Since A, B are of the same order 3 x 3, subtraction of 4A and 3B is defined.
4A - 3B = \(4\left[ \begin{matrix} 5 & 4 & -2 \\ \frac { 1 }{ 2 } & \frac { 3 }{ 4 } & \sqrt { 2 } \\ 1 & 9 & 4 \end{matrix} \right] -3\left[ \begin{matrix} -7 & 4 & -3 \\ \frac { 1 }{ 4 } & \frac { 7 }{ 2 } & 3 \\ 5 & -6 & 9 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 20 & 16 & -8 \\ 2 & 3 & 4\sqrt { 2 } \\ 4 & 36 & 16 \end{matrix} \right] +\left[ \begin{matrix} 21 & -12 & 9 \\ -\frac { 3 }{ 4 } & -\frac { 21 }{ 2 } & -9 \\ -15 & 18 & -27 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 41 & 4 & 1 \\ \frac { 5 }{ 4 } & -\frac { 15 }{ 2 } & 4\sqrt { 2 } -9 \\ -11 & 54 & -11 \end{matrix} \right] \)

Since A and B have same order 3 x 3, 2A + B is defined.
We have 2A + B = \(2\left[ \begin{matrix} 7 & 8 & 6 \\ 1 & 3 & 9 \\ -4 & 3 & -1 \end{matrix} \right] +\left[ \begin{matrix} 4 & 11 & -3 \\ -1 & 2 & 4 \\ 7 & 5 & 0 \end{matrix} \right] =\left[ \begin{matrix} 14 & 16 & 12 \\ 2 & 6 & 18 \\ -8 & 6 & -2 \end{matrix} \right] +\left[ \begin{matrix} 4 & 11 & -3 \\ -1 & 2 & 4 \\ 7 & 5 & 0 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 18 & 27 & 9 \\ 1 & 8 & 22 \\ -1 & 11 & -2 \end{matrix} \right] \)

It is not possible to add A and B because they have different orders.

If A = \(\left[ \begin{matrix} \sqrt { 7 } & -3 \\ -\sqrt { 5 } & 2 \\ \sqrt { 3 } & -5 \end{matrix} \right] \)
-A\(\left[ \begin{matrix}- \sqrt { 7 } &3 \\ \sqrt { 5 } & -2 \\- \sqrt { 3 } & 5 \end{matrix} \right] \)
Transpose od -A = (-A)^T = \(\left[ \begin{matrix} -\sqrt { 7 } & +\sqrt { 5 } & -\sqrt { 3 } \\ +3 & -2 &+ 5 \end{matrix} \right] \)

The information is represented in the form of a 3 x 2 matrix as follows
A = \(\left( \begin{matrix} 23 & 18 \\ 47 & 36 \\ 15 & 16 \end{matrix} \right) \)
The entry in the second row and first column represent that there are 47 men workers in factory II.

3x²+ kx + 81 = 0
Let the roots be ∝ and ∝2
∝ + ∝2 = \(\frac{-k}{3}\) ....(1)
∝ + ∝2 = \(\frac{81}{3}\)
⇒ k³ = 27
⇒ k³ = 3...(3)
3 + 32 = \(\frac{-k}{3}\)
⇒ (3 + 9)3 = -k
⇒ k = -36.

x² - x - 20 = 0
Here, a = 1, b = -1, c = -20
Now, Δ = b² - 4ac
Δ = (-1)² - 4(1) (-20) = 81
Here, Δ = 81 > 0. So, the equation will have real and unequal roots

9x² - 12x + 4 = 0
\(\frac { { 9x }^{ 2 } }{ 9 } -\frac { 12 }{ 9 } x+\frac { 4 }{ 9 } =0\)
\({ x }^{ 2 }-\frac { 4 }{ 3 } x=\frac { -4 }{ 9 } \)
\({ x }^{ 2 }-\frac { 4 }{ 3 } x+{ \left( \frac { -2 }{ 3 } \right) }^{ 2 }=\frac { -4 }{ 9 } +{ \left( \frac { -2 }{ 3 } \right) }^{ 2 }\)
\({ \left( x-\frac { 2 }{ 3 } \right) }^{ 2 }=\frac { -4 }{ 9 } +\frac { 4 }{ 9 } =0\)
\(x-\frac { 2 }{ 3 } =0,x-\frac { 2 }{ 3 } =0\)
\(x=\frac { 2 }{ 3 } =0,x-\frac { 2 }{ 3 } =0\)
x = \(\frac { 2 }{ 3 } ,\frac { 2 }{ 3 } \)

7x² + ax + 2 = 0
\(x=\frac { -a\pm \sqrt { { a }^{ 2 }-56 } }{ 2\times 7 } \)
\(\alpha =\frac { -a\pm \sqrt { { a }^{ 2 }-56 } }{ 14 } ,\beta =\frac { -a-\sqrt { { a }^{ 2 }-56 } }{ 14 } \)
\(\beta -\alpha =\frac { -a-\sqrt { { a }^{ 2 }-56 } +a-\sqrt { { a }^{ 2 }-56 } }{ 14 } \)
\(=\frac { -2\sqrt { { a }^{ 2 }-56 } }{ 14 } =\frac { -13 }{ 7 } \) (given)
\(\Rightarrow \frac { -\sqrt { { a }^{ 2 }-56 } }{ 7 } =\frac { -13 }{ 7 } \)
\(\Rightarrow -\sqrt { { a }^{ 2 }-56 } =-13\)
Squaring on both sides.
a²-56 =169
a² = 225
a = 土15

Let a number be x.
Its reciprocal is \(\frac{1}{x}\)
\(x-\frac { 1 }{ x } =\frac { 24 }{ 5 } \)
\(\frac { { x }^{ 2 }-1 }{ x } =\frac { 24 }{ 5 } \)
5x²-5-24x = 0 ⇒ 5x²-24x-5 = 0
5x²-25x+x-5 = 0
5x(x -5)+1(x-5) = 0
(5x+1)(x-5) = 0
x = \(\frac{1}{5},5\)
∴ The number is \(\frac{-1}{5}\) or 5.