Let A be a 2 × 2 matrix with det(A) = -1 and det((A + I)(adj(A) + I)) = 4. Then the sum of the diago

Q: Why can't we just assume $A$ is a diagonal matrix to solve this?

That is a tempting shortcut! While it might work for specific cases, it is dangerous. In JEE Advanced, you must prove your result holds for the general case. By using the general matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, we ensure our logic is robust and mathematically sound, regardless of the specific values of $b$ and $c$.

Q: Why does the expression simplify so cleanly to $A + \text{adj}(A)$?

It feels like magic, doesn't it? It happens because the identity matrix $I$ commutes with everything. When you expand $(A+I)(\text{adj}(A)+I)$, the term $A \cdot \text{adj}(A)$ becomes $-I$. This $-I$ perfectly cancels the $+I$ term that arises from $I \cdot I$. It is a beautiful example of how algebraic structure simplifies complexity.

Q: Why is the answer $\pm 2$ but the option only shows $2$?

Great observation! The equation $(a+d)^2 = 4$ yields two possible values for the trace: $2$ and $-2$. In multiple-choice questions, you must select the value that appears in the options. Since $-2$ is not provided, $2$ is the intended answer. Always check your options after solving!

Question

Let $A$ be a $2 	imes 2$ matrix with $\det(A) = -1$ and $\det((A + I)(	ext{adj}(A) + I)) = 4$. Then the sum of the diagonal elements of $A$ can be :

Options:

(A) -1

(B) 2

(C) 1

(D) $-\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We start with a $2 	imes 2$ matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$. Our goal is to find the trace, which is defined as $a + d$.

We are given the expression $\det((A + I)(	ext{adj}(A) + I)) = 4$. The presence of the adjoint matrix, $	ext{adj}(A)$, is a significant clue.

Whenever you see a matrix and its adjoint together, your mind should immediately jump to the fundamental identity:

$$A \cdot 	ext{adj}(A) = \det(A) \cdot I$$

We are given that $\det(A) = -1$, so our identity simplifies to $A \cdot 	ext{adj}(A) = -I$. Keep this in your pocket; it is the key that unlocks the problem.

### The Algebraic Expansion

Now, let's expand the complex expression $(A + I)(	ext{adj}(A) + I)$ using the distributive law:

$$(A + I)(	ext{adj}(A) + I) = A \cdot 	ext{adj}(A) + A \cdot I + I \cdot 	ext{adj}(A) + I \cdot I$$

Remember that the identity matrix $I$ acts as the multiplicative identity. Thus, $A \cdot I = A$, $I \cdot 	ext{adj}(A) = 	ext{adj}(A)$, and $I \cdot I = I$.

Substituting these into our expansion, the expression simplifies to:

$$A \cdot 	ext{adj}(A) + A + 	ext{adj}(A) + I$$

### The Geometric Collapse

This is where the magic happens. By substituting our secret partner identity $A \cdot 	ext{adj}(A) = -I$ into the expression, we get:

$$-I + A + 	ext{adj}(A) + I$$

The $-I$ and the $+I$ are like two opposing forces that perfectly cancel each other out. We are left with an incredibly simple result:

$$A + 	ext{adj}(A)$$

### The Final Reveal

Now, let's write out $A + 	ext{adj}(A)$ explicitly. For $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$, the adjoint is $	ext{adj}(A) = \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.

Adding these together, we obtain:

$$A + 	ext{adj}(A) = \begin{bmatrix} a+d & 0 \ 0 & a+d \end{bmatrix}$$

This is a diagonal matrix. The determinant of a diagonal matrix is simply the product of its diagonal elements:

$$\det(A + 	ext{adj}(A)) = (a+d)(a+d) = (a+d)^2$$

We are given that this determinant equals $4$. Therefore, $(a+d)^2 = 4$, which implies $a+d = \pm 2$.

Given the context of such problems, we identify the trace as **$2$**. We have conquered the problem by uncovering the beautiful, underlying symmetry of matrices.

Animated Solution for Mathematics - Matrices and Determinants: Let $A$ be a $2 \times 2$ matrix with $det (A) = - 1$ and $det ((A + I) (adj (A) + I)) = 4$ . Then the sum of the diagonal elements of $A$ can be :

Select Answer:

Visualized Solution

Defining Matrix $A$

The Fundamental Adjoint Property

Substituting the Determinant

Expanding the Given Expression

Applying the Distributive Law

Simplifying the Expansion

Substituting the Adjoint Product

Cancellation of Terms

Visualizing the Adjoint Matrix

Adding the Matrices

Applying the Determinant Condition

Evaluating the Determinant

Solving for the Trace

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Algebraic Expansion

The Geometric Collapse

The Final Reveal

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