In matrix algebra, when dealing with conditions involving products of matrices, be sure to manipulate and combine the equations carefully. Using properties like associativity and distributivity of matrix multiplication can help simplify expressions, as demonstrated in this problem.
The correct answer is: (B) A + B.
We are given two matrices \( A \) and \( B \) such that:
\( A(AB) = A \cdot B \)
This simplifies to:\( A^2B = AB \)
Since \( AB = B \), this becomes:\( A^2B = B \)
- From \( BA = A \), multiply both sides on the right by \( B \):\( (BA)B = A \cdot B \)
This simplifies to:\( AB^2 = AB \)
Since \( AB = B \), this becomes:\( AB^2 = B \)
Now, we can add the two equations \( A^2B = B \) and \( AB^2 = B \):\( A^2B + AB^2 = B + B \)
This simplifies to:\( A^2 + B^2 = A + B \)
Therefore, the correct answer is (B) A + B.If \[ A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix} \] then find \( A^{-1} \). Hence, solve the system of linear equations: \[ x - 2y = 10, \] \[ 2x - y - z = 8, \] \[ -2y + z = 7. \]
You are given a dipole of charge \( +q \) and \( -q \) separated by a distance \( 2l \). A sphere 'A' of radius \( R \) passes through the centre of the dipole as shown below and another sphere 'B' of radius \( 2R \) passes through the charge \( +q \). Then the electric flux through the sphere A is