The given matrix equation is:
\[\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \cdot A \cdot \begin{pmatrix} -3 & 2 \\ 5 & -3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.\]
Let \[A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}.\] Substituting A into the equation, we compute step-by-step.
Step 1: Simplify the right product \[A \cdot \begin{pmatrix} -3 & 2 \\ 5 & -3 \end{pmatrix}.\]
\[A \cdot \begin{pmatrix} -3 & 2 \\ 5 & -3 \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} -3 & 2 \\ 5 & -3 \end{pmatrix}.\]
Multiply the matrices:
\[\begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} -3 & 2 \\ 5 & -3 \end{pmatrix} = \begin{pmatrix} -3a + 5b & 2a - 3b \\ -3c + 5d & 2c - 3d \end{pmatrix}.\]
Step 2: Multiply by \[\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}.\]
Now multiply:
\[\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \cdot \begin{pmatrix} -3a + 5b & 2a - 3b \\ -3c + 5d & 2c - 3d \end{pmatrix}.\]
Perform the matrix multiplication:
\[\begin{pmatrix} 2(-3a + 5b) + 1(-3c + 5d) & 2(2a - 3b) + 1(2c - 3d) \\ 3(-3a + 5b) + 2(-3c + 5d) & 3(2a - 3b) + 2(2c - 3d) \end{pmatrix}.\]
Simplify each term: - First row, first column:
\[2(-3a + 5b) + (-3c + 5d) = -6a + 10b - 3c + 5d.\]
- First row, second column:
\[2(2a - 3b) + (2c - 3d) = 4a - 6b + 2c - 3d.\]
- Second row, first column:
\[3(-3a + 5b) + 2(-3c + 5d) = -9a + 15b - 6c + 10d.\]
- Second row, second column:
\[3(2a - 3b) + 2(2c - 3d) = 6a - 9b + 4c - 6d.\]
Thus, the resulting matrix is:
\[\begin{pmatrix} -6a + 10b - 3c + 5d & 4a - 6b + 2c - 3d \\ -9a + 15b - 6c + 10d & 6a - 9b + 4c - 6d \end{pmatrix}.\]
Step 3: Equate to identity matrix.
We equate:
\[\begin{pmatrix} -6a + 10b - 3c + 5d & 4a - 6b + 2c - 3d \\ -9a + 15b - 6c + 10d & 6a - 9b + 4c - 6d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.\]
From this, solve the system of equations:
Solve this system, and you find:
\[A = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}.\]
Conclusion: The matrix A is:
\[\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}.\]
Match List-I with List-II
List-I (Matrix) | List-II (Inverse of the Matrix) |
---|---|
(A) \(\begin{bmatrix} 1 & 7 \\ 4 & -2 \end{bmatrix}\) | (I) \(\begin{bmatrix} \tfrac{2}{15} & \tfrac{1}{10} \\[6pt] -\tfrac{1}{15} & \tfrac{1}{5} \end{bmatrix}\) |
(B) \(\begin{bmatrix} 6 & -3 \\ 2 & 4 \end{bmatrix}\) | (II) \(\begin{bmatrix} \tfrac{1}{5} & -\tfrac{2}{15} \\[6pt] -\tfrac{1}{10} & \tfrac{7}{30} \end{bmatrix}\) |
(C) \(\begin{bmatrix} 5 & 2 \\ -5 & 4 \end{bmatrix}\) | (III) \(\begin{bmatrix} \tfrac{1}{15} & \tfrac{7}{30} \\[6pt] \tfrac{2}{15} & -\tfrac{1}{30} \end{bmatrix}\) |
(D) \(\begin{bmatrix} 7 & 4 \\ 3 & 6 \end{bmatrix}\) | (IV) \(\begin{bmatrix} \tfrac{2}{15} & -\tfrac{1}{15} \\[6pt] \tfrac{1}{6} & \tfrac{1}{6} \end{bmatrix}\) |
A quantity \( X \) is given by: \[ X = \frac{\epsilon_0 L \Delta V}{\Delta t} \] where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( L \) is the length,
- \( \Delta V \) is the potential difference,
- \( \Delta t \) is the time interval.
The dimension of \( X \) is the same as that of: