-1/5

Let A be any square matrix. Then, 
\(A=\frac { 1 }{ 2 } \left( A+{ A }^{ T } \right) +\frac { 1 }{ 2 } \left( A-{ A }^{ T } \right) \)
= P + Q (say),
where, \(P=\frac { 1 }{ 2 } \left( A+{ A }^{ T } \right) \)
and \(Q=\frac { 1 }{ 2 } \left( A-{ A }^{ T } \right) \)
Now, \({ P }^{ T }=\left[ \frac { 1 }{ 2 } \left( A+{ A }^{ T } \right) \right] ^{ T }\)
\(=\frac { 1 }{ 2 } \left( A+{ A }^{ T } \right) ^{ T }\quad \left[ \because \quad \left( KA \right) ^{ T }=K.{ A }^{ T } \right] \)
\(\Rightarrow \ { P }^{ T }=\frac { 1 }{ 2 } \left[ { A }^{ T }+\left( { A }^{ T } \right) ^{ T } \right] \) \(\left[ \because \ \left( A+B \right) ^{ T }={ A }^{ T }+{ B }^{ T } \right] \)
\(\Rightarrow \ { P }^{ T }=\frac { 1 }{ 2 } \left( { A }^{ T }+A \right) \)  \(\left[ \because \ \left( { A }^{ I } \right) ^{ T }=A \right] \)
\(\Rightarrow { P }^{ T }=\frac { 1 }{ 2 } \left( A{ +A }^{ T } \right) =P\)
\(\therefore \)  P is symmetric matrix.
Also, \({ Q }^{ T }=\frac { 1 }{ 2 } \left( A-{ A }^{ T } \right) ^{ T }\)
\(=\frac { 1 }{ 2 } \left( A-{ A }^{ T } \right) ^{ T }\)
\(=\frac { 1 }{ 2 } \left[ { A }^{ T }-\left( { A }^{ T } \right) ^{ T } \right] \)
\(=\frac { 1 }{ 2 } \left[ { A }^{ T }-A \right] \)
\({ \Rightarrow Q }^{ T }=\frac { 1 }{ 2 } \left[ A-{ A }^{ T } \right] =-Q\)
\(\therefore \)  Q is skew symmetric matrix.
Thus, A = P + Q, where P is a symmetric matrix and Q is a skew symmetric matrix.
Hence, A is expressible as the sum of a symmetric and a skew symmetric matrix.
Uniqueness : If possible, let A = R + S, where R is symmetric and S is skew symmetric, then
A^T = (R + S)^T = R^T + S^T
\(\Rightarrow \)  A^T = R - S  (\(\because \)  R^T = R and S^T = - S)
Now, A = R + S and A^T = R - S
\(\Rightarrow R=\frac { 1 }{ 2 } \left[ A+{ A }^{ T } \right] =P\)
\(S=\frac { 1 }{ 2 } \left[ A-{ A }^{ T } \right] =Q\)
Hence, A is uniquely expressible as the sum of a symmetric and a skew symmetric matrix.

We have, \(A=\left[ \begin{matrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{matrix} \right] \)
\({ A }^{ T }=\left[ \begin{matrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{matrix} \right] \)
\(\Rightarrow A+{ A }^{ T }=\left[ \begin{matrix} 2cos\theta & 0 \\ 0 & cos\theta \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
\(\Rightarrow 2cos\theta =1\)
\(\Rightarrow cos\theta =\frac { 1 }{ 2 } \Rightarrow \theta =\cos ^{ -1 }{ \left( \frac { 1 }{ 2 } \right) } \)
\(\because \quad \theta =\frac { \pi }{ 3 } \)

We have, \(A=\left[ \begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix} \right] \) and \(B=\left[ \begin{matrix} -1 & 4 \\ 0 & 2 \end{matrix} \right] \)
\(AB=\left[ \begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix} \right] \left[ \begin{matrix} -1 & 4 \\ 0 & 2 \end{matrix} \right] \)
\(AB=\left[ \begin{matrix} -2+0 & 8+2 \\ -3+0 & 12+8 \end{matrix} \right] \)
\(\left[ \begin{matrix} -2 & 10 \\ -3 & 20 \end{matrix} \right] \)
and \(BA=\left[ \begin{matrix} -1 & 4 \\ 0 & 2 \end{matrix} \right] \left[ \begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 10 & 15 \\ 6 & 8 \end{matrix} \right] \)
\(\therefore\)   \(AB\neq BA\)

We have, \(\left[ \begin{matrix} { x }^{ 2 } \\ { y }^{ 2 } \end{matrix} \right] -3\left[ \begin{matrix} x \\ 2y \end{matrix} \right] =\left[ \begin{matrix} -2 \\ -9 \end{matrix} \right] \)
\(\Rightarrow\)  x² - 3x = - 2 and y² - 6y = - 9
\(\Rightarrow\)  x² - 3x + 2 = 0 and y² - 6y + 9 = 0
\(\Rightarrow\)  x² - 2x - x + 2 = 0 and y² - 3y - 3y + 9 = 0
\(\Rightarrow\)  x(x - 2) - 1(x - 2) = 0 and y(y - 3) - 3(y - 3) = 0
\(\Rightarrow\)  (x - 2)(x - 1) = 0 and (y - 3)(y - 3) = 0
\(\therefore\)   x = 1, 2 and y = 3, 3

We have, \(\left[ \begin{matrix} 2x+y & x-y \\ x-z & x+y+z \end{matrix} \right] =\left[ \begin{matrix} 10 & -1 \\ 2 & 8 \end{matrix} \right] \)
\(\Rightarrow\) 2x + y = 10, x - y = - 1
x - z = 2 and x + y + z = 8
\(\therefore\) 2(y - 1) + y = 10  \(\Rightarrow\)  2y + y + 2 = 10
\(\Rightarrow\)3y = 12  \(\Rightarrow\)  y = 4
\(\therefore\)  x = 3
3 - z,  \(\Rightarrow\)  z = 1
\(\therefore\)  x = 3, y = 4, z = 1

Any square matrix can be expressed
\(A=\frac{1}{2}\left(A+A^{\prime}\right)+\frac{1}{2}\left(A-A^{\prime}\right)\)
we know that (A + A′) is a symmetric matrix and (A – A′) is a skew symmetric matrix. Since for any matrix A, (kA)′ = kA′, it follows that \(\frac{1}{2}\left(A+A^{\prime}\right)\) is symmetric matrix and \(\frac{1}{2}\left(A-A^{\prime}\right)\) is skew symmetric matrix. Thus, any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix.

A square matrix \(A=[{ a }_{ ij }] \)is skew symmetric
if \({ a }_{ ij }=-{ a }_{ ji }\forall i,j \)
Let \(i=j\Rightarrow { a }_{ ii }=-{ a }_{ ii }\Rightarrow { 2a }_{ ii }=0\Rightarrow { a }_{ ii }=0\Rightarrow 0  \)
All the diagonal elements of a skew symmetric matrix are always zero.

\(AA'= \left[ \begin{matrix} 1 & 2 & 3 \end{matrix} \right] \left[ \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right] =\left[ 1+4+9 \right] =\left[ 14 \right]\)

Let numbers be x, y, z then
x + y + z = -1
2y + 3z = 5
x + y – z = -1
\(x=-\frac { 7 }{ 2 } \), y = \(\frac { 5 }{ 2 } \), z = 0

Getting \({ A }^{ 2 }=\left( \begin{matrix} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{matrix} \right) \)
and \({ A }^{ 3 }=\left( \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right) \)
\(\therefore\)  A³ - 4A² - 3A + 11I
\(=\left( \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right) -\left( \begin{matrix} 36 & 28 & 20 \\ 4 & 16 & 4 \\ 32 & 36 & 36 \end{matrix} \right) -\left( \begin{matrix} 3 & 9 & 6 \\ 6 & 0 & -3 \\ 3 & 6 & 9 \end{matrix} \right) +\left( \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) =0\)
Alternative Method :
\(A=\left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] \)
\({ A }^{ 2 }=\left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] \left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{matrix} \right] \)
\({ A }^{ 3 }=\left[ \begin{matrix} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{matrix} \right] \left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right] \)
\(\therefore\)  A³ - 4A² - 4A + 11A
\(=\left[ \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right] -4\left[ \begin{matrix} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{matrix} \right] -3\left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] +\left[ \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right] -\left[ \begin{matrix} 36 & 28 & 20 \\ 4 & 16 & 4 \\ 32 & 36 & 36 \end{matrix} \right] -\left[ \begin{matrix} 3 & 9 & 6 \\ 6 & 0 & -3 \\ 3 & 6 & 9 \end{matrix} \right] +\left[ \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
= 0

\(p I+q A=\left[\begin{array}{ll} p & 0 \\ 0 & p \end{array}\right]+\left[\begin{array}{rr} 0 & q \\ -q & q \end{array}\right]=\left[\begin{array}{cc} p & q \\ -q & p+q \end{array}\right] \)
\(\text {Now }(p I+q A)^{2}=\left[\begin{array}{cc} p & q \\ -q & p+q \end{array}\right]\left[\begin{array}{cc} p & q \\ -q & p+q \end{array}\right] \)
\(\Rightarrow \ p^{2}-q^{2}=0 \text { and } 2 p q+q^{2}=1 ; 2 p q+p^{2}=1 \)
\(\Rightarrow \ p^{2}=q^{2} \Rightarrow p=\pm q \)
\(\text {When } \quad p=q \Rightarrow 2 q^{2}+q^{2}=1 \Rightarrow 3 q^{2}=1\)
\(\Rightarrow \quad q=\pm \frac{1}{\sqrt{3}}=p\)
\(\text {When } p=-q \Rightarrow-2 q^{2}+q^{2}=1 \Rightarrow q^{2}=-1 \text { (not possible) }\)

(i) \(A=\frac { 1 }{ 2 } (A+A')+\frac { 1 }{ 2 } (A-A');\) 
where \(A=\frac { 1 }{ 2 } (A+A')\) is symmetric
and \(\frac { 1 }{ 2 } (A-A')\) is skew symmetric Proceed.
\(\text { Let } A=\left[\begin{array}{rr} 3 & 5 \\ 1 & -1 \end{array}\right] \text { , then } A^{\prime}=\left[\begin{array}{ll} 3 & 1 \\ 5 & -1 \end{array}\right]\)
\(\text { Now, } A+A^{\prime}=\left[\begin{array}{rr} 3 & 5 \\ 1 & -1 \end{array}\right]+\left[\begin{array}{rr} 3 & 1 \\ 5 & -1 \end{array}\right]=\left[\begin{array}{rr} 6 & 6 \\ 6 & -2 \end{array}\right]\)
\(\text { Let } P=\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{rr} 6 & 6 \\ 6 & -2 \end{array}\right]=\left[\begin{array}{rr} 3 & 3 \\ 3 & -1 \end{array}\right]\)
\(\text { Now, } P^{\prime}=\left[\begin{array}{rr} 3 & 3 \\ 3 & -1 \end{array}\right]=P\)
\(\text { Thus, }P=\frac{1}{2}\left(A+A^{\prime}\right) \text { is a symmetric matrix. }\)
\(\text { Now, } A-A^{\prime}=\left[\begin{array}{rr} 3 & 5 \\ 1 & -1 \end{array}\right]-\left[\begin{array}{cc} 3 & 1 \\ 5 & -1 \end{array}\right]=\left[\begin{array}{rr} 0 & 4 \\ -4 & 0 \end{array}\right]\)
\(\text { Let } Q=\frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{rr} 0 & 4 \\ -4 & 0 \end{array}\right]=\left[\begin{array}{rr} 0 & 2 \\ -2 & 0 \end{array}\right]\)
\(\text { Now, } Q^{\prime}=\left[\begin{array}{rr} 0 & 2 \\ -2 & 0 \end{array}\right]=-Q\)
Thus \(Q=\frac{1}{2}\left(A-A^{\prime}\right) \text { is a skew-symmetric matrix. }\)
\(\text { Representing } A \text { as the sum of } P \text { and } Q \text { : }\)
\(P+Q=\left[\begin{array}{rr} 3 & 3 \\ 3 & -1 \end{array}\right]+\left[\begin{array}{rr} 0 & 2 \\ -2 & 0 \end{array}\right]=\left[\begin{array}{rr} 3 & 5 \\ 1 & -1 \end{array}\right]=A\)
(ii) \(\text { Let } A=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right], \text { then } A^{\prime}=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]\)
\(\text { Now, } A+A^{\prime}=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]+\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]=\left[\begin{array}{rrr} 12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6 \end{array}\right]\)
\(\text { Now, } P^{\prime}=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]=P\)
Thus,\(P=\frac{1}{2}\left(A+A^{\prime}\right) \text { is a symmetric matrix. }\)
\(\text { Now, } A-A^{\prime}=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]+\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\)
\(\text { Let } Q=\frac{1}{2}\left(A-A^{\prime}\right)=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\)
\(\text { Now, } Q^{\prime}=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]=-Q\)
Thus \(Q=\frac{1}{2}\left(A-A^{\prime}\right) \text { is a skew-symmetric matrix. }\)
\(\text { Representing } A \text { as the sum of } P \text { and } Q \text { : }\)
\(P+Q=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]+\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]=A\)
(iii) Let \(A=\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]\), then \(A^{\prime}=\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]\)
Now, \(A+A=\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]+\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]=\left[\begin{array}{ccc}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{array}\right]\)
\(P=\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{ccc}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{array}\right]=\left[\begin{array}{ccc}3 & \frac{1}{2} & -\frac{-5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2\end{array}\right]\)
Now, \(P^{\prime}=\left[\begin{array}{ccc}3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2\end{array}\right]=\left[\begin{array}{ccc}3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2\end{array}\right]=P\)
Thus, \(P=\frac{1}{2}\left(A+A^{\prime}\right)\) is a symmetric matrix. 
Now, \(A-A^{\prime}=\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]-\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]=\left[\begin{array}{ccc}0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0\end{array}\right]\)
Let \(Q=\frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{ccc}0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0\end{array}\right]=\left[\begin{array}{ccc}0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0\end{array}\right]\)
Now, \(Q=\left[\begin{array}{ccc}0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0\end{array}\right]=\left[\begin{array}{ccc}0 & -\frac{5}{2} & -\frac{3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0\end{array}\right]=-Q\)
Thus, \(Q=\frac{1}{2}\left(A-A^{\prime}\right)\) is a skew -symmetric matrix.
Representing A as the sum of P and Q
\(P+Q=\left[\begin{array}{ccc} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -22 & ] \end{array}\right]+\left[\begin{array}{ccc} 0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0 \end{array}\right]=\left[\begin{array}{ccc} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{array}\right]=A\)
(iv)
\(\text { Let } A=\left[\begin{array}{rr} 1 & 5 \\ -1 & 2 \end{array}\right] \text { , then } A^{\prime}=\left[\begin{array}{lr} 1 & -1 \\ 5 & 2 \end{array}\right]\)
\(\text { Now } A+A^{\prime}=\left[\begin{array}{rr} 1 & 5 \\ -1 & 2 \end{array}\right]+\left[\begin{array}{rr} 1 & -1 \\ 5 & 2 \end{array}\right]=\left[\begin{array}{ll} 2 & 4 \\ 4 & 4 \end{array}\right]\)
\(\text { Let } P=\frac{1}{2}\left(A+A^{\prime}\right)=\left[\begin{array}{ll} 1 & 2 \\ 2 & 2 \end{array}\right]\)
\(P=\frac{1}{2}\left(A+A^{\prime}\right) \text { is a symmetric matrix. }\)
\(\text { Now, } A-A^{\prime}=\left[\begin{array}{rr} 1 & 5 \\ -1 & 2 \end{array}\right]-\left[\begin{array}{rr} 1 & -1 \\ 5 & 2 \end{array}\right]=\left[\begin{array}{rr} 0 & 6 \\ -6 & 0 \end{array}\right]\)
\(\text { Let } Q=\frac{1}{2}\left(A-A^{\prime}\right)=\left[\begin{array}{rr} 0 & 3 \\ -3 & 0 \end{array}\right]\)
\(\text { Now, } Q^{\prime}=\left[\begin{array}{rr} 0 & -3 \\ 3 & 0 \end{array}\right]=-Q\)
\(Q=\frac{1}{2}\left(A-A^{\prime}\right) \text { is a skew-symmetric matrix. }\)
\(\text { Representing } A \text { as the sum of } P \text { and } Q \text { : }\)
\(P+Q=\left[\begin{array}{ll} 1 & 2 \\ 2 & 2 \end{array}\right]+\left[\begin{array}{rr} 0 & 3 \\ -3 & 0 \end{array}\right]=\left[\begin{array}{rr} 1 & 5 \\ -1 & 2 \end{array}\right]=A\)

We have:  \(A=\begin{bmatrix} 1 & 4 \\ 3 & 7 \end{bmatrix}.\)
\(\therefore\ A\prime =\begin{bmatrix} 1 & 3 \\ 4 & 7 \end{bmatrix}\)
\( \therefore \ A-A\prime =\begin{bmatrix} 1 & 4 \\ 3 & 7 \end{bmatrix}+\begin{bmatrix} 1 & 3 \\ 4 & 7 \end{bmatrix}\)
\(=\begin{bmatrix} 1 & 4 \\ 3 & 7 \end{bmatrix}+\begin{bmatrix} -1 & -3 \\ -4 & -7 \end{bmatrix}\)
\(=\begin{bmatrix} 1-1 & 4-3 \\ 3-4 & 7-7 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} ...(1)\)
Now    \((A-A\prime )\prime =\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}=-\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\)
Hence,  \(A-A\prime \)  is skew-symmetric matrix.

We have:   \(A=\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7 \end{matrix} \right] \) and\(B=\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix} \right] .\)
\(\therefore \ 2A+3B=2\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7 \end{matrix} \right] +3\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 6 & 0 & 0 \\ 0 & -10 & 0 \\ 0 & 0 & 14 \end{matrix} \right] +\left[ \begin{matrix} -3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 12 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 6-3 & 0+0 & 0+0 \\ 0+0 & -10+6 & 0+0 \\ 0+0 & 0+0 & 14+12 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 26 \end{matrix} \right] \) =diag.\(\left[ \begin{matrix} 3, & -4, & 26 \end{matrix} \right] .\)

We have: \(3\begin{bmatrix} x & y \\ z & w \end{bmatrix}=\begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix}+\begin{bmatrix} 4 & x+y \\ z+w & 3 \end{bmatrix},\)
\(\Rightarrow \begin{bmatrix} 3x & 3y \\ 3z & 3w \end{bmatrix}=\begin{bmatrix} x+4 & x+y+6 \\ z+w-1 & 2w+3 \end{bmatrix},\)
Equating corresponding elements:
3x = x + y  \(\Rightarrow \) 2x = 4 \(\Rightarrow \) x = 2
3y = x + y + 6 \(\Rightarrow \) 2y = 2 + 6 \(\Rightarrow \) 2y = 8 \(\Rightarrow \) y = 4
3w = 2w + 3 \(\Rightarrow \) w = 3   and  
3z = z + w - 1\(\Rightarrow \) 2z = 3 - 1 = 2 \(\Rightarrow \) z = 1.
Hence, x = 2, y = 4, z = 1 and w = 3.

We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural numbers whose product is 18.
The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6), and (6, 3)
Hence, the possible orders of a matrix having 18 elements are:
1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, and 6 × 3
(1, 5) and (5, 1) are the ordered pairs of natural numbers whose product is 5.
Hence, the possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1.

We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the ordered pairs of natural numbers whose product is 24.
The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), and  (6, 4)
Hence, the possible orders of a matrix having 24 elements are:
1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, and 6 × 4
(1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13.
Hence, the possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1.