Solution:

Here,
A = \(\begin{bmatrix} 3 & 4 \\ 6 & 8 \end{bmatrix}\)
B = \(\begin{bmatrix} 2 & 5 \\ 4 & 3 \end{bmatrix}\)
(A - B)T = ?
Now,
or, (A - B)T= \(\left(\begin{bmatrix} 3 & 4 \\ 6 & 8 \end{bmatrix} - \begin{bmatrix} 2 & 5 \\ 4 & 3 \end{bmatrix}\right)\)T
or, (A - B)T= \(\left(\begin{bmatrix} 3-2 & 4-5 \\ 6-4 & 8-3 \end{bmatrix}\right)\)T
or, (A - B)T= \(\left(\begin{bmatrix} 1& -1\\ 2& 5\end{bmatrix}\right)\)T
\(\therefore\) (A - B)T= \(\begin{bmatrix} 1& 2\\ -1& 5\end{bmatrix}\)



Solution:

Given,
A = \(\begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix}\)
B = \(\begin{bmatrix} 6 & 9 \\ 12 & 15 \end{bmatrix}\)
\(\frac{1}{2}\)A + \(\frac{1}{3}\)B = ?
Now,
or, \(\frac{1}{2}\)A + \(\frac{1}{3}\)B = \(\frac{1}{2}\)\(\begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix}\) + \(\frac{1}{3}\)\(\begin{bmatrix} 6 & 9 \\ 12 & 15 \end{bmatrix}\)
= \(\begin{bmatrix} 2 \times\ \frac{1}{2} & 4 \times\ \frac{1}{2} \\ 6 \times\ \frac{1}{2} & 8 \times\ \frac{1}{2} \end{bmatrix}\) +\(\begin{bmatrix} 6 \times\ \frac{1}{3} & 9 \times\ \frac{1}{3} \\ 12 \times\ \frac{1}{3} & 15 \times\ \frac{1}{3} \end{bmatrix}\)
= \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\) + \(\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}\)
= \(\begin{bmatrix} 3 & 5 \\ 7 & 9 \end{bmatrix}\)

Solution:

Given,
A = \(\begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix}\)
B = \(\begin{bmatrix} 6 & 9 \\ 12 & 15 \end{bmatrix}\)
2A + 3B = ?
Now,
or, 2A + 3B = 2\(\begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix}\) + 3\(\begin{bmatrix} 6 & 9 \\ 12 & 15 \end{bmatrix}\)
or, 2A + 3B = \(\begin{bmatrix} 2 \times\ 2 & 4 \times\ 2 \\ 6 \times\ 2 & 8 \times\ 2 \end{bmatrix}\) + \(\begin{bmatrix} 6 \times\ 3 & 9 \times\ 3 \\ 12 \times\ 3 & 15 \times\ 3 \end{bmatrix}\)
or, 2A + 3B = \(\begin{bmatrix} 4 & 8 \\ 12 & 16 \end{bmatrix}\) + \(\begin{bmatrix} 18 & 27 \\ 36 & 45 \end{bmatrix}\)
or, 2A + 3B = \(\begin{bmatrix} 4+18 & 8+27 \\ 12+36 & 16+45 \end{bmatrix}\)
\(\therefore\) 2A + 3B = \(\begin{bmatrix} 22 & 35 \\ 48 & 61 \end{bmatrix}\)

Solution:

Given,
Equating corresponding value
or, 8x = 16
or, x = \(\frac{16}{8}\)
\(\therefore\) x = 2
Then,
or, x + y = 5
or, 2 + y = 5
or, y = 5 - 2
\(\therefore\) y = 3

So, the value of x is 2 and y is 3.

Solution:

Here,
x - y = 4 ....................(i)
x + y = 8 ....................(ii)
Adding equation (i) and (ii), we get,
x - y = 4
\(\underline{x + y = 8}\)
2x = 12
x = \(\frac{12}{2}\)
\(\therefore\) x = 6
Now,
Substituting the value of x in equation (ii)
or, x + y = 8
or, 6 + y = 8
or, y = 8 - 6
\(\therefore\) y = 2
So, the value of x is 6 and y is 2.

Solution:

Here,
A = \(\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\)
B = \(\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}\)
C = \(\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\)
A + B - 2C = ?
Now,
or, A + B - 2C = \(\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\) +\(\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}\) - 2\(\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\)
or, A + B - 2C = \(\begin{bmatrix} 3+2 & 0+3 \\ 0+4 & 3+5 \end{bmatrix}\) - \(\begin{bmatrix} 2\times2 & 3\times2 \\ 1\times2 & 2\times2 \end{bmatrix}\)
or, A + B - 2C = \(\begin{bmatrix} 5 & 3\\ 4 & 8 \end{bmatrix}\) -\(\begin{bmatrix} 4 & 6 \\ 2 & 4 \end{bmatrix}\)
or, A + B - 2C = \(\begin{bmatrix} 5-4 & 3-6 \\ 4-2 & 8-4 \end{bmatrix}\)
\(\therefore\) A + B - 2C = \(\begin{bmatrix} 1 & -3 \\ 2 & 4 \end{bmatrix}\)

Solution:

Given,
A = \(\begin{bmatrix} 2 & 5 \\ 3 & 6 \\ 8 & 10 \end{bmatrix}\)
B = \(\begin{bmatrix} 1 & 4 \\ 2 & 10 \\ 8 & 12 \end{bmatrix}\)
A - B = ?
Now,
or, A - B = \(\begin{bmatrix} 2 & 5 \\ 3 & 6 \\ 8 & 10 \end{bmatrix}\) -\(\begin{bmatrix} 1 & 4 \\ 2 & 10 \\ 8 & 12 \end{bmatrix}\)
or, A - B = \(\begin{bmatrix} 2-1 & 5-4 \\ 3-2 & 6-10 \\ 8-8 & 10-12 \end{bmatrix}\)
\(\therefore\) A - B = \(\begin{bmatrix} 1 & 1 \\ 1 & -4 \\ 0 & -2 \end{bmatrix}\)

Solution:

Here,
3x - 4y = 1 .................... (i)
28 = 14y .................... (ii)
Equating the corresponding value
or, 14y = 28
or, y = \(\frac{28}{14}\)
or, y = 2
Now,
Substituting the value of y in equation (i)
or, 3x - 4y = 1
or, 3x - 4 \(\times\) 2 = 1
or, 3x - 8 = 1
or, 3x = 1 + 8
or, 3x = 9
or, x = 3
\(\therefore\) The value of x is 3 and y is 2.

Solution:

Here,
X = \(\begin{bmatrix} 4 & 5 \\ 6 & 8 \end{bmatrix}\)
Y = \(\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}\)
X + Y = ?
Now,
or, X + Y = \(\begin{bmatrix} 4 & 5 \\ 6 & 8 \end{bmatrix}\) +\(\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}\)
or, X + Y = \(\begin{bmatrix} 4+2 & 5+3 \\ 6+4 & 8+5 \end{bmatrix}\)
\(\therefore\) X + Y = \(\begin{bmatrix} 6 & 8 \\ 10 & 13 \end{bmatrix}\)

Solution:

Here,
A = \(\begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix}\)
B = \(\begin{bmatrix} 6 & 9 \\ 12 & 15 \end{bmatrix}\)
4A - 3B = ?
Now,
or, 4A - 3B = 4\(\begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix}\) - 3\(\begin{bmatrix} 6 & 9 \\ 12 & 15 \end{bmatrix}\)
or, 4A - 3B = \(\begin{bmatrix} 8 & 16 \\ 24 & 32 \end{bmatrix}\) - \(\begin{bmatrix} 18 & 27 \\ 36 & 45 \end{bmatrix}\)
or, 4A - 3B = \(\begin{bmatrix} 8-18 & 16-27 \\ 24-36 & 32-45 \end{bmatrix}\)
\(\therefore\) 4A - 3B = \(\begin{bmatrix} -10& -11\\ -12& -13\end{bmatrix}\)


Solution:

Given:
\(\begin{bmatrix} 2x + 3y & 8 \\ 6 & 4 \end{bmatrix}\) = \(\begin{bmatrix} 5 & 8 \\ y + 5 & 4 \end{bmatrix}\)
Now,
2x + 3y = 5 ...............(i)
y + 5 = 6 .................(ii)
Equating the corresponding value
or, y + 5 = 6
or, y - 6 -5
or, y = 1
Again,
Substituting the value of y in equation (i)
or, 2x + 3y = 5
or, 2x + 3 \(\times\) 1 = 5
or, 2x = 5 - 3
or, 2x = 2
or, x = 1
\(\therefore\) The value of x is 1 and y is 1.

Solution:

Given,
A = \(\begin{bmatrix} 2 & 3 & 4 \\ 5 & 6 & 7 \end{bmatrix}\)
B = \(\begin{bmatrix} 2 & 3 \\ 0 & 8 \\ 9 & 10 \end{bmatrix}\)
C = \(\begin{bmatrix} 3 & -4 & 6 \\ 7 & 2 & 5 \\ 1 & 9 & 8 \end{bmatrix}\)
A + BT = ?
Now,
or, B = \(\begin{bmatrix} 2 & 3 \\ 0 & 8 \\ 9 & 10 \end{bmatrix}\)
\(\therefore\) BT = \(\begin{bmatrix} 2 & 0 & 9 \\ 3 & 8 & 10 \end{bmatrix}\)

Then,
or, A + BT = \(\begin{bmatrix} 2 & 3 & 4 \\ 5 & 6 & 7 \end{bmatrix}\) + \(\begin{bmatrix} 2 & 0 & 9 \\ 3 & 8 & 10 \end{bmatrix}\)
or, A + BT = \(\begin{bmatrix} 2+2 & 3+0 & 4+9 \\ 5+3 & 6+8 & 7+10 \end{bmatrix}\)
\(\therefore\), A + BT = \(\begin{bmatrix} 4 & 3 & 13 \\ 8 & 14 & 17 \end{bmatrix}\)