Let `a=[(1,-1),(2, -1)] and B=[(a,1),(b,-1)]` are two matrices. If `(A+B)^(2)=A^(2)+B^(2)`, then the value of `3a+4b` is equal to

Options

A. 15
B. 17
C. 19
D. 21

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that the equation \( (A + B)^2 = A^2 + B^2 \) holds true. Let's break this down step by step.

### Step 1: Define the Matrices
We have:
\[ A = \begin{pmatrix} 1 & -1 \\ 2 & -1 \end{pmatrix}, \quad B = \begin{pmatrix} a & 1 \\ b & -1 \end{pmatrix} \]

### Step 2: Calculate \( A + B \)
To find \( A + B \):
\[ A + B = \begin{pmatrix} 1 + a & -1 + 1 \\ 2 + b & -1 - 1 \end{pmatrix} = \begin{pmatrix} 1 + a & 0 \\ 2 + b & -2 \end{pmatrix} \]

### Step 3: Calculate \( (A + B)^2 \)
Now we compute \( (A + B)^2 \):
\[ (A + B)^2 = (A + B)(A + B) = \begin{pmatrix} 1 + a & 0 \\ 2 + b & -2 \end{pmatrix} \begin{pmatrix} 1 + a & 0 \\ 2 + b & -2 \end{pmatrix} \]
Calculating the product:
- First row, first column:
\[ (1 + a)(1 + a) + 0(2 + b) = (1 + a)^2 \]
- First row, second column:
\[ (1 + a)(0) + 0(-2) = 0 \]
- Second row, first column:
\[ (2 + b)(1 + a) + (-2)(2 + b) = (2 + b)(1 + a) - 2(2 + b) = (2 + b)(1 + a - 2) = (2 + b)(a - 1) \]
- Second row, second column:
\[ (2 + b)(0) + (-2)(-2) = 4 \]

Thus,
\[ (A + B)^2 = \begin{pmatrix} (1 + a)^2 & 0 \\ (2 + b)(a - 1) & 4 \end{pmatrix} \]

### Step 4: Calculate \( A^2 \) and \( B^2 \)
Now we calculate \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & -1 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 2 & -1 \end{pmatrix} \]
Calculating the product:
- First row, first column:
\[ 1 \cdot 1 + (-1) \cdot 2 = 1 - 2 = -1 \]
- First row, second column:
\[ 1 \cdot (-1) + (-1) \cdot (-1) = -1 + 1 = 0 \]
- Second row, first column:
\[ 2 \cdot 1 + (-1) \cdot 2 = 2 - 2 = 0 \]
- Second row, second column:
\[ 2 \cdot (-1) + (-1) \cdot (-1) = -2 + 1 = -1 \]

Thus,
\[ A^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \]

Next, we calculate \( B^2 \):
\[ B^2 = B \cdot B = \begin{pmatrix} a & 1 \\ b & -1 \end{pmatrix} \begin{pmatrix} a & 1 \\ b & -1 \end{pmatrix} \]
Calculating the product:
- First row, first column:
\[ a^2 + 1 \cdot b = a^2 + b \]
- First row, second column:
\[ a \cdot 1 + 1 \cdot (-1) = a - 1 \]
- Second row, first column:
\[ b \cdot a + (-1) \cdot b = ab - b \]
- Second row, second column:
\[ b \cdot 1 + (-1)(-1) = b + 1 \]

Thus,
\[ B^2 = \begin{pmatrix} a^2 + b & a - 1 \\ ab - b & b + 1 \end{pmatrix} \]

### Step 5: Set Up the Equation
Now we set up the equation \( (A + B)^2 = A^2 + B^2 \):
\[ \begin{pmatrix} (1 + a)^2 & 0 \\ (2 + b)(a - 1) & 4 \end{pmatrix} = \begin{pmatrix} -1 + (a^2 + b) & a - 1 \\ ab - b & -1 + (b + 1) \end{pmatrix} \]

### Step 6: Equate the Matrices
From the first row, first column:
\[ (1 + a)^2 = -1 + a^2 + b \quad \text{(1)} \]
From the second row, second column:
\[ 4 = b \quad \text{(2)} \]
From the second row, first column:
\[ (2 + b)(a - 1) = ab - b \quad \text{(3)} \]

### Step 7: Solve the Equations
From equation (2), we have:
\[ b = 4 \]

Substituting \( b = 4 \) into equation (1):
\[ (1 + a)^2 = -1 + a^2 + 4 \]
\[ (1 + a)^2 = a^2 + 3 \]
Expanding the left side:
\[ 1 + 2a + a^2 = a^2 + 3 \]
Cancelling \( a^2 \) from both sides:
\[ 1 + 2a = 3 \]
\[ 2a = 2 \quad \Rightarrow \quad a = 1 \]

### Step 8: Calculate \( 3a + 4b \)
Now we can find \( 3a + 4b \):
\[ 3a + 4b = 3(1) + 4(4) = 3 + 16 = 19 \]

### Final Answer
The value of \( 3a + 4b \) is \( \boxed{19} \).