Let A = [[[x+1], [x+2], [x+3]], [[x], [x+3], [x+3]], [[x], [x+2], [x+4]]], where [t] denotes the gre

Q: Wait, why can't we just expand the determinant directly? It's a 3x3 matrix, how hard can it be?

While you technically could, expanding it directly involves multiplying terms like $[x+1][x+3][x+4]$, which creates a cubic polynomial in terms of $[x]$. This is a classic trap in JEE Advanced problems designed to waste your time and increase the probability of calculation errors. By using row operations first, we reduce the complexity to a simple linear equation. Always look for symmetry and simplification before jumping into brute-force expansion!

Q: Is the property $[x+n] = [x]+n$ always true? What if $x$ is a negative number or a fraction?

That is a fantastic question! Yes, the property $[x+n] = [x]+n$ holds true for any real number $x$ as long as $n$ is an integer. The greatest integer function effectively 'shifts' the floor of the number by the integer $n$. Whether $x$ is positive, negative, or fractional, this property remains a fundamental tool in your algebraic arsenal. Trust the property; it is mathematically robust.

Q: Why is the final answer an interval $[62, 63)$ and not just $x=62$?

Remember that we solved for $k$, where $k = [x]$. We found $k=62$, which means $[x] = 62$. By the definition of the greatest integer function, $[x] = 62$ implies that $x$ must be at least $62$ but strictly less than $63$. If $x$ were $63$, the greatest integer would be $63$, not $62$. Thus, any value in the interval $[62, 63)$ satisfies the condition. This is why we express the solution as an interval.

Question

Let $A = \begin{pmatrix} [x+1] & [x+2] & [x+3] \ [x] & [x+3] & [x+3] \ [x] & [x+2] & [x+4] \end{pmatrix}$, where $[t]$ denotes the greatest integer less than or equal to $t$. If $\det(A) = 192$, then the set of values of $x$ is the interval:

Options:

(A) $[68, 69)$

(B) $[62, 63)$

(C) $[65, 66)$

(D) $[60, 61)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The greatest integer function, denoted by $[t]$, is often used to intimidate students, but it obeys a beautiful, simple rule: $[x+n] = [x]+n$ for any integer $n$. This is our golden key.

By applying this property, we strip away the complexity, turning $[x+1]$ into $[x]+1$, $[x+2]$ into $[x]+2$, and so on.

### The Algebraic Transformation

To simplify the expression, let us define a new variable, $k = [x]$.

Suddenly, the matrix $A$ transforms from a collection of brackets into a clean, manageable algebraic structure:

$$A = \begin{pmatrix} k+1 & k+2 & k+3 \ k & k+3 & k+3 \ k & k+2 & k+4 \end{pmatrix}$$

The fog has lifted. We are no longer dealing with functions; we are dealing with a standard matrix of linear terms.

### The Art of Row Operations

Many students would rush to expand the determinant now, but that is where the trap lies. We want to be surgical by creating zeros.

Perform the row operation $R_1 	o R_1 - R_2$. Subtracting the second row from the first, we get:

$$R_1 	o (k+1-k, \ k+2-(k+3), \ k+3-(k+3)) = (1, -1, 0)$$

Now, perform the operation $R_3 	o R_3 - R_2$. We get:

$$R_3 	o (k-k, \ k+2-(k+3), \ k+4-(k+3)) = (0, -1, 1)$$

We have transformed the dense matrix into a sparse, elegant one:

$$\det(A) = \begin{vmatrix} 1 & -1 & 0 \ k & k+3 & k+3 \ 0 & -1 & 1 \end{vmatrix}$$

### The Final Collapse

Expanding along the first row is now a breeze:

$$\det(A) = 1 \cdot \begin{vmatrix} k+3 & k+3 \ -1 & 1 \end{vmatrix} - (-1) \cdot \begin{vmatrix} k & k+3 \ 0 & 1 \end{vmatrix} + 0$$

This simplifies to:

$$1[(k+3)(1) - (k+3)(-1)] + 1[(k)(1) - (k+3)(0)] = (k+3 + k+3) + k = 3k + 6$$

The entire determinant has collapsed into the simple linear expression $3k + 6$.

### Solving for $k$

We are given that $\det(A) = 192$. Setting our expression equal to this value:

$$3k + 6 = 192$$

Subtracting $6$ gives $3k = 186$, and dividing by $3$ yields:

$$k = 62$$

### The Final Interval

We have found $k=62$, which means $[x] = 62$. By the definition of the greatest integer function, this implies that $x$ must be in the range:

**$x \in

Animated Solution for Mathematics - Matrices and Determinants: Let $A = [x + 1] [x] [x] [x + 2] [x + 3] [x + 2] [x + 3] [x + 3] [x + 4]$ , where $[t]$ denotes the greatest integer less than or equal to $t$ . If $det (A) = 192$ , then the set of values of $x$ is the interval:

Select Answer:

Visualized Solution

Analyze the Matrix $A$

The Property of $[x + n]$

Substitution $[x] = k$

Row Operation $R_{1} \to R_{1} - R_{2}$

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

Simplified Determinant Form

Expand the Determinant

Solve for $k$

Find the Interval for $x$

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Algebraic Transformation

The Art of Row Operations

The Final Collapse

Solving for $k$

The Final Interval

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Select Answer:

Visualized Solution

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

The Property of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [x+n]

Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [x]=k

Row Operation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​→R1​−R2​

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

Simplified Determinant Form

Expand the Determinant

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Find the Interval for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Algebraic Transformation

The Art of Row Operations

The Final Collapse

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

The Final Interval

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