Let A = [[m, n], [p, q]], d = |A| ≠ 0, |A - d(adj A)| = 0. Then (1) (1+d)² = (m+q)² (2) 1+d² = (m+q)

Q: Why can we just divide by $d$ at the end? What if $d$ was zero?

The problem explicitly states $d \neq 0$. In JEE, always check the constraints! If $d$ were zero, the matrix $A$ would be singular, and the adjoint would have a different relationship with the zero matrix. Since $d$ is non-zero, it is a safe, valid algebraic operation.

Q: I got lost in the algebra. Is there a trick to avoid expanding everything?

Algebra is like a dance. You can't skip steps, but you can look for patterns. Notice how $(1+d)$ appears repeatedly? Don't expand it immediately; keep it as a block to see if it cancels out later. This is a hallmark of JEE-level problems—they reward structural thinking over brute force.

Q: Why did we replace $np$ with $mq - d$?

Look at the options. They only contain $m, q,$ and $d$. If you keep $n$ and $p$ in your final equation, you'll never reach the answer. The substitution is the bridge between the general matrix and the specific form required by the options.

Question

Let $A = \begin{bmatrix} m & n \ p & q \end{bmatrix}, d = |A| 
eq 0, |A - d(	ext{adj} A)| = 0$. Then

Options:

(A) $(1+d)^2 = (m+q)^2$

(B) $1+d^2 = (m+q)^2$

(C) $(1+d)^2 = m^2 + q^2$

(D) $1+d^2 = m^2 + q^2$

Sigma Percentile · Accepted Answer

Welcome, warriors of the JEE. Today, we are not just solving a matrix problem; we are peeling back the layers of algebraic elegance.

When you look at a matrix $A = \begin{bmatrix} m & n \ p & q \end{bmatrix}$ and a condition like $|A - d(	ext{adj } A)| = 0$, it is easy to feel overwhelmed. But I want you to take a deep breath. This is not a test of your endurance; it is a test of your ability to see the hidden symmetry in the numbers.

### Analyzing the Anatomy of the Matrix

We begin with the matrix $A$. Its determinant, $d = mq - np$, is the heartbeat of this problem.

The problem gives us a gift: $d 
eq 0$. This is not just a condition; it is a permission slip. It tells us that we can divide by $d$ later without fear.

Now, consider the adjoint. For a $2 	imes 2$ matrix, the adjoint is a beautiful, simple transformation: swap the diagonal elements $m$ and $q$, and negate the off-diagonal elements $n$ and $p$.

$$ 	ext{adj } A = \begin{bmatrix} q & -n \ -p & m \end{bmatrix} $$

### Constructing the Beast

The expression $A - d(	ext{adj } A)$ looks intimidating, but let us break it down. We multiply the adjoint by the scalar $d$, which distributes to every element:

$$ d(	ext{adj } A) = \begin{bmatrix} dq & -dn \ -dp & dm \end{bmatrix} $$

Now, we subtract this from $A$. The magic happens here. Look at the top-right element: $n - (-dn) = n + dn = n(1+d)$.

Look at the bottom-left: $p - (-dp) = p + dp = p(1+d)$. Suddenly, the factor $(1+d)$ emerges like a lighthouse in the fog. We have transformed the matrix into:

$$ \begin{bmatrix} m-dq & n(1+d) \ p(1+d) & q-dm \end{bmatrix} $$

### The Determinant Dance

We are told the determinant of this new matrix is zero. So, we set the product of the diagonals minus the product of the off-diagonals to zero:

$$ (m-dq)(q-dm) - [n(1+d)][p(1+d)] = 0 $$

This simplifies to:

$$ (m-dq)(q-dm) - np(1+d)^2 = 0 $$

Now, expand the first part: $mq - dm^2 - dq^2 + d^2mq$. Grouping the terms, we get $mq - d(m^2 + q^2) + d^2mq$.

### The Elegant Substitution

Remember our definition $d = mq - np$? This implies $np = mq - d$. This is the key that unlocks the door.

Substitute this into our equation:

$$ mq - d(m^2 + q^2) + d^2mq - (mq - d)(1+d)^2 = 0 $$

Now, expand $(mq - d)(1+d)^2$. It becomes $(mq - d)(1 + 2d + d^2) = mq + 2dmq + d^2mq - d - 2d^2 - d^3$.

When we subtract this from our previous expression, the $mq$ terms and the $d^2mq$ terms vanish into thin air. We are left with:

$$ -d(m^2 + q^2) - 2dmq + d(1+d)^2 = 0 $$

### The Grand Finale

Factor out the $-d$ from the first two terms:

$$ -d(m^2 + q^2 + 2mq) + d(1+d)^2 = 0 $$

Recognize the perfect square? $m^2 + q^2 + 2mq = (m+q)^2$. Our equation is now:

$$ -d(m+q)^2 + d(1+d)^2 = 0 $$

Since $d 
eq 0$, we divide by $d$ and rearrange to find:

$$ (1+d)^2 = (m+q)^2 $$

We have arrived. It was not a battle of brute force, but a dance of substitution and simplification. You have mastered the logic. Keep this clarity, and no problem will ever be too complex for you.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = [m p n q], d = ∣ A ∣ \neq = 0, ∣ A - d (adj A) ∣ = 0$ . Then

Select Answer:

Visualized Solution

Define Matrix $A$ and its Determinant $d$

Find the Adjoint of Matrix $A$

Construct the Matrix $d (adj A)$

Compute the Matrix $A - d (adj A)$

Set the Determinant to Zero

Expand the First Product

Substitute $n p = m q - d$

Simplify the Equation

Final Algebraic Grouping

Conclusion

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Anatomy of the Matrix

Constructing the Beast

The Determinant Dance

The Elegant Substitution

The Grand Finale

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Visualized Solution

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