If `[(1,lamda,1)][(1,3,2),(0,5,1),(0,3,2)][(lamda),(1),(-2)]=O` then `lamda=`

Options

A. `-1`
B. `-1//2`
C. `1//2`
D. `1`

Answer

To solve the equation given by the matrix multiplication, we will follow these steps:

1. **Define the Matrices**:
We have three matrices:
- Matrix A: \( A = \begin{pmatrix} 1 & \lambda & 1 \end{pmatrix} \) (1x3 matrix)
- Matrix B: \( B = \begin{pmatrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{pmatrix} \) (3x3 matrix)
- Matrix C: \( C = \begin{pmatrix} \lambda \\ 1 \\ -2 \end{pmatrix} \) (3x1 matrix)

We need to find \( \lambda \) such that \( A \cdot B \cdot C = O \) (the zero matrix).

2. **Multiply Matrices A and B**:
First, we multiply matrix A by matrix B:
\[ A \cdot B = \begin{pmatrix} 1 & \lambda & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{pmatrix} \]

The resulting matrix will be:
\[ = \begin{pmatrix} 1 \cdot 1 + \lambda \cdot 0 + 1 \cdot 0 & 1 \cdot 3 + \lambda \cdot 5 + 1 \cdot 3 & 1 \cdot 2 + \lambda \cdot 1 + 1 \cdot 2 \end{pmatrix} \]
Simplifying this gives:
\[ = \begin{pmatrix} 1 & 3 + 5\lambda + 3 & 2 + \lambda + 2 \end{pmatrix} = \begin{pmatrix} 1 & 5\lambda + 6 & \lambda + 4 \end{pmatrix} \]

3. **Multiply the Result by Matrix C**:
Now, we multiply the result by matrix C:
\[ \begin{pmatrix} 1 & 5\lambda + 6 & \lambda + 4 \end{pmatrix} \cdot \begin{pmatrix} \lambda \\ 1 \\ -2 \end{pmatrix} \]
This results in:
\[ = 1 \cdot \lambda + (5\lambda + 6) \cdot 1 + (\lambda + 4) \cdot (-2) \]
Simplifying this gives:
\[ = \lambda + 5\lambda + 6 - 2\lambda - 8 = (1 + 5 - 2)\lambda + (6 - 8) = 4\lambda - 2 \]

4. **Set the Result Equal to Zero**:
We set the result equal to zero:
\[ 4\lambda - 2 = 0 \]

5. **Solve for \( \lambda \)**:
Solving for \( \lambda \):
\[ 4\lambda = 2 \implies \lambda = \frac{2}{4} = \frac{1}{2} \]

Thus, the value of \( \lambda \) is \( \frac{1}{2} \).