Q.If $A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$, then $A^{-1}$ is
(A) $\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$
(B) $\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$
(C) $\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
(D) $\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

The key insight here is that $A$ is a diagonal matrix — all non-diagonal entries are zero. For such matrices, inversion is beautifully simple: you just invert each diagonal element individually. No row operations, no cofactors, no fuss.

Why does this work? Think about what a diagonal matrix does when it multiplies a vector: it scales each coordinate independently by the corresponding diagonal entry. The inverse must undo that scaling, so it scales each coordinate by the reciprocal. If the original scaling factor is $d$, the inverse scaling factor is $1/d$. That’s the whole story.

Let’s walk through it step by step.

1. Identify the structure.

   $A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ is diagonal. Its diagonal entries are $a_{11} = -1$, $a_{22} = 1$, $a_{33} = 1$.

2. Apply the diagonal inverse rule.

   For a diagonal matrix $D = \text{diag}(d_1, d_2, \dots, d_n)$, the inverse is $D^{-1} = \text{diag}(1/d_1, 1/d_2, \dots, 1/d_n)$, provided no $d_i = 0$. Here none are zero, so:

3. Notice the result. The inverse turned out to be exactly the same as $A$. That’s because each diagonal entry is its own reciprocal: $(-1)^{-1} = -1$ and $1^{-1} = 1$. So $A$ is an involutory matrix — a matrix that is its own inverse.

4. Check against the options.
   • (A) has all $-1$ on the diagonal — that would be the inverse of $\text{diag}(-1, -1, -1)$, not our $A$.
   • (B) has $1, -1, -1$ — that’s the inverse of $\text{diag}(1, -1, -1)$.
   • (C) has $-1, -1, 1$ — that’s the inverse of $\text{diag}(-1, -1, 1)$.
   • (D) has $-1, 1, 1$ — exactly what we computed.

Let’s go through each option one by one, understanding the reasoning behind each property.

1. Option (A): $(A + B)^{-1} = B^{-1} + A^{-1}$ This looks tempting if you’re used to the distributive law, but matrix inversion does not distribute over addition. To check, multiply $(A+B)$ by $(B^{-1} + A^{-1})$:

That’s $A B^{-1} + B A^{-1} + 2I$, which is not $I$ in general. So this is false.

2. Option (B): $(AB)^{-1} = B^{-1} A^{-1}$

   This is the reversal law for inverses of products.

   Check: $(AB)(B^{-1}A^{-1}) = A (B B^{-1}) A^{-1} = A I A^{-1} = A A^{-1} = I$.

   Similarly, $(B^{-1}A^{-1})(AB) = I$. So this is correct.

   Tip

   Remember the order: when you invert a product, the order reverses — like taking off shoes and socks: first off goes the last thing you put on.

3. Option (C): $\text{adj}(A) = |A| A^{-1}$

   This is a standard formula from the definition of the inverse: $A^{-1} = \frac{\text{adj}(A)}{|A|}$, provided $|A| \neq 0$.

   Rearranging gives $\text{adj}(A) = |A| A^{-1}$. So this is correct.

4. Option (D): $|A^{-1}| = |A|^{-1}$

   Since $A A^{-1} = I$, taking determinants: $|A| \cdot |A^{-1}| = |I| = 1$.

   Hence $|A^{-1}| = \frac{1}{|A|} = |A|^{-1}$. So this is correct.

Only option (A) fails the test.

When dealing with matrices, finding the inverse can often be a lengthy process involving determinants and adjoints. However, for special types of matrices, this process simplifies significantly. A diagonal matrix is one such case.

A diagonal matrix is a square matrix where all the entries outside the main diagonal are zero. For example, the given matrix $A$ is a diagonal matrix because its only non-zero elements are $A_{11}=2$, $A_{22}=3$, and $A_{33}=5$.

The fundamental definition of an inverse matrix $A^{-1}$ is that when multiplied by the original matrix $A$, it yields the identity matrix $I$. That is, $AA^{-1} = I$. The identity matrix $I$ is also a diagonal matrix with all diagonal elements equal to 1.

Consider a general diagonal matrix $D = \text{diag}(d_1, d_2, \dots, d_n)$. If its inverse $D^{-1}$ is also a diagonal matrix, say $D^{-1} = \text{diag}(x_1, x_2, \dots, x_n)$, then their product $DD^{-1}$ would be:

For this product to be the identity matrix $I = \text{diag}(1, 1, \dots, 1)$, we must have $d_i x_i = 1$ for all $i=1, \dots, n$. This implies $x_i = \frac{1}{d_i}$.

This shows that the inverse of a diagonal matrix is simply another diagonal matrix where each diagonal element is the reciprocal of the corresponding element in the original matrix. This property holds true as long as all diagonal elements $d_i$ are non-zero, which ensures the matrix is invertible.

Now, let's apply this understanding to the given problem.

1. Identify the matrix type:

   The given matrix is $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix}$.

   This is a diagonal matrix because all its non-diagonal elements are zero. The diagonal elements are $d_1 = 2$, $d_2 = 3$, and $d_3 = 5$.

2. Apply the inverse property for diagonal matrices:

   Since $A$ is a diagonal matrix, its inverse $A^{-1}$ will also be a diagonal matrix. Each diagonal element of $A^{-1}$ will be the reciprocal of the corresponding diagonal element of $A$.

3. Calculate the reciprocal diagonal elements:

   The reciprocals of the diagonal elements are:

   • For $d_1 = 2$, the reciprocal is $\frac{1}{2}$.
   • For $d_2 = 3$, the reciprocal is $\frac{1}{3}$.
   • For $d_3 = 5$, the reciprocal is $\frac{1}{5}$.

4. Construct the inverse matrix:

   Placing these reciprocals on the main diagonal and keeping all other elements zero, we get:

5. Compare with the given options:

   Comparing our calculated $A^{-1}$ with the provided options:

   (A) $\begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{5} \end{bmatrix}$

   (B) $\dfrac{1}{30}\begin{bmatrix} 15 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 6 \end{bmatrix} = \begin{bmatrix} \frac{15}{30} & 0 & 0 \\ 0 & \frac{10}{30} & 0 \\ 0 & 0 & \frac{6}{30} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{5} \end{bmatrix}$

   (C) $\dfrac{1}{30}\begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix}$

   (D) $\dfrac{1}{30}\begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{5} \end{bmatrix}$

   Both option (A) and option (B) represent the same matrix, which is our calculated $A^{-1}$. In multiple-choice questions, sometimes equivalent forms are given. Since (A) is the most direct representation, it is typically the intended answer.

We start with the equation $A^2 - 3A + I = O$. This is a matrix polynomial that looks very much like a scalar quadratic. The trick is to treat it as a relation that lets us express $A^{-1}$ as a linear combination of $A$ and $I$.

1. Rewrite the equation to isolate $I$. From $A^2 - 3A + I = O$, bring the $I$ term to the other side:

2. Factor $A$ on the left. Since matrix multiplication is not commutative in general, but here we are factoring $A$ out (and $A$ commutes with itself), we can write:

This is valid because $A$ and $I$ always commute.

3. Multiply both sides by $A^{-1}$ (which exists, as we are told $A^{-1}$ is defined). Left-multiply by $A^{-1}$:

4. Solve for $A^{-1}$. Multiply both sides by $-1$:

5. Compare with the given form $A^{-1} = xA + yI$.

   We have $A^{-1} = (-1)A + 3I$. So $x = -1$ and $y = 3$.

6. Compute $x + y$.