Question

plus two base question, please help me​

Answer 1

I've attached the images.

Answer 2

★ Solution :-

Given ,

• $\left[\begin{array}{cc} \sf 2a+b&\sf a-2b\\\sf 5c-d&\sf 4c+3d\end{array}\right]=\left[\begin{array}{cc}\sf 4 &\sf -3\\\sf 11&\sf 24\end{array}\right]$

We need to find

• a , b , c , d

Here , by comparing both the matrices

→ 2a + b = 4

→ 2a = b - 4

→ a = $\sf \dfrac{b-4}{2}$

Assuming as eq 1

$\rule{200}2$

→ a - 2b = - 3

→ a = - 3 + 2b

Assuming as eq 2

Now , from eq 1 ,2

→ a = - 3 + 2b = $\sf\dfrac{b-4}{2}$

→ 2(-3 + 2b) = b - 4

→ - 6 + 2b - b = - 4

→ b = - 4 + 6

→ b = 2

Now , substituting b = 2 in eq 2

→ a = - 3 + 2(2)

→ a = - 3 + 4

→ a = 1

$\rule{200}2$

→ 5c - d = 11

→ 5c - 11 = d

Assuming as eq 2

Now , substituting the value of d in 4c + 3d = 24

→ 4c + 3(5c - 11) = 24

→ 4c + 15c - 33 = 24

→ 19c = 57

→ c = $\sf \dfrac{57}{19}$

→ c = 3

Now , substuting value of c in eq 2

→ 5(3) - 11 = d

→ 15 - 11 = d

→ d = 4

Hence ,

• a = 1
• b = 2
• c = 3
• d = 4