Matrices - Class 12 Commerce Applied Mathematics - Extra Questions

Given that A is of order $$1 \times 3$$ B is of order $$4 \times 3$$ .

Now, the number of columns in A and the number of rows in B are not equal.

So, the matrix product AB is not defined.

$${ A }_{ ij }=\dfrac { 1 }{ 2 } |-3i+j|$$

Find 2A and 3B,then subtract them.

C = $$\dfrac{1}{2}$$ (B - A) = $$\begin{bmatrix}1 & -2 & 5/2\\ -1/2 & 1 & 3/2\\ 1/2 & 1/2 & 1 \end{bmatrix}$$

Here, $$A=\begin{pmatrix} 4 & -2\\5&-9\end{pmatrix}$$ and   $$B=\begin{pmatrix} 8&2\\-1&-3\end{pmatrix}$$

$$\begin{array}{l} X+\left[ { \begin{array} { *{ 20 }{ c } }4 & 6 \\ { -3 } & 7 \end{array} } \right] =\left[ { \begin{array} { *{ 20 }{ c } }3 & { -6 } \\ 5 & { -8 } \end{array} } \right]  \\ X=\left[ { \begin{array} { *{ 20 }{ c } }3 & { -6 } \\ 5 & { -8 } \end{array} } \right] -\left[ { \begin{array} { *{ 20 }{ c } }4 & 6 \\ { -3 } & 7 \end{array} } \right]  \\ X=\left[ { \begin{array} { *{ 20 }{ c } }{ 3-4 } & { -6-6 } \\ { 5+3 } & { -8-7 } \end{array} } \right]  \\ hence, \\ X=\left[ { \begin{array} { *{ 20 }{ c } }{ -1 } & { -12 } \\ 8 & { -15 } \end{array} } \right]  \end{array}$$

Since this matrix has all its elements zero except for those in the diagonal from top left to bottom right, this type of matrix is called Diagonal matrix. 

$$A=\begin{bmatrix} 5 & 7 \\ 9 & 4 \end{bmatrix}B=\begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}$$

Let $$A = [a_{ij}]$$ be a $$2\times 2$$ matrix such that $$a_{ij} = \dfrac {|-3i + j|}{2}$$ . Then,
$$a_{11} = \dfrac {|-3 + 1|}{2} = 1, a_{12} = \dfrac {|-3 + 2|}{2} = \dfrac {1}{2}, a_{21} = \dfrac {|-6 + 1|}{2} = \dfrac {5}{2}, a_{22} = \dfrac {|-6 + 2|}{2} = 2$$
$$\therefore = \begin{bmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{bmatrix} = \begin{bmatrix}1 & 1/2\\ 5/2 & 2\end{bmatrix}$$

$$a_{ij}=i+j$$

$$A = \begin{bmatrix}0 & a & -3\\ 2 & 0 & -1\\ b & 1 & 0\end{bmatrix}$$

Given for a matrix  $$A = [a_{ij}]$$ is a $$2\times 2$$ matrix such that $$a_{ij} = i + 2j$$ .

Given for  $$A$$ being skew-symmetric and $$n\in N$$ , such that $$(A^{n})^{T} = \lambda A^{n}$$

Let $$A=[a_{ij}]$$ be a skew-symmetric matrix. Then,

$$Given\quad \\ B=\left[ \begin{matrix} 1 & 3 \\ -2 & 5 \end{matrix} \right] \quad and\quad C=\left[ \begin{matrix} -2 & 5 \\ 3 & 4 \end{matrix} \right]$$

$$A-C=\left[ \begin{matrix} 2 \\ 5 \end{matrix} \right] +\left[ \begin{matrix} 6 \\ -2 \end{matrix} \right] =\left[ \begin{matrix} 2-6 \\ 5+2 \end{matrix} \right] =\left[ \begin{matrix} -4 \\ 7 \end{matrix} \right]$$

(i) It's a matrix of order 2
(ii) It's a matrix of order 1 x 3
(iii) It's a matrix of order 3 x 1
(iv) It's a matrix of order 3 x 2
(v) It's a matrix of order  2 x 3
(vi) It's a matrix of order 2 x 3

$$B-A = [4\ \ -5]- [8\ \ -3] = [4-8\ \ -5-(-3) ] = [ -4\ \ -2]$$

(i) Given $$a_{i j}=2 i-j$$  
Therefore matrix of order 
$$2 \times 2$$ is $$\left[\begin{array}{ll}1 & 0 \\ 3 & 2\end{array}\right]$$

(ii) Given $$a_{i j}=i . j$$  
Therefore matrix of order 
$$2 \times 2$$ is $$\left[\begin{array}{ll}1 & 2 \\ 2 & 4\end{array}\right]$$

$$3A = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix}\quad\\\therefore 3A-C= \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix}-\begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} =\begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix}_{ 2\times 2 }$$

$$6 \left[ \begin{matrix} 3 \\ -2 \end{matrix} \right] -2 \left[ \begin{matrix} -8 \\ 1 \end{matrix} \right]  = \left[ \begin{matrix} 18 \\ -12 \end{matrix} \right] -\left[ \begin{matrix} -16 \\ 2 \end{matrix} \right] =\left[ \begin{matrix} 34 \\ -14 \end{matrix} \right]$$

$$A -B=\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}-\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}  = \begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix}_{2 \times 2}$$

$$A-B= \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}-\begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}=\begin{bmatrix} 4-1 & 1-0 \\ 2+2 & 3-1 \end{bmatrix}=\begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix}$$

$$\textbf{Step 1: Write A as sum of symmetric and skew-symmetric parts}$$

We have
$$\quad           A = \begin{pmatrix}4 & 2 & -3 \\ 1 & 3 & -6 \\ -5 & 0 & -7\end{pmatrix}$$
$$\therefore\quad A' = \begin{pmatrix}4 & 1 & -5 \\ 2 & 3 & 0 \\ -3 & -6 & -7\end{pmatrix}$$
Then
$$\quad           A+A' = \begin{pmatrix}4& 2 & -3 \\ 1 & 3 & -6 \\ -5 & 0 & 7\end{pmatrix} + \begin{pmatrix}4& 1 & -5 \\ 2 & 3 & 0 \\ -3 & -6 & -7\end{pmatrix} = \begin{pmatrix}8& 3 & -8 \\ 3 & 6 & -6 \\ -8 & -6 & 0\end{pmatrix} \quad              ....(1)$$

$$\quad           A-A' = \begin{pmatrix}4& 2 & -3 \\ 1 & 3 & -6 \\ -5 & 0 & 7\end{pmatrix} - \begin{pmatrix}4& 1 & -5 \\ 2 & 3 & 0 \\ -3 & -6 & -7\end{pmatrix} = \begin{pmatrix}0 & 1 & 2 \\ -1 & 0 & -6 \\ -2 & 6 & 14\end{pmatrix}\quad               ....(2)$$

Adding $$(1)$$ and $$(2)$$ , we get

$$\quad 2A = \begin{pmatrix}8& 3 & -8 \\ 3 & 6 & -6 \\ -8 & -6 & 0\end{pmatrix}+\begin{pmatrix}0 & 1 & 2 \\ -1 & 0 & -6 \\ -2 & 6 & 14\end{pmatrix}$$

$$\therefore\quad A = \begin{pmatrix}4 & 3/2 & -4 \\ 3/2 & 3 & -3 \\ -4 & -3 & 0\end{pmatrix} + \begin{pmatrix}0 & 1/2 & 1 \\ -1/2 & 0 & -3 \\ -1 & 3 & 7\end{pmatrix}$$

Let $$A$$ be the given matrix.
Then $$\quad A = \displaystyle\frac{1}{2}(A+A')+\displaystyle\frac{1}{2}(A-A')$$
Now $$\quad (A+A')' = A' +(A')' = A'+A$$
$$\therefore A+A'$$ is symmetric.
or $$\quad \displaystyle\frac{1}{2}(A+A')$$ is symmetric.
Also, $$\quad (A-A')' = A' - (A')' = A'-A = -(A-A')$$
$$\therefore\quad A-A'$$ is anti-symmetric.
or $$\quad \displaystyle\frac{1}{2}(A-A')$$ is anti-symmetric.
$$\therefore \quad A = \displaystyle\frac{1}{2}(A+A') + \displaystyle\frac{1}{2}(A-A') =$$ symmetric matrix $$+$$ anti-symmetric matrix

In general  a $$3 \times 4$$ matrix is given by,
$$\displaystyle A=\left[ \begin{matrix} { a }_{ 11 } \\ { a }_{ 21 } \\ { a }_{ 31 } \end{matrix}\begin{matrix} { a }_{ 12 } \\ { a }_{ 22 } \\ { a }_{ 32 } \end{matrix}\begin{matrix} { a }_{ 13 } \\ { a }_{ 23 } \\ { a }_{ 33 } \end{matrix}\begin{matrix} { a }_{ 14 } \\ { a }_{ 24 } \\ { a }_{ 34 } \end{matrix} \right]$$

(i) Let  $$\displaystyle A=\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$$ , then $$\quad { A }^{ \prime  }=\begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}$$

$$\begin{bmatrix} a_{11}&a_{12} \\a_{21}&a_{22}\end{bmatrix}$$

Let $$A=\left( \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \end{matrix} \right)$$
$${ a }_{ ij }=\left| 2i-3j \right|$$
$${ a }_{ 11 }=\left| 2\left( 1 \right) -\left( 3 \right) \left( 1 \right)  \right| =\left| 2-3 \right| =\left| -1 \right| =1$$
$${ a }_{ 12 }=\left| 2\left( 1 \right) -\left( 3 \right) \left( 2 \right)  \right| =\left| 2-6 \right| =\left| -4 \right| =4$$
$${ a }_{ 13 }=\left| 2\left( 1 \right) -3\left( 3 \right)  \right| =\left| 2-9 \right| =\left| -7 \right| =7$$
$${ a }_{ 21 }=\left| 2\left( 2 \right) -3\left( 1 \right)  \right| =\left| 4-3 \right| =1$$
$${ a }_{ 22 }=\left| 2\left( 2 \right) -3\left( 2 \right)  \right| =\left| 4-6 \right| =\left| -2 \right| =2$$
$${ a }_{ 23 }=\left| 2\left( 2 \right) -3\left( 3 \right)  \right| =\left| 4-9 \right| =\left| -5 \right| =5$$
Therefore, $$A=\left( \begin{matrix} 1 & 4 & 7 \\ 1 & 2 & 5 \end{matrix} \right)$$

$${\textbf{Step -1: Writing generalized concept.}}$$

                  $${\text{We know that, if a matrix has m rows and n columns, then it has mn elements}}{\text{.}}$$

                   $${\text{And the order of a matrix is given by, (Number of rows)}} \times ({\text{Number of columns)}}$$

                   $${\text{So if a matrix has mn elements, its order can be any ordered pair with product equal to mn}}$$

$${\textbf{Step -2: Writing possible orders.}}$$

                   $${\text{Given, the matrix has 8 elements}}{\text{.}}$$

                   $$\therefore {\text{ Its order will be any ordered pair of natural numbers with product equal to 8}}{\text{.}}$$

                   $$\Rightarrow {\text{ Order  = 1 $\times$ 8, 2 $\times$ 4, 4 $\times$ 2 and 8 $\times$ 1}}$$

$$\mathbf{{\text{Thus, the possible orders a matrix with 8 elements can have are  \{ 1}} \times {\text{8, 2}} \times {\text{4, 4}} \times {\text{2, 8}} \times {\text{1\} }}.}$$

$$A=\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix},\quad { A }^{ T }=\begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}\\ P=\frac { 1 }{ 2 } (A+{ A }^{ T })=\frac { 1 }{ 2 } \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix}=\begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}\\ { P }^{ T }=\begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}\\ $$

Any square matrix $$A$$ can be expressed as,

$$A = \begin{bmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{bmatrix} \quad a_{11} = \dfrac {1}{1} = 1\quad a_{21} = \dfrac {2}{1} = 2$$

Here $$A'=-A$$ and $$B'=-B$$ given as both $$A$$ and $$B$$ are skew symmetric. 

Here $$D = \begin{vmatrix} 5 & -6 & 4 \\ 7 & 4 & -3 \\ 2 & 1 & 6\end{vmatrix}$$
Apply $$C_1 - 2C_1, C_3 - 6C_2$$
$$\therefore D = \begin{vmatrix} 17 & -6 & 40\\ -1 & 4 & -27\\ 0 & 1 & 0\end{vmatrix} = -\begin{vmatrix} 17 & 40 \\ -1 & -27 \end{vmatrix} = -[-459 + 40] = 419 \neq 0$$
Since $$D \neq 0$$ , therefore the equations are consistent.
$$D_1 = \begin{vmatrix} 15 & -6 & 4 \\ 19 & 4 & -3\\ 46 & 1 & 6\end{vmatrix} = 15(27) - 19(-40) + 46(2) = 1257$$
Similarly, $$D_2 = 1676, D_3 = 2514$$
$$\therefore$$ $$\dfrac{x}{1257} = \dfrac{y}{1676} = \dfrac{z}{2514} = \dfrac{1}{419} = \therefore x = 3, y = 4, z = 6$$

Let us consider the matrices as A , B , C  which are the $$1 \, \times \, 3 , 3 \, \times \, 3 , 3 \, \times \, 1$$   so that the matrix ABC will be $$1 \, \times \, 1$$ Evidently they are comfortable for multiplication 
ABC = A(BC)  OR (AB ) C 
(BC) =  $$\begin{bmatrix}ax + hy + gz \\ hx + by + fz  \\ gx + fy + cz \end{bmatrix}_{3 \, \times \, 1}$$
ABC' = A (BC)  n= $$[ x , y , z]_{1 \, \times \, 3 , }$$ ] $$\begin{bmatrix}ax + hy + gz \\ hx + by + fz  \\ gx + fy + cz \end{bmatrix}_{3 \, \times \, 1}$$
$$= [ x(ax \, + \, hy \, + \, gz) \, + \, y (hx  \, + \, by \, fz ) \, + \, z (gx \, + \, fy \, + \, cz )]_{1 \, \times \, 1, }$$
$$[ax^2  \, + \, by^2 \, cz^2  \, + \, 2fyz \, + \, 2gzx \, + \, 2hxy ]_{1 \, \times \, 1}$$
similarly we can find ABC as (AB)C are AB will be $$1 \, \times \, 3 ,$$ and C being $$3 \, \times \, 1$$ (AB) C will br $$1 \, \times \, 1$$ matrix which will be same as above. 

$$A_\alpha\,A_\beta \, = \,\begin{bmatrix}cos \, \alpha& sin \,\alpha\\ -sin \, \alpha  & cos \, \alpha \end{bmatrix} \begin{bmatrix}cos \, \beta& sin \,\beta\\ -sin \, \beta  & cos \, \beta \end{bmatrix}$$
$$= \,\begin{bmatrix}cos \, (\alpha + \beta)& sin \,(\alpha + \beta)\\ -sin \, (\alpha + \beta)  & cos \, (\alpha + \beta)\end{bmatrix} \, = \, A_{\alpha + \beta}$$
Also $$A_\beta \cdot A_\alpha \, = \,  A_{\alpha + \beta}$$ which can be shown  as above .    Hence proved .

A =  $$\begin{bmatrix}  3 & -4 \\ 1  & -1\end{bmatrix}$$ then $$A^1 \, = \, \begin{bmatrix}1 + 2k   &  - 4k \\ k    & 1 -2k\end{bmatrix}_{k = 1}$$
Also $$A^2$$ = AA
= $$\begin{bmatrix}  3 & -4 \\ 1  & -1\end{bmatrix}$$    $$\begin{bmatrix}  3 & -4 \\ 1  & -1\end{bmatrix}$$ =  $$\begin{bmatrix}  9-4 & -12+4 \\ 3-1  & -4+1\end{bmatrix}$$
or $$A^2$$ =  $$\begin{bmatrix} 5 & -8 \\ 2  & -3\end{bmatrix}$$ =  $$\begin{bmatrix} 1+22 & -4.2 \\ 2  & 1-2.2\end{bmatrix}_{k \, = \, 2}$$
Now assume that $$A^k \, = \, \begin{bmatrix}1 + 2k   &  - 4k \\ k    & 1 -2k\end{bmatrix}_{k = k}$$
$$\therefore A^{k + 1} \, = \, AA^k \, = \,  \begin{bmatrix}  3 & -4 \\ 1  & -1\end{bmatrix}  \begin{bmatrix}1 + 2k   &  - 4k \\ k    & 1 -2k\end{bmatrix}$$
= $$\begin{bmatrix}3 + 6k - 4   & -12k -4 + 8k   \\ 1 + 2k - k    & -4k -1 + 2k\end{bmatrix}$$
= $$\begin{bmatrix}3 + 2k    & -4k -4    \\ 1 + k  & -2k -1 \end{bmatrix}$$
= $$\begin{bmatrix}1 + 2(k + 1)    & -4(k +1) \\ 1 + k  & 1-2{k +1} \end{bmatrix}$$
we observe that our assumption is true for k = k + 1 and it was true when k = 1 or 2. Hence it is true for all +ve integral values for k .

$$Let\quad A=\quad \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}\\ \\ { A }^{ 1 }=\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}=A$$

Let, $$A = \begin {bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end {bmatrix}$$

$$AC+B^2-10C$$

Given  $$A=\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$$ .

Given

A is a skew symmetric matrix.

AB=BA

Let $$A = \begin{pmatrix} a_{11} & a_{12} & .... & a_{1n} \\ a_{21} & a_{22} & .... & a_{2n} \\ .\\.\\.\\a_{m_1} & a_{m_2} & .... & a_{mn} \end{pmatrix} = \begin{pmatrix} 1 \times a_{11} & 1 \times a_{12} & ... & 1 \times a_{1n} \\ 1 \times a_{21} & 1 \times a_{22} & ... & 1 \times a_{2n} \\ .\\.\\.\\ 1 \times a_{m_1} & 1 \times a_{m_2} & ... & 1 \times a_{mn} \end{pmatrix}$$

$$A^2=I$$

$$\begin{array}{l} A=\left[ { \begin{array} { *{ 20 }{ c } }3 & 5 \\ 7 & { -9 } \end{array} } \right] \, \, \, \, \, \, \, \, \, and\, \, \, \, \, \, \, B=\left[ { \begin{array} { *{ 20 }{ c } }6 & { -4 } \\ 2 & 3 \end{array} } \right]  \\ \left( { 4A-3B } \right) =4A+\left( { -3B } \right)  \\ Now, \\ 4A=\left[ { \begin{array} { *{ 20 }{ c } }{ 4\times 3 } & { 4\times 5 } \\ { 4\times 7 } & { 4\times \left( { -9 } \right)  } \end{array} } \right] = \\ =\left[ { \begin{array} { *{ 20 }{ c } }{ 12 } & { 20 } \\ { 28 } & { -36 } \end{array} } \right]  \\ And, \\ -3B=\left( { -3 } \right) \times B=\left[ { \begin{array} { *{ 20 }{ c } }{ \left( { -3 } \right) \times 6 } & { \left( { -3 } \right) \times \left( { -4 } \right)  } \\ { \left( { -3 } \right) \times 2 } & { \left( { -3 } \right) \times 3 } \end{array} } \right]  \\ =\left[ { \begin{array} { *{ 20 }{ c } }{ -18 } & { 12 } \\ { -6 } & { -9 } \end{array} } \right]  \\ \therefore \left( { 4A-3B } \right) =4A+\left( { -3B } \right)  \\ =\left[ { \begin{array} { *{ 20 }{ c } }{ 12 } & { 20 } \\ { 28 } & { -36 } \end{array} } \right] +\left[ { \begin{array} { *{ 20 }{ c } }{ -18 } & { 12 } \\ { -6 } & { -9 } \end{array} } \right]  \\ =\left[ { \begin{array} { *{ 20 }{ c } }{ 12+\left( { -18 } \right)  } & { 20+12 } \\ { 28+\left( { -6 } \right)  } & { -36+\left( { -9 } \right)  } \end{array} } \right]  \\ =\left[ { \begin{array} { *{ 20 }{ c } }{ -6 } & { 32 } \\ { 22 } & { -45 } \end{array} } \right]  \\ Hence, \\ \left( { 4A-3B } \right) =\left[ { \begin{array} { *{ 20 }{ c } }{ -6 } & { 32 } \\ { 22 } & { -45 } \end{array} } \right]  \\  \end{array}$$

We have,

Let $$A = \begin{bmatrix} 2 & 5 & -1 \\ 3 & 1 & 5 \\ 7 & 6 & 9 \end{bmatrix}$$

$$a_{ij}=\dfrac{|2i-3j|}{2}$$

$$a_{ij}$$	$$i=\,,j=$$	$$a_{ij}=\dfrac{|2i-3j|}{2}$$
$$a_{11}$$	$$i=1,\,j=1$$	$$a_{11}=\dfrac{|2(1)-3(1)|}{2}=\dfrac{|-1|}{2}=\dfrac{1}{2}$$
$$a_{12}$$	$$i=1,\,j=2$$	$$a_{12}=\dfrac{|2(1)-3(2)|}{2}=\dfrac{|-4|}{2}=\dfrac{4}{2}=2$$
$$a_{21}$$	$$i=2,\,j=1$$	$$a_{21}=\dfrac{|2(2)-3(1)|}{2}=\dfrac{|1|}{2}=\dfrac{1}{2}$$
$$a_{22}$$	$$i=2,\,j=2$$	$$a_{22}=\dfrac{|2(2)-3(2)|}{2}=\dfrac{|-2|}{2}=\dfrac{2}{2}=1$$

$$\Rightarrow$$   Required matrix $$A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}=\begin{bmatrix}\dfrac{1}{2}&2\\\dfrac{1}{2}&1\end{bmatrix}$$

$$a_{ij}=\dfrac{(i-2j)^2}{2}$$

$$a_{ij}$$	$$i=\,,j=$$	$$a_{ij}=\dfrac{(i-2j)^2}{2}$$
$$a_{11}$$	$$i=1,\,j=1$$	$$a_{11}=\dfrac{(1-2(1))^2}{2}=\dfrac{(-1)^2}{2}=\dfrac{1}{2}$$
$$a_{12}$$	$$i=1,\,j=2$$	$$a_{12}=\dfrac{(1-2(2))^2}{2}=\dfrac{(-3)^2}{2}=\dfrac{9}{2}$$
$$a_{21}$$	$$i=2,\,j=1$$	$$a_{21}=\dfrac{(2-2(1))^2}{2}=0$$
$$a_{22}$$	$$i=2,\,j=2$$	$$a_{22}=\dfrac{(2-2(2))^2}{2}=\dfrac{(-2)^2}{2}=\dfrac{4}{2}=2$$

$$\Rightarrow$$   Required matrix $$A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}=\begin{bmatrix}\dfrac{1}{2}&\dfrac{9}{2}\\0&2\end{bmatrix}$$

$$\Rightarrow$$   $$a_{ij}=\dfrac{(i-j)^2}{2}$$

$$a_{ij}=2i-j$$

$$\Rightarrow$$   $$a_{ij}=\dfrac{(i+j)^2}{2}$$

$$a_{ij}=e^{2ix}\sin xj$$

$$a_{ij}$$	$$i=\,,j=$$	$$a_{ij}=e^{2ix}\sin xj$$
$$a_{11}$$	$$i=1,\,j=1$$	$$a_{11}=e^{2\times 1\times x}\sin(x\times 1)=e^{2x}\sin(x)$$
$$a_{12}$$	$$i=1,\,j=2$$	$$a_{12}=e^{2\times 1\times x}\sin(x\times 2)=e^{2x}\sin(2x)$$
$$a_{21}$$	$$i=2,\,j=1$$	$$a_{21}=e^{2\times 2\times x}\sin(x\times 1)=e^{4x}\sin(x)$$
$$a_{22}$$	$$i=2,\,j=2$$	$$a_{22}=e^{2\times 2\times x}\sin(x\times 2)=e^{4x}\sin(2x)$$

$$\Rightarrow$$   Required matrix $$A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}=\begin{bmatrix}e^{2x}\sin(x)&e^{2x}\sin(2x)\\e^{4x}\sin(x)&e^{4x}\sin(2x)\end{bmatrix}$$

$$a_{ij}=\dfrac{(2i+j)^2}{2}$$

$$a_{ij}$$	$$i=\,,j=$$	$$a_{ij}=\dfrac{(2i+j)^2}{2}$$
$$a_{11}$$	$$i=1,\,j=1$$	$$a_{11}=\dfrac{(2(1)+1)^2}{2}=\dfrac{(3)^2}{2}=\dfrac{9}{2}$$
$$a_{12}$$	$$i=1,\,j=2$$	$$a_{12}=\dfrac{(2(1)+2)^2}{2}=\dfrac{(4)^2}{2}=\dfrac{16}{2}=8$$
$$a_{21}$$	$$i=2,\,j=1$$	$$a_{21}=\dfrac{(2(2)+1)^2}{2}=\dfrac{(5)^2}{2}=\dfrac{25}{2}$$
$$a_{22}$$	$$i=2,\,j=2$$	$$a_{22}=\dfrac{(2(2)+2)^2}{2}=\dfrac{(6)^2}{2}=\dfrac{36}{2}=18$$

$$\Rightarrow$$   Required matrix $$A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}=\begin{bmatrix}\dfrac{9}{2}&8\\\dfrac{25}{2}&18\end{bmatrix}$$

Let $$A =[a_{ij}]$$ be a $$2\times 2$$ matrix such that $$a_{ij} = \dfrac {(i + j)^{2}}{2}$$ . 

Here, $$a_{ij}=2i+\dfrac{i}{j}$$

$$A=\begin{bmatrix}2&4\\3&2\end{bmatrix}$$ and $$B=\begin{bmatrix}1&3\\-2&5\end{bmatrix}$$

$$B=\begin{bmatrix}1&3\\-2&5\end{bmatrix}$$ and $$C=\begin{bmatrix}-2&5\\3&4\end{bmatrix}$$

$$A=\begin{bmatrix}2&4\\3&2\end{bmatrix}$$ and $$B=\begin{bmatrix}-2&5\\3&4\end{bmatrix}$$

The symmetric matrix is of type, $$\begin{bmatrix}x&a&b \\a&y&c \\b&c&z \end{bmatrix}$$ and skew-symmetric matrix is of type  $$\begin{bmatrix}0&f&g \\-f&0&h \\-g&-h&0 \end{bmatrix}$$

Given $$A,B$$ are both symmetric matrices of same order.

$$\begin{bmatrix} 9& -1 & 4\\ -2 & 1 & 3\end{bmatrix} = A + \begin{bmatrix} 1& 2 & -1\\ 0 &4  &9 \end{bmatrix}$$               [ Given ]

$$A = \begin{bmatrix}0 & 1 & -2\\ -1 & 0 & 3\\ x & -3 & 0\end{bmatrix}$$

Given for  a $$2\times 2$$ matrix $$A = [a_{ij}]$$ , whose elements $$a_{ij}$$ are given by $$a_{ij} = \left\{\begin{matrix}\dfrac {|-3i + j|}{2}, & \ \text{if}\ i\neq j\\ (i + j)^{2}, & \ \text{if}\ i = j\end{matrix}\right.$$

$$A=\left[ \begin{matrix} 3 & 1 & 2 \\ 1 & 2 & -3 \end{matrix} \right] \\ B=\left[ \begin{matrix} -2 & 0 & 4 \\ 5 & -3 & 2 \end{matrix} \right] \\ now,\quad \\ 2A\quad =\quad \left[ \begin{matrix} 6 & 2 & 4 \\ 2 & 4 & -6 \end{matrix} \right] \quad \\ thus,\\ 2A-B\quad =\quad \quad \left[ \begin{matrix} 6-(-2) & 2-0 & 4-4 \\ 2-5 & 4-(-3) & -6-2 \end{matrix} \right] =\left[ \begin{matrix} 8 & 2 & 0 \\ -3 & 7 & -8 \end{matrix} \right] \\ $$

$$Given\quad \\ A=\left[ \begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix} \right] ,\quad B=\left[ \begin{matrix} 1 & 3 \\ -2 & 5 \end{matrix} \right] \quad and\quad C=\left[ \begin{matrix} -2 & 5 \\ 3 & 4 \end{matrix} \right]$$

Given $$A=\begin {bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix},$$    $$B=\begin {bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}$$ and  $$C=\begin {bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix}$$

Given: $$a_{ij}=\dfrac{1}{2}|-3i+j|;2\times 3$$ matrix

We have, $$A= \begin{bmatrix} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{bmatrix}$$ then, $$A{'}= \begin{bmatrix} 2 & 7 & 1 \\ 4 & 3 & -2 \\ -6 & 5 & 4 \end{bmatrix}$$
Hence $$\dfrac{A+A^{'}}{2}=\dfrac{1}{2} \begin{bmatrix} 4 & 11 & -5 \\ 11 & 6 & 3 \\ -5 & 3 & 8 \end{bmatrix}=\begin{bmatrix} 2 & \dfrac { 11 }{ 2 }  & \dfrac { 5 }{ 2 }  \\ \dfrac { 11 }{ 2 }  & 3 & \dfrac { 3 }{ 2 }  \\ \dfrac { -5 }{ 2 }  & \dfrac { 3 }{ 2 }  & 4 \end{bmatrix}$$

And $$\dfrac{A-A^{'}}{2} \begin{bmatrix} 0 & -3 & -7 \\ 3 & 0 & 7 \\ 7 & -7 & 0 \end{bmatrix}=\begin{bmatrix} 0 & -\dfrac { 3 }{ 2 } - & \dfrac { 7 }{ 2 }  \\ \dfrac { 3 }{ 2 }  & 0 & \dfrac { 7 }{ 2 }  \\ \dfrac { 7 }{ 2 }  & -\dfrac { 7 }{ 2 }  & 0 \end{bmatrix}$$

Therefore,

$$\dfrac{A+A^{'}}{2} + \dfrac{A-A^{'}}{2}=\begin{bmatrix} 2 & \dfrac { 11 }{ 2 }  & \dfrac { -5 }{ 2 }  \\ \dfrac { 11 }{ 2 }  & 3 & \dfrac { 3 }{ 2 }  \\ \dfrac { -5 }{ 2 }  & \dfrac { 3 }{ 2 }  & 4 \end{bmatrix}+\begin{bmatrix} 0 & \dfrac { -3 }{ 2 }  & \dfrac { -7 }{ 2 }  \\ \dfrac { 3 }{ 2 }  & 0 & \dfrac { 7 }{ 2 }  \\ \dfrac { 7 }{ 2 }  & \dfrac { -7 }{ 2 }  & 0 \end{bmatrix}=\begin{bmatrix} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 0 & -2 & 4 \end{bmatrix}=A$$

We know that, the notation, namely $$A[a_v]_{m \times n}$$ indicates that $$A$$ is a matrix order $$m \times n$$ , also $$1 \le i \le m, 1\le j\le n; i,j \in N$$ .
Here, $$A=[a_v]_{m \times n}$$
$$\Rightarrow A=\dfrac{(i-2j)^2}{2}, 1 \le i \le 2, 1\le j\le 2 ....(i)$$
$$\therefore a_{11}=\dfrac{(1-2)^2}{2}=\dfrac{1}{2}$$
$$\therefore a_{12}=\dfrac{(1-2 \times 2)^2}{2}=\dfrac{9}{2}$$
$$\therefore a_{21}=\dfrac{(2-2 \times 1)^2}{2}=0$$
$$\therefore a_{22}=\dfrac{(2-2 \times 2^2)^2}{2}=2$$
Thus, $$A={ \begin{bmatrix} \dfrac { 1 }{ 2 }  & \dfrac { 9 }{ 2 }  \\ 0 & 0 \end{bmatrix} }_{ 2\times 2 }$$

Since, $$A[a_v]_{m \times n}$$ $$1 \le i \le m, 1\le j\le n; i,j \in N$$ .
$$\therefore A=[e^{ix} \sin fx]_{3 \times 2}; \  1 \le i \le 3, 1\le j\le 2$$
$$\|Rightarrow a_{11}=e^{lx}.\sin 1.x=e^x \sin x$$
$$a_{12}=e^{lx}. \sin 2x=e^x \sin 2x$$
$$a_{21}=e^{2x}. \sin 1x=e^{2x} \sin x$$
$$a_{22}=e^{2x}. \sin 2x=e^{2x} \sin 2x$$
$$a_{31}=e^{3x}. \sin 1x=e^{3x} \sin x$$
$$a_{32}=e^{3x}. \sin 2x=e^{3x} \sin 2x$$
$$\therefore A= { \begin{bmatrix} { e }^{ x }\sin { x }  & { e }^{ x }\sin { 2x }  \\ { e }^{ 2x }\sin { x }  & { e }^{ 2x }\sin { 2x }  \\ { e }^{ 3x }\sin { x }  & { e }^{ 3x }\sin { 2x }  \end{bmatrix} }_{ 3\times 2 }$$

We know that, the notation, namely $$A[a_v]_{m \times n}$$ indicates that $$A$$ is a matrix order $$m \times n$$ , also $$1 \le i \le m, 1\le j\le n; i,j \in N$$ .
Here, $$A=[a_v]_{m \times n}=|-2i+3j| , \ 1 \le i \le 2, 1\le j\le 2$$
$$\therefore a_{11}=|-2 \times 1+3 \times 1|=1$$
$$a_{12}=|-2 \times 1+3 \times 2|=4 \ [\because |-1|=1]$$
$$a_{21}=|-2 \times 1+3 \times 1|=1$$
$$a_{22}=|-2 \times 1+3 \times 2|=2$$
$$\therefore A ={ \begin{bmatrix} \dfrac { 1 }{ 2 }  & \dfrac { 9 }{ 2 }  \\ 0 & 0 \end{bmatrix} }_{ 2\times 2 }$$

$$A=\begin{bmatrix} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{bmatrix}$$

$$A = [a_{ij}]_{3 \times 3} = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$  

$$A_{21} = (-1)^{ 2 + 1} M_{21} = - \begin{vmatrix}a_{12} & a_{13} \\ a_{32} & a_{33}  \end{vmatrix}$$
$$= -(a_{12} a_{33} - a_{13}a_{32} )$$  
$$= - a_{12} a_{33} + a_{13}a_{32}$$   

$$A_{22} = (-1)^{2 + 2} M_{22} = - \begin{vmatrix}a_{11} & a_{13} \\ a_{31} & a_{33} \end{vmatrix}$$  
$$= a_{11} a_{33} - a_{13} a_{31}$$  

$$A_{23} = (-1)^{2 + 3} M_{23} = - \begin{vmatrix} a_{11} & a_{12} \\ a_{31} & a_{32}\end{vmatrix}$$  
$$= - (a_{11} a_{32} - a_{12} a_{31})$$  
$$= - a_{11} a_{32} + a_{12} a_{31}$$  

$$\therefore \, a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23}$$  

$$= a_{11} (- a_{12} a_{33} +  a_{13}a_{32}) + a_{12} (a_{11} a_{33} - a_{13} a_{31}) + a_{13} ( - a_{11} a_{32} + a_{12} a_{31})$$  

$$= -a_{11} a_{12} a_{33} +  a_{11} a_{13}a_{32} + a_{11} a_{12} a_{33} - a_{12} a_{13} a_{31} -  a_{11} a_{13} a_{32} + a_{12} a_{13}  a_{31})$$  
$$= 0$$  

$$A = [a_{ij}]_{3 \times 3} = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$  

$$A_{11} = (-1)^{ 1 + 1} M_{11} = - \begin{vmatrix}a_{22} & a_{23} \\ a_{32} & a_{33}  \end{vmatrix}$$  

$$A_{12} = (-1)^{1 + 2} M_{12} = - \begin{vmatrix}a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix}$$  

$$A_{13} = (-1)^{1 + 3} M_{13} = - \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32}\end{vmatrix}$$  

$$\therefore \, a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}$$  

$$= a_{11} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33}  \end{vmatrix} - a_{12} \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33}  \end{vmatrix} + a_{3} \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32}  \end{vmatrix}$$  

$$= \begin{vmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} = |A|$$  

$$a_{ij}=\frac{1}{2}\left| -3i+j \right|$$
$$A=\left[ \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } & { a }_{ 14 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } & { a }_{ 24 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } & { a }_{ 34 } \end{matrix} \right] _{3\times 4}$$
= $$\left[ \begin{matrix} 1 & \frac { 1 }{ 2 }  & 0 & \frac { 1 }{ 2 }  \\ \frac { 5 }{ 2 }  & 2 & \frac { 3 }{ 2 }  & 1 \\ 4 & \frac { 7 }{ 2 }  & 3 & \frac { 5 }{ 2 }  \end{matrix} \right] _{3\times 4}$$

$$3A=\left[ \begin{matrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{matrix} \right] _{ 3\times 3 }5B=\left[ \begin{matrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{matrix} \right]_{3 \times 3}$$
$$3A-5B=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] =0$$

$$A=A=\begin{bmatrix} 6 & 2 \\ 5 & 4 \end{bmatrix} \therefore A^T =A=\begin{bmatrix} 6 & 5 \\ 2 & 4 \end{bmatrix}$$

Given, $$B=b_{ij}$$ whose elements are

$$A=\left[ \begin{matrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{matrix} \right]$$
$$A'=\left[ \begin{matrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{matrix} \right]$$
$$(A+A')=\left[ \begin{matrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{matrix} \right]$$
$$B=\frac{1}{2}(A+A')=\left[ \begin{matrix} 3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{matrix} \right]$$
$$B'=\left[ \begin{matrix} 3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{matrix} \right] =B$$

$$B_{11} =1\times 1=1$$

$$A=\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{matrix} \right] \quad A'=\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{matrix} \right]$$
$$(A+A')=\left[ \begin{matrix} 12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6 \end{matrix} \right]$$
$$B=\frac{1}{2}(A+A')=\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{matrix} \right]$$
$$B'=\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{matrix} \right] =B$$
since $$B'=B; B=\frac{1}{2}(A+A')$$ is a symmetric matrix 
$$(A-A')=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] =0$$
$$\therefore C=\frac{1}{2}(A-A')=0=C'$$
$$\therefore C=C'  \quad \therefore C=\frac{1}{2}(A-A')$$ is s skew-symmetric matrix 
$$B+C=\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{matrix} \right] +\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] =\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{matrix} \right]$$
$$\therefore B+C=A$$
Hence proved.

$$\left(i\right) a_{ij} = \dfrac{\left(i + 2j\right)^2}{2i}$$
$$a_{11} = \dfrac{\left(1 + 2 \times 1\right)^2}{2 \times 1} = \dfrac{3^2}{2} = \dfrac{9}{2}$$
$$a_{12} = \dfrac{\left(1 + 2 \times 2\right)^2}{2 \times 1} = \dfrac{5^2}{2} = \dfrac{25}{2}$$
$$a_{21} = \dfrac{\left(2 + 2 \times 1\right)^2}{2 \times 2} = \dfrac{4^2}{4} = 4$$
$$a_{22} = \dfrac{\left(2 + 2 \times 1\right)^2}{2 \times 2} = \dfrac{6^2}{4} = 9$$
$$\therefore$$  Required matrix = $$\begin{bmatrix}9/2 &  25/2  \\ 4 & 9 \end{bmatrix}$$

$$a_{ij} = 2i-3j$$
$$a_{11} = 2 \times 1 – 3 \times 1 = 2-3 = -1$$
$$a_{12}=2 \times 1 -3 \times 2 = 2-6 =-4$$
$$a_{21} = 2\times 2 -3 \times 1 = 4 -3 = 1$$
$$a_{22}=2\times 2 -3 \times 2 = 4-6=-2$$
Required matrix = $$\begin{bmatrix} -1 & -4  \\ 1 & -2  \end{bmatrix}$$

$$\rightarrow$$ A square matrix A is defined as skew-symmetric if and only if it is the opposite of its transpose

Given, $$A=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ and $$B=\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$$ , 

$$a_{ij} = \dfrac{2i-j}{3i+j}$$  
$$a_{11} = \dfrac{2 \times 1 -1}{3 \times 1 +1} = \dfrac{2-1}{3+1} = \dfrac{1}{4}$$
$$a_{12} = \dfrac{2 \times 1 -2}{3 \times 1 +2} = \dfrac{2-2}{3+2} = 0$$
$$a_{21} = \dfrac{2 \times 2 -1}{3 \times 2 +1} = \dfrac{4-1}{6+1} = \dfrac{3}{7}$$
$$a_{22} = \dfrac{2 \times 2 -2}{3 \times 2 +2} =\dfrac{2}{8} =\dfrac{1}{4}$$
$$\therefore$$ Required matrix = $$\begin{bmatrix} a_{11} & a_{12}  \\ a_{21} & a_{21}  \end{bmatrix}$$
$$\begin{bmatrix} 1/4 & 0  \\ 3/7 & 1/4  \end{bmatrix}$$

$$A=\begin{bmatrix} 0 & -1 \\ 4 & -3 \end{bmatrix}\qquad B=\begin{bmatrix} -5 \\ 6 \end{bmatrix}\\ 3A={ \begin{bmatrix} 0 & -3 \\ 12 & -9 \end{bmatrix} }_{ 2\times 2 }\qquad 2B={ \begin{bmatrix} -10 \\ 12 \end{bmatrix} }_{ 2\times 1 }$$

Since A is   $$m \, \times  \, n$$ and AB defined therefore B should be   $$n \, \times  \, p$$ because the number of columns of A should be equal to number of rows of B.
Again B is row $$n \, \times  \, p$$ and A is $$m \, \times  \, n$$ .
And since BA is also defined therefore p should be equal to m by the same argument as above i.e., p = m
$$\therefore$$ B is $$m \, \times  \, n$$ matrix.

A = 3 $$\times$$ 2 and B = 3 $$\times$$ 3. Since the number of columns of A is not equal to number of rows of B as such the matrix AB is not defined. But the matrix BA is defined because the number of columns in B is equal to the number of rows in A each being equal to 3.
BA = $$\begin{bmatrix} 0&  1& 0\\  0&  2& 1\\  2&  3& 1\end{bmatrix} \, \begin{bmatrix} 1& 2\\  3& 0\\  4& 1\end{bmatrix}$$
=  $$\begin{bmatrix} 0.1+1.3+0.4& 0.2+1.0+0.1\\  0.1+2.3+1.4& 0.2+2.0+1.1 \\ 2.1+3.3+1.4& 2.2+3.0+1.1 \end{bmatrix} \, = \, \begin{bmatrix} 3& 0\\ 10& 1\\  15& 5\end{bmatrix}$$
Verification for the Product to be correct.
We have proved above that 
BA =  $$\begin{bmatrix} 0&  1& 0\\  0&  2& 1\\  2&  3& 1\end{bmatrix} \, \begin{bmatrix} 1& 2\\  3& 0\\  4& 1\end{bmatrix} \, = \, \begin{bmatrix} 3& 0\\  10& 1\\  15& 5\end{bmatrix}$$
sum     2    6    2                                28   6
Multiply the row of sum i.e. 2,6,2 with the columns of 2nd matrix and the result should come out to be row of sum of product matrix on the right i.e. 28,6 as shown below.
i.e. 2.1 + 6.3 + 2.4 = 2 + 18 + 8 = 28
and 2.2 + 6.0 + 2.1 = 4 + 0 + 2 = 6
Hence correct.

Let us choose two non-zero matrices A and B as under
$$A \, = \, \begin{bmatrix} 1& 1\\  3& 3\end{bmatrix}_{2 \, \times \, 2} \, and \, B \, = \, \begin{bmatrix} -1& 1\\  1& -1\end{bmatrix}_{2 \, \times \, 2}$$
$$AB \, = \, \begin{bmatrix} 1(-1)+1.1& 1.1+1(-1)\\  3(-1)+3.1& 3(1)+3(-1)\end{bmatrix}\, = \,\begin{bmatrix} 0& 0\\  0& -0\end{bmatrix} \, = \, O$$
Thus we observe that the matrix AB is null matrix whereas neither A nor B is a null matrix.
$$BA \, = \, \begin{bmatrix} 2& 2\\  -2& -2\end{bmatrix} \, \neq \, O$$

$$\begin{bmatrix} 3x+4y & 2 & x-2y \\ a+b & 2a-b & -1 \end{bmatrix}=\begin{bmatrix} 2 & 2 & 4 \\ 5 & -5 & -1 \end{bmatrix}\quad$$

A matrix is symmetric if A'=A and skew symmetric if A '=A

  $$\begin{array}{l} A=\left( { \begin{array} { *{ 20 }{ c } }1 & 2 \\ 3 & 4 \end{array} } \right) ,\, \, B\left( { \begin{array} { *{ 20 }{ c } }3 & 8 \\ 7 & 2 \end{array} } \right) \\ 2x=B-A \\ 2x=\left( { \begin{array} { *{ 20 }{ c } }3 & 8 \\ 7 & 2 \end{array} } \right) -\left( { \begin{array} { *{ 20 }{ c } }1 & 2 \\ 3 & 4 \end{array} } \right)  \\ 2x=\left( { \begin{array} { *{ 20 }{ c } }2 & 6 \\ 4 & { -2 } \end{array} } \right)  \\ x=\frac { 1 }{ 2 } \left( { \begin{array} { *{ 20 }{ c } }2 & 6 \\ 4 & { -2 } \end{array} } \right) \to x=\left( { \begin{array} { *{ 20 }{ c } }1 & 3 \\ 2 & { -1 } \end{array} } \right)  \end{array}$$

Given $$B=\left[ \begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix} \right]$$