Q.If A = $\begin{bmatrix} 3 & 4 \\ 5 & 2 \end{bmatrix}$ and 2A + B is a null matrix, then B is equal to :
(A) $\begin{bmatrix} 6 & 8 \\ 10 & 4 \end{bmatrix}$
(B) $\begin{bmatrix} -6 & -8 \\ -10 & -4 \end{bmatrix}$
(C) $\begin{bmatrix} 5 & 8 \\ 10 & 3 \end{bmatrix}$
(D) $\begin{bmatrix} -5 & -8 \\ -10 & -3 \end{bmatrix}$

The core idea here is straightforward: a null matrix (or zero matrix) is one where every entry is $0$. The equation $2A + B = 0$ means that when you add the matrix $2A$ to the matrix $B$, you get a matrix of all zeros. So $B$ must be the additive inverse of $2A$ — that is, $B = -2A$.

This is exactly like solving $2x + y = 0$ for a number $y$: you get $y = -2x$. The only difference is that here $x$ and $y$ are matrices, and the arithmetic is done entry by entry.

Let’s work through it step by step.

1. Write down the given matrix $A$.

   $A = \begin{bmatrix} 3 & 4 \\ 5 & 2 \end{bmatrix}$

2. Find $2A$ by scalar multiplication.

   Multiply every entry of $A$ by $2$:

   $2A = \begin{bmatrix} 2 \times 3 & 2 \times 4 \\ 2 \times 5 & 2 \times 2 \end{bmatrix} = \begin{bmatrix} 6 & 8 \\ 10 & 4 \end{bmatrix}$

3. Use the given condition: $2A + B = 0$.

   Here $0$ means the $2 \times 2$ null matrix $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.

   So $B$ must satisfy: $B = 0 - 2A = -2A$.

4. Compute $B = -2A$.

   Multiply every entry of $2A$ by $-1$:

   $B = -\begin{bmatrix} 6 & 8 \\ 10 & 4 \end{bmatrix} = \begin{bmatrix} -6 & -8 \\ -10 & -4 \end{bmatrix}$

5. Match with the options. The result $\begin{bmatrix} -6 & -8 \\ -10 & -4 \end{bmatrix}$ is exactly option (B).

The key here is understanding how determinants behave under two operations: taking the inverse of a matrix, and multiplying a matrix by a scalar. These are separate properties, and you apply them one after the other.

First, recall the fundamental rule: for any invertible square matrix $A$, the determinant of its inverse is the reciprocal of the determinant of $A$. That is, $|A^{-1}| = \frac{1}{|A|}$. This makes sense because $A \cdot A^{-1} = I$, and taking determinants gives $|A| \cdot |A^{-1}| = |I| = 1$.

Second, when you multiply a matrix by a scalar $k$, every entry in the matrix gets multiplied by $k$. For an $n \times n$ matrix, this means each of the $n$ rows (or columns) is scaled by $k$, so the determinant gets multiplied by $k$ a total of $n$ times. Hence $|kA| = k^n |A|$.

Now we combine these. We want $|4A^{-1}|$. Here $A$ is $2 \times 2$, so $n = 2$, and the scalar is $k = 4$.

Let’s work it step by step:

1. Find $|A^{-1}|$.

   Since $|A| = 2$, we have $|A^{-1}| = \frac{1}{|A|} = \frac{1}{2}$.

2. Apply the scalar multiplication property.

   For a $2 \times 2$ matrix $B$, $|4B| = 4^2 |B| = 16 |B|$. Here $B = A^{-1}$, so $|4A^{-1}| = 16 \cdot |A^{-1}|$.

3. Substitute the value from step 1.

   $|4A^{-1}| = 16 \cdot \frac{1}{2} = 8$.

The key idea here is scalar multiplication of a determinant. When you multiply a matrix by a constant $k$, every element in the matrix gets multiplied by $k$. But the determinant is a sum of products of elements — so each term in the determinant expansion picks up a factor of $k$ for each row (or column). For an $n \times n$ matrix, that means the determinant gets multiplied by $k^n$.

Also, remember that the determinant of a matrix and its transpose are always equal: $|A'| = |A|$. So the transpose here is a red herring — it doesn’t change the value.

Let’s walk through it step by step.

1. Start with what’s given.

   $A$ is a $3 \times 3$ matrix, and $|A| = 5$. We need $|2A'|$, where $A'$ is the transpose of $A$.

2. Handle the transpose first.

   Since $|A'| = |A|$, we have $|A'| = 5$. So $|2A'| = |2B|$ where $B = A'$ is just another $3 \times 3$ matrix with determinant $5$.

3. Apply scalar multiplication rule.

   For any $n \times n$ matrix $M$, $|kM| = k^n |M|$. Here $n = 3$ and $k = 2$, so:

4. Check the sign. The options include negative numbers, but scalar multiplication by a positive constant never flips the sign of the determinant. Only swapping rows or multiplying a row by a negative constant can change the sign. So $-10$ and $-40$ are impossible here.