\(\left( \begin{matrix} 4 & -6 \\ 3 & -3 \end{matrix} \right) =\left( \begin{matrix} 1 & 2 \\ 0 & 3 \end{matrix} \right) \left( \begin{matrix} 2 & -4 \\ 1 & -1 \end{matrix} \right)\)
Alternative Method :
We have \(\left( \begin{matrix} 4 & 2 \\ 3 & 3 \end{matrix} \right) =\left( \begin{matrix} 1 & 2 \\ 0 & 3 \end{matrix} \right) \left( \begin{matrix} 2 & 0 \\ 1 & 1 \end{matrix} \right)\)
Applying \({ C }_{ 2 }\rightarrow { C }_{ 2 }+2{ C }_{ 1 }\)
\(\Rightarrow \ \left( \begin{matrix} 4 & -6 \\ 3 & -3 \end{matrix} \right) =\left( \begin{matrix} 1 & 2 \\ 0 & 3 \end{matrix} \right) \left( \begin{matrix} 2 & -4 \\ 1 & -1 \end{matrix} \right)\)

Let A be a skew-symmetric matrix. Then by definition \({ A }^{ \prime }=-A\)
\(\Rightarrow\) the (i, j)^th element of \({ A }^{ \prime }\) = the (i, j)^th element of (- A)
\(\Rightarrow\) the (j, i)^th element of A = - the (i, j)^th element of A
For the diagonal elements i = j \(\Rightarrow\) the (i, j)the element of A = - the (i, j)^th element of A.
\(\Rightarrow\)  the (i, j)^th element of A = 0
Hence the diagonal elements are all zero.

If A is symmetric matrix then
\(A={ A }^{ \prime }\)
\(\Rightarrow \left[ \begin{matrix} 0 & b & -2 \\ 3 & 1 & 3 \\ 2a & 3 & -1 \end{matrix} \right] =\left[ \begin{matrix} 0 & 3 & 2a \\ b & 1 & 3 \\ -2 & 3 & -1 \end{matrix} \right] \)
\(\therefore\)  By equality of matrices, 
b = 3 and a = - 1

\(k=-\frac { 3 }{ 2 }\)
Alternative Method :
Since diagonal elements in a skew symmetric matrix are zero.
We can compare diagonal elements to zero.
i.e., 2k + 3 = 0
⇒  2k = - 3
∴ \(k=-\frac { 3 }{ 2 }\)

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

λ=6
Alternate Method :
Given \({ A }^{ 2 }=\lambda A,\)
where \(A=\left[ \begin{matrix} 3 & -3 \\ -3 & 3 \end{matrix} \right] \)
\({ A }^{ 2 }=\left[ \begin{matrix} 3 & -3 \\ -3 & 3 \end{matrix} \right] \left[ \begin{matrix} 3 & -3 \\ -3 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 9+9 & -9-9 \\ -9-9 & 9+9 \end{matrix} \right] =\left[ \begin{matrix} 18 & -18 \\ -18 & 18 \end{matrix} \right] \)
Since \({ A }^{ 2 }=\lambda A  \Rightarrow \left[ \begin{matrix} 18 & -18 \\ -18 & 18 \end{matrix} \right] =\left[ \begin{matrix} 3\lambda & -3\lambda \\ -3\lambda & 3\lambda \end{matrix} \right] \)
\(\Rightarrow \ 18=3\lambda \ \Rightarrow \lambda =6\)

x + y + z = 10
Alternate Method :
\(\left[ \begin{matrix} x-y & 2y \\ 2y+z & x+y \end{matrix} \right] =\left[ \begin{matrix} 1 & 4 \\ 9 & 5 \end{matrix} \right]\)
By equating,   x - y = 1 ...(i)
and  2y = 4 ⇒  y = 2 .. (ii)
Put y = 2 in (i) ⇒  x - 2 = 1 ⇒  x = 3
2y + z = 9
⇒  2(2) + z = 9
⇒  z = 9 - 4 ⇒  z = 5
∴∴   x + y + z = 3 + 2 + 5 = 10

x - y = 8
Alternate Method :
\(2\left[ \begin{matrix} 3 & 4 \\ 5 & x \end{matrix} \right] =\left[ \begin{matrix} 1 & y \\ 10 & 5 \end{matrix} \right] =\left[ \begin{matrix} 7 & 0 \\ 10 & 5 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} 6 & 8 \\ 10 & 2x \end{matrix} \right] =\left[ \begin{matrix} 1 & y \\ 0 & 5 \end{matrix} \right] =\left[ \begin{matrix} 7 & 0 \\ 10 & 5 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} 7 & 8+y \\ 10 & 2x+5 \end{matrix} \right] =\left[ \begin{matrix} 7 & 0 \\ 10 & 5 \end{matrix} \right]\)
By equating,
8 + y = 0  ⇒  y = - 8
and 2x + 5 = 5 ⇒  x = 0
∴  x - y = 0 - (- 8)
= 8

x + y + z = 0
Alternate method :
\(\left[ \begin{matrix} xy & 4 \\ z+6 & x+y \end{matrix} \right] =\left[ \begin{matrix} 8 & w \\ 0 & 6 \end{matrix} \right] \)
By equating z + 6 = 0 and x + y = 6
⇒  z = - 6, x + y = 6
x + y + z = 6 - 6
⇒  x + y + z = 0

\([ac+bd]+[{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }]=[ac+bd+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }] \)

Matrix is skew symmetric, \(A=\left[ \begin{matrix} 0 & -1 & 2 \\ 1 & 0 & -3 \\ -2 & 3 & 0 \end{matrix} \right] A'=\left[ \begin{matrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ 2 & -3 & 0 \end{matrix} \right] =-\left[ \begin{matrix} 0 & -1 & 2 \\ 1 & 0 & -3 \\ -2 & 3 & 0 \end{matrix} \right] =-A \)
As A' = -A, so matrix A is skew-symmetric.

Possible orders of a matrix with 18 elements are
\(1\times 18,2\times 9,3\times 6,6\times 3,9\times 2,18\times 1.\)

Given \(A=\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix};kA\begin{bmatrix} k & -k \\ -k & k \end{bmatrix} \)
\({ A }^{ 2 }=\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}=\begin{bmatrix} 1+1 & -1-1 \\ -1-1 & 1+1 \end{bmatrix}\) [multiplying row by column]
\({ A }^{ 2 }=kA\Rightarrow \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \)
\(\Rightarrow k=2A\)
On comparing with Eq. (ii), we got k = 2

\({ A }^{ T }-{ B }^{ T }=\left[ \begin{matrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{matrix} \right] -\left[ \begin{matrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{matrix} \right] =\left[ \begin{matrix} 3+1 & 4-1 \\ -1-2 & 2-2 \\ 0-1 & 1-3 \end{matrix} \right] =\left[ \begin{matrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{matrix} \right] \)

\(\left[ \begin{matrix} x & +3y & y \\ 7 & -x & 4 \end{matrix} \right] =\begin{bmatrix} 4 & -1 \\ 0 & 4 \end{bmatrix}\)
\(\Rightarrow x+3y=4;y=-1;7-x=0 \)
\(\Rightarrow x=7,y=-1\)

We shall prove the result by using principle of mathematical induction
\(\mathrm{P}(n): \text { If } \mathrm{A}=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \text {, then } \mathrm{A}^n=\left[\begin{array}{cc} \cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta \end{array}\right], n \in \mathbf{N}\)
\(P(1): A=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \text {, so } A^1=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\)
Therefore, the result is true for n = 1.
Let the result be true for n = k. So
\(\mathrm{P}(k): \mathrm{A}=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \text {, then } \mathrm{A}^k=\left[\begin{array}{cc} \cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta \end{array}\right]\)
Now, we prove that the result holds for n = k +1
\( \mathrm{A}^{k+1} =\mathrm{A} \cdot \mathrm{A}^k=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc} \cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta \end{array}\right] \)
\(=\left[\begin{array}{cc} \cos \theta \cos k \theta-\sin \theta \sin k \theta & \cos \theta \sin k \theta+\sin \theta \cos k \theta \\ -\sin \theta \cos k \theta+\cos \theta \sin k \theta & -\sin \theta \sin k \theta+\cos \theta \cos k \theta \end{array}\right]\)
\(=\left[\begin{array}{cc} \cos (\theta+k \theta) & \sin (\theta+k \theta) \\ -\sin (\theta+k \theta) & \cos (\theta+k \theta) \end{array}\right]=\left[\begin{array}{cc} \cos (k+1) \theta & \sin (k+1) \theta \\ -\sin (k+1) \theta & \cos (k+1) \theta \end{array}\right]\)
Therefore, the result is true for n = k + 1. Thus by principle of mathematical induction, we have \(A^n=\left[\begin{array}{rr} \cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta \end{array}\right]\), holds for all natural numbers.

\(\left[ \left( \begin{matrix} 1 & \omega & { \omega }^{ 2 } \\ \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & \omega \end{matrix} \right) +\left( \begin{matrix} \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & \omega \\ \omega & { \omega }^{ 2 } & 1 \end{matrix} \right) \right] \left[ \begin{matrix} 1 \\ \omega \\ { \omega }^{ 2 } \end{matrix} \right] =\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\( \left[ \left( \begin{matrix} 1 & \omega & { \omega }^{ 2 } \\ \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & \omega \end{matrix} \right) +\left( \begin{matrix} \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & \omega \\ \omega & { \omega }^{ 2 } & 1 \end{matrix} \right) \right] \left[ \begin{matrix} 1 \\ \omega \\ { \omega }^{ 2 } \end{matrix} \right] =\left( \begin{matrix} 1+\omega & \omega +{ \omega }^{ 2 } & { \omega }^{ 2 }+1 \\ \omega +{ \omega }^{ 2 } & { \omega }^{ 2 }+1 & 1+\omega \\ { \omega }^{ 2 }+\omega & 1+{ \omega }^{ 2 } & \omega +1 \end{matrix} \right) \left[ \begin{matrix} 1 \\ \omega \\ { \omega }^{ 2 } \end{matrix} \right] \)
\(=\left( \begin{matrix} -{ \omega }^{ 2 } & -1 & -\omega \\ -1 & -\omega & -{ \omega }^{ 2 } \\ -1 & -\omega & -{ \omega }^{ 2 } \end{matrix} \right) \left[ \begin{matrix} 1 \\ \omega \\ { \omega }^{ 2 } \end{matrix} \right] \quad \quad \left[ \because \quad 1+\omega +{ \omega }^{ 2 }=0 \right] \)
\(=\left( \begin{matrix} -{ \omega }^{ 2 } & -\omega & -{ \omega }^{ 3 } \\ -1 & -{ \omega }^{ 2 } & -{ \omega }^{ 4 } \\ -1 & -{ \omega }^{ 2 } & -{ \omega }^{ 4 } \end{matrix} \right) =\left[ \begin{matrix} -{ \omega }^{ 2 } & -\omega & -1 \\ -1 & -{ \omega }^{ 2 } & -{ \omega } \\ -1 & -{ \omega }^{ 2 } & -{ \omega } \end{matrix} \right] \quad \left[ \because { \omega }^{ 3 }=1\quad \& \quad { \omega }^{ 4 }=\omega \quad \right] \)
 \(=\left[ \begin{matrix} -0 \\ -0 \\ -0 \end{matrix} \right] =\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \) 
which is true.

We have: \(A=\begin{bmatrix} 2 & 4 \\ 5 & 6 \end{bmatrix}.\)
\(\therefore\ A\prime =\begin{bmatrix} 2 & 5 \\ 4 & 6 \end{bmatrix}.\)
\(\therefore \ A+A\prime =\begin{bmatrix} 2 & 4 \\ 5 & 6 \end{bmatrix}+\begin{bmatrix} 2 & 5 \\ 4 & 6 \end{bmatrix}\)
\( =\begin{bmatrix} 2+2 & 4+5 \\ 5+4 & 6+6 \end{bmatrix}=\begin{bmatrix} 4 & 9 \\ 9 & 12 \end{bmatrix}...(1)\)
Now \((A+A\prime )\prime =\begin{bmatrix} 4 & 9 \\ 9 & 12 \end{bmatrix}=A+A\prime \)   [Using (1)]
Hence,   \(A+A\prime \)   is a symmetric matrix.

We have: \(x\begin{bmatrix} 2x & 2 \\ 3 & x \end{bmatrix}+2\begin{bmatrix} 8 & 5x \\ 4 & 4x \end{bmatrix}=2\begin{bmatrix} { x }^{ 2 }+8 & 24 \\ 10 & 6x \end{bmatrix}.\) 
\(x\begin{bmatrix} 2x & 2 \\ 3 & x \end{bmatrix}+2\begin{bmatrix} 8 & 5x \\ 4 & 4x \end{bmatrix}=2\begin{bmatrix} { x }^{ 2 }+8 & 24 \\ 10 & 6x \end{bmatrix}.\)
\(\Rightarrow \ \begin{bmatrix} { 2x }^{ 2 } & 2x \\ 3x & { x }^{ 2 } \end{bmatrix}+\begin{bmatrix} 16 & 10x \\ 8 & 8x \end{bmatrix}=\begin{bmatrix} { 2x }^{ 2 }+16 & 48 \\ 20 & 12x \end{bmatrix}\)
\(\Rightarrow \ \begin{bmatrix} { 2x }^{ 2 }+16 & 12x \\ 3x+8 & { x }^{ 2 }+8x \end{bmatrix}=\begin{bmatrix} { 2x }^{ 2 }+16 & 48 \\ 20 & 12x \end{bmatrix}\)
Comparing, \(12x=48,\quad 3x+8=20,\quad and\quad { x }^{ 2 }+8x=12x.\) 
All these give \(x=4\)

(i) In the given matrix, the number of rows is 3 and the number of columns is 4. Therefore, the order of the matrix is 3 x 4.
(ii) Since the order of the matrix is 3 x 4, there are 3 x 4 = 12 elements in it.
(iii) \({ a }_{ 13 }=19,{ a }_{ 21 }=35,{ a }_{ 33 }=-5,{ a }_{ 24 }=12,{ a }_{ 23 }=\frac { 5 }{ 2 } .\)

x +y + z = 9,
x + z = 5,
y + z = 7, on solving we get
x = 2, y = 4, z = 3

We shall prove the result by using principle of mathematical induction.
Let \(P\left( n \right) { :A }^{ n }=\left[ \begin{matrix} { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \\ { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \\ { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \end{matrix} \right] \)
Now, \(P\left( 1 \right) { :A }^{ 1 }=\left[ \begin{matrix} { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \\ { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \\ { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \end{matrix} \right] =\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] \)
The result is true for n = 1.
Let the result be true for n = k.
So, \({ A }^{ k }=\left[ \begin{matrix} { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \end{matrix} \right] \)
Now, we prove that P(k + 1) is true.
Now, A^k+1 = A. A^k
\(=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \\ { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \\ { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \\ { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \\ { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \end{matrix} \right] \)
= A^k+1
Hence, it is true n = k + 1.
Hence, by principle of mathematical induction P(n) is true for all \(n\epsilon N\)

Let the cost of 'A' variety pen be Rs. x, the cost of 'B' variety pen be Rs. y and the cost of 'C' variety pen be Rs. z.
According to the question,
x + y + z = 21
4x + 3y + 2z = 60
and 6x + 2y + 3z = 70
It can be written as,
\(\left[ \begin{matrix} 1 & 1 & 1 \\ 4 & 3 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 21 \\ 60 \\ 70 \end{matrix} \right] \)
i.e.,  AX = B,
where \(A=\left[ \begin{matrix} 1 & 1 & 1 \\ 4 & 3 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)
\(X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)
and \(B=\left[ \begin{matrix} 21 \\ 60 \\ 70 \end{matrix} \right] \)
\(\left| A \right| =\left| \begin{matrix} 1 & 1 & 1 \\ 4 & 3 & 2 \\ 6 & 2 & 3 \end{matrix} \right| \)
= 1(9 - 4) - 1(12 - 12) + 1(8 - 18)
= 5 - 10 = - 5 \(\neq \) 0.
\(\Rightarrow\)  A^-1 exists.
C₁₁ = (- 1)² (9 - 4) = 5
C₁₂ = (- 1)³ (12-12) = 0
C₁₃ = (- 1)⁴ (8 - 18) = - 10
C₂₁ = - (3 - 2) = - 1
C₂₂ = - 3
C₂₃ = - (2 - 6) = 4
C₃₁ = (2 - 3) = - 1
C₃₂ = - (2 - 4) = 2
C₃₃ = 3 - 4 = - 1
\(adjA=\left[ \begin{matrix} 5 & -1 & -1 \\ 0 & -3 & 2 \\ -10 & 4 & -1 \end{matrix} \right] \)
\({ A }^{ -1 }=\frac { 1 }{ \left| A \right| } adjA\)
Also  AX = B
Premultiplying by A^-1
\(\Rightarrow\)   X = A^-1B
\(\Rightarrow\)   X = \(\frac { 1 }{ \left| A \right| } \left( adjA \right) B\)
\(=\frac { 1 }{ \left| A \right| } \left[ \begin{matrix} 5 & -1 & -1 \\ 0 & -3 & 2 \\ -10 & 4 & -1 \end{matrix} \right] \left[ \begin{matrix} 21 \\ 60 \\ 70 \end{matrix} \right] \)
\(=\frac { -1 }{ 5 } \left[ \begin{matrix} 105-60-70 \\ -180+140 \\ -210+240-70 \end{matrix} \right] \)
\(=\frac { -1 }{ 5 } \left[ \begin{matrix} -25 \\ -40 \\ -40 \end{matrix} \right] \)
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 8 \\ 8 \end{matrix} \right] \)
\(\therefore\)   x = 5, y = 8, z = 8
\(\therefore\)  cost of 'A' variety pen = Rs. 5
'B' variety pen = Rs. 8
'C' variety pen = Rs. 8

\(3A-5B=3\left[ \begin{matrix} \frac { 2 }{ 3 } & 1 & \frac { 5 }{ 3 } \\ \frac { 1 }{ 3 } & \frac { 2 }{ 3 } & \frac { 4 }{ 3 } \\ \frac { 7 }{ 3 } & 2 & \frac { 2 }{ 3 } \end{matrix} \right] -5\left[ \begin{matrix} \frac { 2 }{ 3 } & \frac { 3 }{ 5 } & 1 \\ \frac { 1 }{ 5 } & \frac { 2 }{ 5 } & \frac { 4 }{ 5 } \\ \frac { 7 }{ 5 } & \frac { 6 }{ 5 } & \frac { 2 }{ 5 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{matrix} \right] -\left[ \begin{matrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 2-2 & 3-3 & 5-5 \\ 1-1 & 2-2 & 4-4 \\ 7-7 & 6-6 & 2-2 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] .\)