If `[(2a+b,a-2b),(5c-d,4c+3d)]=[(4,-3),(11,24)]` , then value of `a+b-c +2d` is :

Options

A. 8
B. 10
C. 4
D. `-8`

Answer

To solve the problem, we need to compare the components of the given matrices and solve for the variables \(a\), \(b\), \(c\), and \(d\).

Given:
\[ \begin{pmatrix} 2a + b & a - 2b \\ 5c - d & 4c + 3d \end{pmatrix} = \begin{pmatrix} 4 & -3 \\ 11 & 24 \end{pmatrix} \]

We can equate the corresponding elements of the matrices:

1. \(2a + b = 4\) (Equation 1)
2. \(a - 2b = -3\) (Equation 2)
3. \(5c - d = 11\) (Equation 3)
4. \(4c + 3d = 24\) (Equation 4)

### Step 1: Solve for \(a\) and \(b\)

From Equation 1:
\[ 2a + b = 4 \quad \text{(1)} \]

From Equation 2:
\[ a - 2b = -3 \quad \text{(2)} \]

Now, we can express \(b\) from Equation 1:
\[ b = 4 - 2a \quad \text{(5)} \]

Substituting (5) into (2):
\[ a - 2(4 - 2a) = -3 \]
\[ a - 8 + 4a = -3 \]
\[ 5a - 8 = -3 \]
\[ 5a = 5 \]
\[ a = 1 \]

Now substituting \(a = 1\) back into (5) to find \(b\):
\[ b = 4 - 2(1) = 4 - 2 = 2 \]

### Step 2: Solve for \(c\) and \(d\)

Now we will solve for \(c\) and \(d\) using Equations 3 and 4.

From Equation 3:
\[ 5c - d = 11 \quad \text{(3)} \]

From Equation 4:
\[ 4c + 3d = 24 \quad \text{(4)} \]

We can express \(d\) from Equation 3:
\[ d = 5c - 11 \quad \text{(6)} \]

Substituting (6) into (4):
\[ 4c + 3(5c - 11) = 24 \]
\[ 4c + 15c - 33 = 24 \]
\[ 19c - 33 = 24 \]
\[ 19c = 57 \]
\[ c = 3 \]

Now substituting \(c = 3\) back into (6) to find \(d\):
\[ d = 5(3) - 11 = 15 - 11 = 4 \]

### Step 3: Calculate \(a + b - c + 2d\)

Now we have:
- \(a = 1\)
- \(b = 2\)
- \(c = 3\)
- \(d = 4\)

We need to find:
\[ a + b - c + 2d \]
Substituting the values:
\[ 1 + 2 - 3 + 2(4) = 1 + 2 - 3 + 8 \]
\[ = 3 - 3 + 8 = 8 \]

Thus, the final answer is:
\[ \boxed{8} \]