Q.The inverse of the matrix $\begin{bmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 5 \end{bmatrix}$ is
(A) $\begin{bmatrix} 0 & 0 & 3 \\ 0 & 2 & 0 \\ 5 & 0 & 0 \end{bmatrix}$
(B) $\begin{bmatrix} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{5} \end{bmatrix}$
(C) $\begin{bmatrix} -\frac{1}{3} & 0 & 0 \\ 0 & -\frac{1}{2} & 0 \\ 0 & 0 & -\frac{1}{5} \end{bmatrix}$
(D) $\begin{bmatrix} -3 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -5 \end{bmatrix}$

Concept: Inverse of a diagonal matrix — just take the reciprocal of each diagonal entry.

Reasoning:

1. The given matrix is diagonal: all off-diagonal entries are zero.
2. For a diagonal matrix $D = \text{diag}(d_1, d_2, \dots, d_n)$, its inverse (if it exists) is $D^{-1} = \text{diag}(1/d_1, 1/d_2, \dots, 1/d_n)$, provided no $d_i = 0$.
3. Here $d_1 = 3$, $d_2 = 2$, $d_3 = 5$. Their reciprocals are $\frac{1}{3}, \frac{1}{2}, \frac{1}{5}$. …

The key insight here is that a diagonal matrix — one with non-zero entries only on its main diagonal — has a beautifully simple inverse. Why? Because when you multiply two diagonal matrices, you just multiply the corresponding diagonal entries. So to “undo” a diagonal matrix, you need a matrix that, when multiplied, gives back the identity matrix (1’s on the diagonal, 0’s elsewhere). The natural choice is to take the reciprocal of each diagonal entry.

Let’s see this in action.

1. Recall the definition of an inverse.

   For a square matrix $A$, its inverse $A^{-1}$ satisfies $A \cdot A^{-1} = I$, where $I$ is the identity matrix. For a $3 \times 3$ identity, $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

2. Write the given matrix.

   $A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 5 \end{bmatrix}$.

   Notice that all off-diagonal entries are zero. This is a diagonal matrix.

3. Guess the form of the inverse.

   Since multiplying diagonal matrices is just entry-wise multiplication, the inverse should also be diagonal. Let’s denote it as $B = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix}$.

4. Multiply $A$ and $B$.

   $A \cdot B = \begin{bmatrix} 3a & 0 & 0 \\ 0 & 2b & 0 \\ 0 & 0 & 5c \end{bmatrix}$.

   For this to equal $I$, we need:

   • $3a = 1 \implies a = \frac{1}{3}$
   • $2b = 1 \implies b = \frac{1}{2}$
   • $5c = 1 \implies c = \frac{1}{5}$

5. Write the inverse.

   So $A^{-1} = \begin{bmatrix} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{5} \end{bmatrix}$. …

Method: Inverting a diagonal matrix by reciprocals

Use this shortcut whenever the matrix has non-zero entries only on its main diagonal.

Steps

Step 1: Confirm the matrix is diagonal.

Every off-diagonal entry must be $0$. Only then does the shortcut apply.

Step 2: Check invertibility.

A diagonal matrix is invertible exactly when no diagonal entry is $0$, since $\det D=d_1d_2\cdots d_n$.

Step 3: Replace each diagonal entry by its reciprocal. …

Common Mistakes

Mistake 1: Negating the entries instead of reciprocating.

Why it's wrong: $-3\times3=-9\neq1$, so the negated matrix (option D) is not the inverse. Correct approach: replace each $d_i$ by $\tfrac1{d_i}$, giving $\tfrac13,\tfrac12,\tfrac15$.

Mistake 2: Using negative reciprocals.

Why it's wrong: $-\tfrac13\times3=-1\neq1$ (option C fails). Correct approach: the reciprocal is positive for a positive entry. …