)$

$=\alpha \cdot(\beta \cdot A) .$

If $A$ is any matrix, then the transpose of $A$, denoted $A^t$, is obtained as follows:

Let, $A=\left[a_{i j}\right]$ and $A^t=\left[b_{i j}\right]$.

Then, $b_{i j}=a_{j i}$.

If $A$ is a matrix of order $n \times m$, then $A^t$ is a matrix of order $m \times n$.

1)$\quad$ A = ${\displaystyle \left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{3}}\\ 3& \frac{1}{\sqrt{3}}& \frac{1}{\sqrt{5}}\\ 5& \frac{1}{\sqrt{5}}& \frac{1}{\sqrt{7}}\end{array}\right]}$

${\displaystyle A^t=\left[\begin{array}{ccc}\frac{1}{2}& 3& 5\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{3}}& \frac{1}{\sqrt{5}}\\ \frac{1}{\sqrt{3}}& \frac{1}{\sqrt{5}}& \frac{1}{\sqrt{7}}\end{array}\right]}$

2. A= ${\displaystyle \left[\begin{array}{ccc}i& 2i& 1+2i\\ 3i& 2i& 2+3i\\ 4+5i& 3& 4\end{array}\right]}$

${\displaystyle A^t=\left[\begin{array}{ccc}i& 3i& 4+5i\\ 2i& 2i& 3\\ 1+2i& 2+3i& 4\end{array}\right]}$

1. $(A+B)^t=A^t+B^t.$

Proof:

Let, $A=\left[a_{i j}\right]$ and $B=\left[b_{i j}\right]$

$(A+B)^t=\left(\left[a_{i j}\right]+\left[b_{i j}\right]\right)^t=\left(\left[a_{i j}+b_{i j}\right]\right)^t$

$=\left[a_{j i}+b_{j i}\right]$

$=\left[a_{j i}\right]+\left[b_{j i}\right]$

$=A^t+B^t .$

$\therefore(A+B)^t =A^t+B^t .$

2. For any scalar $\alpha$, any matrix $A,(\alpha A)^t=\alpha \cdot A^t$.

Proof:

Let, $A=[a_{ij}]$

$(\alpha.A)^t = (\alpha .[a_{ij}])^t=([\alpha . a_{ij}])^t$

$=[\alpha . a_{ij}]=\alpha.[a_{ij}]$

$=\alpha. A^t$

Definition : $A$ is called a symmetric matrix if $A=A^t$ and a matrix $A$ is called a skew-symmetric matrix if $A=-A^t$.

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

$A^t=\left[\begin{array}{lll}1 & 2 & 3 \\2 & 3 & 4 \\3 & 4 & 5\end{array}\right]$

$\quad \text { Note that } A=A^t.$

$\quad \therefore A \text { is symmetric. }$

Let, $A$ be an upper triangular matrix $x$. Then $A$ cannot be symmetric unless $A$ is a diagonal matrix. In particular, every diagonal matrix is skew symmetric.

If $A$ is any, square matrix, then $A+A^t$ is a symmetric matrix.

Proof:

Let, $A=\left[a_{ij}\right]$. Then, $A^t=\left[a_{j i}\right]$.

$A+A^t =\left[a_{i j}\right]+\left[a_{j i }\right] =\left[a_{i j}+a_{j i}\right] \longrightarrow(1)$

$\left(A+A^t\right)^t =\left(\left[a_{i j}+a_{j i}\right]\right)^t =\left[a_{j i}+a_{i j}\right]$

$=\left[a_{i j}+a_{j i}\right] \quad =A+A^t \quad \text { (From (1) }$

Thus, $A+A^t$ is symmetric.

If $A$ is any square matrix, then $A-A^t$ is a skew-symmetric matrix.

Proof:

Let, $A=\left[a_{i j}\right]$. Then $A^t=\left[a_{j i}\right]$.

$A - A^t =\left[a_{i j}\right]-\left[a_{j i}\right]$

$=\left[a_{i j}\right]+\left[-a_{j i}\right] =\left[a_{i j}-a_{j i}\right] \rightarrow(1) .$

$\left(A-A^t\right)^t =\left(\left[a_{i j}-a_{j i}\right]\right)^t$

$=\left[-a_{j i}+a_{i j}\right]$

$=\left[-\left(a_{i j}-a_{j i}\right)\right]$

$=-\left[a_{i j}-a_{j i}\right]$

$=-\left(A-A^t\right) \quad(\text { From (1)). }$

Thus, $A-A^t$ is a skew-symmetric matrix.

Given any square matrix $A$, it can be written as a sum of a symmetric matrix and a skew-symmetic matrix.

Proof. Let $A$ be any square matrix.

Claim: $A=B+C$ when $B$ is a symmetric matrix and $C$ is a skew-symmetric matrix

Let, $B=\left(\frac{A+A^t}{2}\right)$

$\text { and } C=\left(\frac{A-A^t}{2}\right)^2$

Then, $B$ is a symmetric matrix and $C$ is a skew-symmetric matrix.

$B+C =\frac{\left(A+A^t\right)}{2}+\frac{\left(A-A^t\right)}{2}$

$=\left(\frac{A}{2}+\frac{A^t}{2}\right)+\left(\frac{A}{2}-\frac{A^t}{2}\right)$

$=\frac{A}{2}+\frac{A^t}{2}+\frac{A}{2}-\frac{A^t}{2}$

$=\frac{A}{2}+\frac{A}{2}$

$=\left(\frac{1}{2}+\frac{1}{2}\right) \cdot A$

$=1 \cdot A=A$

Thus, $B+C=A$ or $A=B+C$.

Let, $A$ be a matrix of order $m \times n$ and $B$ be a matrix of order $n \times r$. Then the product

of $A$ & $B$, denoted $A B$, is obtained as follows:

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

$B=\left[b_{i j}\right]_{\substack{ 1 \leqslant i \leqslant n },{\ 1 \leqslant j \leqslant r }}$

Let, $A B=c=\left[c_{i j}\right]$. Then,

$c_{i j}=\sum_{k=1}^n a_{i k} b_{k j} \text {. }$

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

$B=\left[\begin{array}{l}5 \\ 6\end{array}\right]$

$m=2$, $n=2$, $r=1$

order of A = 2 x 2

order of B = 2 x 1

$\therefore\left[\begin{array}{ll}1 & 2 \\3 & 4\end{array}\right] \left[\begin{array}{l}5 \\ 6\end{array}\right]$

$=\left[\begin{array}{ll}\sum_1^2 a_{1k} b_{k1} \\ \sum_1^2 a_{2k} b_{k-1}\end{array}\right]$

$=\left[\begin{array}{ll}a_{11} b_{11} + a_{12} b_{21} \\a_{21} b_{11} + a_{22} b_{21}\end{array}\right]$

$=\left[\begin{array}{c}5+12 \\15+24\end{array}\right]$

$=\left[\begin{array}{l}17 \\39\end{array}\right]$

$A =\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right] _ {\tiny 2 × 2} \\$

$B=\left[\begin{array}{ccc}1 & 3 & 5 \\ 2 & 4 & 6\end{array}\right]_{\tiny 2 × 3}$

$A B =\left[\begin{array}{lll}1+4 & 3+8 & 5+12 \\ 3+8 & 9+16 & 15+24\end{array}\right]$

$=\left[\begin{array}{lll}5 & 11 & 17 \\ 11 & 25 & 39\end{array}\right]$

$A=\left[\begin{array}{lll}1 & 2 & 3 \\4 & 5 & 6 \\7 & 8 & 9\end{array}\right]_{3×3}$

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

$AB =\left[\begin{array}{lll}1+6+15 & 2+8+18 \\4+15+30 & 8+20+36 \\7+24+45 & 14+32+54\end{array}\right]$$=\left[\begin{array}{lll}22 & 28 \\49 & 64 \\76 & 100\end{array}\right]$

For any two square matrices $A$ & $B$ of same order,

$(A B)^t=B^t A^t \text {. }$

Proof:

Let,$A=\left[a_{i j}\right]$ and $B=\left[b_{i j}\right], 1 \leqslant i, j \leqslant n$.

$\text { Let } C=A B=\left[c_{i j}\right]$

where, $c_{i j}=\sum_{k=1}^n a_{i k} b_{k j}$.

$(A B)^t=c^t=\left[c_{i j}\right]^t=\left[c_{j i}\right]=\left[\sum_{k=1}^n a_{j k} b_{k i}\right]$ .

$B^t A^t =\left[b_{i j}\right]^t\left[a_{i j}\right]^t$

$=\left[b_{j i}\right]\left[a_{j i}\right]$

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

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

$=(A B)^t .$

Thus, $(A B)^t =B^t A^t .$

$A =\left[\begin{array}{lll}1 & 2 & 3 \\4 & 5 & 6\end{array}\right]_{2×3}$

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

$A B=\left[\begin{array}{llll}1+10+3 & 2+12+6 & 3+14+12 & 4+16+15 \\4+25+6 & 8+30+12 & 12+35+24 & 16+40+30\end{array}\right]$

$=\left[\begin{array}{llll}14 & 20 & 29 & 35 \\35 & 50 & 11 & 86\end{array}\right]_ {2\times4 }$

$(A B)^t=\left[\begin{array}{ll}14 & 35 \\20 & 50 \\29 & 71 \\38 & 86\end{array}\right]$

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

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

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

$=\left[\begin{array}{lll}1+10+3 & 4+25+6 \\ 2+12+6 & 8+30+12 \\ 3+14+12 & 12+35+24 \\ 4+16+15 & 16+40+30\end{array}\right]$

$=\left[\begin{array}{ll}14 & 35 \\ 20 & 50 \\ 24 & 71 \\ 35 & 86\end{array}\right]$

$(A B)^t=B^t A^t$