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.For a $3 \times 3$ matrix $A$, if $A(\operatorname{adj} A) = \begin{bmatrix} 99 & 0 & 0 \\0 & 99 & 0 \\0 & 0 & 99 \end{bmatrix}$, then $\det(A)$ is equal to:
A block of certain mass is placed on a rough floor. The coefficients of static and kinetic friction between the block and the floor are 0.4 and 0.25 respectively. A constant horizontal force \( F = 20 \, \text{N} \) acts on it so that the velocity of the block varies with time according to the following graph. The mass of the block is nearly (Take \( g = 10 \, \text{m/s}^2 \)):
A wooden block of mass M lies on a rough floor. Another wooden block of the same mass is hanging from the point O through strings as shown in the figure. To achieve equilibrium, the coefficient of static friction between the block on the floor and the floor itself is
The circuit shown in the figure contains two ideal diodes \( D_1 \) and \( D_2 \). If a cell of emf 3V and negligible internal resistance is connected as shown, then the current through \( 70 \, \Omega \) resistance (in amperes) is: