Let k be a positive real number and let A = [[2k-1, 2√k, 2√k], [2√k, 1, -2k], [-2√k, 2k, -1]] and B

Q: Why is the determinant of matrix $B$ zero? I thought I had to calculate it.

Ah, the classic trap! You don't need to calculate it. Because $B^T = -B$, it is skew-symmetric. For any odd-order skew-symmetric matrix, the determinant is always zero. It is a beautiful, deep property of linear algebra that rewards those who observe the structure before diving into the arithmetic.

Q: Why do we use row operations on $A$ instead of just expanding it?

Expanding a $3 \times 3$ matrix with variables like $2\sqrt{k}$ is a recipe for algebraic errors. Row and column operations are your scalpel; they cut through the complexity to reveal the simple structure underneath. Always look for ways to create zeros before you expand.

Q: I got $|A| = (2k+1)^3$. Is that correct?

Spot on! That is the elegant core of this problem. Once you reach that, the rest is just solving a simple equation. You have mastered the hardest part of the journey.

Question

Let $k$ be a positive real number and let $A = \begin{bmatrix} 2k-1 & 2\sqrt{k} & 2\sqrt{k} \ 2\sqrt{k} & 1 & -2k \ -2\sqrt{k} & 2k & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 2k-1 & \sqrt{k} \ 1-2k & 0 & 2\sqrt{k} \ -\sqrt{k} & -2\sqrt{k} & 0 \end{bmatrix}$. If $\det(	ext{adj } A) + \det(	ext{adj } B) = 10^6$, then $[k]$ is equal to [Note : adj M denotes the adjoint of square matrix M and [k] denotes the largest integer less than or equal to k.]

Sigma Percentile · Accepted Answer

### The Symphony of Matrices: A Journey into Elegance

Welcome, future engineers. Today, we are not just solving a problem; we are peeling back the layers of a mathematical onion.

When you first look at a problem involving matrices $A$ and $B$ with variables like $2\sqrt{k}$ and $2k-1$, it is natural to feel a surge of intimidation. The terms look messy, the algebra seems daunting, and the prospect of calculating determinants feels like a long, tedious road.

But here is the secret: Mathematics is rarely about brute force. It is about observation, strategy, and finding the hidden symmetry.

### Phase 1: The Mystery of Matrix $B$

Let us start with Matrix $B$. Many students would immediately jump to the Sarrus rule or cofactor expansion. Stop. Breathe. Look at the matrix.

The diagonal elements are all zero. The elements $b_{ij}$ are the negatives of $b_{ji}$. We have encountered a **skew-symmetric matrix**.

In the world of linear algebra, skew-symmetric matrices are special. Specifically, for any skew-symmetric matrix of odd order—and here, $B$ is a $3 	imes 3$ matrix—the determinant is always zero.

Why? Because:

$$|B| = |B^T| = |-B| = (-1)^n |B|$$

Since $n=3$, we get $|B| = -|B|$, which implies $2|B| = 0$, or $|B| = 0$. Just like that, the entire $B$ component of our equation vanishes. We have simplified our problem by half just by observing the structure.

### Phase 2: The Battle with Matrix $A$

Now, we turn our attention to Matrix $A$. It looks formidable. But remember, we want to create zeros. We want to make the determinant expansion as painless as possible.

We apply the column operation $C_2 	o C_2 - C_3$. Why? Because in the first row, we see $2\sqrt{k}$ in both the second and third columns. Subtracting them gives us a zero.

This is the first crack in the armor. Then, we apply the row operation $R_2 	o R_2 - R_3$. Suddenly, the second column is filled with zeros, save for one element.

We have transformed a complex determinant into a simple expansion problem. This is the power of elementary operations—they are the tools of a master.

### Phase 3: The Algebraic Elegance

As we expand along the second column, the expression simplifies beautifully. We are left with:

$$|A| = -(1+2k) [ (2k-1)(1-2k) - (2\sqrt{k})(4\sqrt{k}) ]$$

Notice the terms inside the bracket. $(2k-1)(1-2k)$ is simply $-(2k-1)^2$. And $2\sqrt{k} 	imes 4\sqrt{k}$ is $8k$.

When we combine these, we get $-(2k-1)^2 - 8k$. Pulling out the negative sign, we are left with $(2k-1)^2 + 8k$, which expands to $4k^2 - 4k + 1 + 8k$, or $4k^2 + 4k + 1$.

This is a perfect square: $(2k+1)^2$. Multiplying by the $(2k+1)$ we factored out earlier, we arrive at the stunning result:

$$|A| = (2k+1)^3$$

The complexity has collapsed into a single, elegant term.

### Phase 4: The Final Convergence

We are almost there. We know that $\det(	ext{adj } A) = |A|^{n-1}$. Since $n=3$, this is $|A|^2$.

Our equation becomes $(|A|)^2 = 10^6$. Substituting our result, we get:

$$((2k+1)^3)^2 = 10^6 \Rightarrow (2k+1)^6 = 10^6$$

Since $k$ is a positive real number, we can equate the bases: $2k+1 = 10$. Solving for $k$, we find $k = 4.5$.

The question asks for the greatest integer less than or equal to $k$, denoted as $[k]$. Thus, $[4.5] = \mathbf{4}$.

You see? What started as a terrifying matrix equation was actually a carefully constructed puzzle waiting for you to find the key. Keep this mindset. When you face a problem, do not just calculate—observe, simplify, and let the elegance of mathematics guide you to the answer.

Enter Numerical Value:

Visualized Solution

Analyze the Given Equation

Analyze Matrix $B$

Determinant of Matrix $B$

Set up $∣ A ∣$

Column Operation $C_{2} \to C_{2} - C_{3}$

Row Operation $R_{2} \to R_{2} - R_{3}$

Expand $∣ A ∣$

Simplify the Expression

Final Value of $∣ A ∣$

Apply Adjoint Property

Substitute and Solve

Find $k$ and $[k]$

The Sigma Insight: Adjoint and Inverse of a Matrix

The Symphony of Matrices

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Let $A = 246 2 + p 6 + 2 p 12 + 3 p 2 + p + q 8 + 3 p + 2 q 20 + 6 p + 3 q$ . If $det (adj (adj (3 A))) = 2^{m} \cdot 3^{n}, m, n \in N$ , then $m + n$ is equal to

Let $a \in R$ and $A$ be a matrix of order $3 \times 3$ such that $det (A) = - 4$ and $A + I = 12 a a 11 102$ , where $I$ is the identity matrix of order $3 \times 3$ . If $det ((a + 1) adj ((a - 1) A))$ is $2^{m} 3^{n}$ , $m, n \in {0, 1, 2, \dots, 20}$ , then $m + n$ is equal to :

Let $B = 11 α 32 α α 34$ , $α > 2$ be the adjoint of a matrix $A$ and $∣ A ∣ = 2$ . Then $[α - 2 α α] B α - 2 α α$ is equal to

Let the matrix $A = 001100010$ and the matrix $B_{0} = A^{49} + 2 A^{98}$ . If $B_{n} = adj (B_{n - 1})$ for all $n \geq 1$ , then $det (B_{4})$ is equal to :

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Let A be a $3 \times 3$ matrix such that $X^{T} A X = 0$ for all nonzero $3 \times 1$ matrices $X = x y z$ . If $A 111 = 14 - 5$ and $A 121 = 04 - 8$ , and $det (a d j (2 (A + I))) = 2^{α} 3^{β} 5^{γ}$ , $α, β, γ \in N,$ then $α^{2} + β^{2} + γ^{2}$ is

Let $A = {a_{ij}}$ be a $3 \times 3$ matrix, where $a_{ij} = ⎩ ⎨ ⎧ (- 1)^{j - i} 2 (- 1)^{i + j} if i < j if i = j if i > j$ then $det (3 Adj (2 A^{- 1}))$ is equal to

Let $A$ be a $3 \times 3$ real matrix. If $det (2 Adj (2 Adj (Adj (2 A)))) = 2^{41}$ , then the value of $det (A^{2})$ equal .

Let A be a $3 \times 3$ matrix such that $A + A^{T} = O$ . If $A 1 - 1 0 = 332, A^{2} 1 - 1 0 = - 3 19 - 24$ and $det (adj (2 adj (A + I))) = (2)^{α} \cdot (3)^{β} \cdot (11)^{γ}, α, β, γ$ are non-negative integers, then $α + β + γ$ is equal to .........

For some $α, β \in R$ , let $A = [α 1 22]$ and $B = [11 1 β]$ be such that $A^{2} - 4 A + 2 I = B^{2} - 3 B + I = O$ . Then $(det (adj (A^{3} - B^{3})))^{2}$ is equal to .........

Enter Numerical Value:

Visualized Solution

Analyze the Given Equation

Analyze Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Determinant of Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Set up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣A∣

Column Operation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​→C2​−C3​

Row Operation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​→R2​−R3​

Expand .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣A∣