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Matrices and Determinants question

2021 · 26 Feb · Shift 1 · Q32
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  5. /2021 · 26 Feb · Shift 1 · Q32

Matrices and Determinants question

2021 · 26 Feb · Shift 1 · Q32

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
The value of ∣(a+1)(a+2)a+21(a+2)(a+3)a+31(a+3)(a+4)a+41∣\left| {\begin{matrix} {(a + 1)(a + 2)} & {a + 2} & 1 \\ {(a + 2)(a + 3)} & {a + 3} & 1 \\ {(a + 3)(a + 4)} & {a + 4} & 1 \\ \end{matrix} } \right|​(a+1)(a+2)(a+2)(a+3)(a+3)(a+4)​a+2a+3a+4​111​​ is :
  1. A
    −-− 2
  2. B
    0
  3. C
    (a + 2)(a + 3)(a + 4)
  4. D
    (a + 1)(a + 2)(a + 3)
View written solutionFree

Correct answer: A

  1. Let
D=∣(a+1)(a+2)a+21(a+2)(a+3)a+31(a+3)(a+4)a+41∣.D=\begin{vmatrix} (a+1)(a+2) & a+2 & 1\\ (a+2)(a+3) & a+3 & 1\\ (a+3)(a+4) & a+4 & 1 \end{vmatrix}.D=​(a+1)(a+2)(a+2)(a+3)(a+3)(a+4)​a+2a+3a+4​111​​.
  1. Observe the row pattern. Define x1=a+2,x2=a+3,x3=a+4.x_1=a+2,\quad x_2=a+3,\quad x_3=a+4.x1​=a+2,x2​=a+3,x3​=a+4. Then the rows become
[x1(x1−1), x1, 1],[x2(x2−1), x2, 1],[x3(x3−1), x3, 1].[x_1(x_1-1),\ x_1,\ 1], \quad [x_2(x_2-1),\ x_2,\ 1], \quad [x_3(x_3-1),\ x_3,\ 1].[x1​(x1​−1), x1​, 1],[x2​(x2​−1), x2​, 1],[x3​(x3​−1), x3​, 1].

So

D=∣x12−x1x11x22−x2x21x32−x3x31∣.D=\begin{vmatrix} x_1^2-x_1 & x_1 & 1\\ x_2^2-x_2 & x_2 & 1\\ x_3^2-x_3 & x_3 & 1 \end{vmatrix}.D=​x12​−x1​x22​−x2​x32​−x3​​x1​x2​x3​​111​​.
  1. Apply the column operation C1→C1+C2.C_1 \to C_1 + C_2.C1​→C1​+C2​. Then xi2−xi+xi=xi2,x_i^2-x_i+x_i=x_i^2,xi2​−xi​+xi​=xi2​, so the determinant becomes
D=∣x12x11x22x21x32x31∣.D=\begin{vmatrix} x_1^2 & x_1 & 1\\ x_2^2 & x_2 & 1\\ x_3^2 & x_3 & 1 \end{vmatrix}.D=​x12​x22​x32​​x1​x2​x3​​111​​.
  1. This is a Vandermonde-type determinant. Reverse the columns to match standard form:
∣x12x11x22x21x32x31∣=−∣1x1x121x2x221x3x32∣.\begin{vmatrix} x_1^2 & x_1 & 1\\ x_2^2 & x_2 & 1\\ x_3^2 & x_3 & 1 \end{vmatrix} = -\begin{vmatrix} 1 & x_1 & x_1^2\\ 1 & x_2 & x_2^2\\ 1 & x_3 & x_3^2 \end{vmatrix}.​x12​x22​x32​​x1​x2​x3​​111​​=−​111​x1​x2​x3​​x12​x22​x32​​​.

The minus sign appears because swapping column 1 and column 3 requires one interchange.

  1. Now use the Vandermonde determinant formula:
∣1x1x121x2x221x3x32∣=(x2−x1)(x3−x1)(x3−x2).\begin{vmatrix} 1 & x_1 & x_1^2\\ 1 & x_2 & x_2^2\\ 1 & x_3 & x_3^2 \end{vmatrix} =(x_2-x_1)(x_3-x_1)(x_3-x_2).​111​x1​x2​x3​​x12​x22​x32​​​=(x2​−x1​)(x3​−x1​)(x3​−x2​).

Here, x2−x1=(a+3)−(a+2)=1,x_2-x_1=(a+3)-(a+2)=1,x2​−x1​=(a+3)−(a+2)=1, x3−x1=(a+4)−(a+2)=2,x_3-x_1=(a+4)-(a+2)=2,x3​−x1​=(a+4)−(a+2)=2, x3−x2=(a+4)−(a+3)=1.x_3-x_2=(a+4)-(a+3)=1.x3​−x2​=(a+4)−(a+3)=1. Hence,

∣1x1x121x2x221x3x32∣=1⋅2⋅1=2.\begin{vmatrix} 1 & x_1 & x_1^2\\ 1 & x_2 & x_2^2\\ 1 & x_3 & x_3^2 \end{vmatrix}=1\cdot 2\cdot 1=2.​111​x1​x2​x3​​x12​x22​x32​​​=1⋅2⋅1=2.

Therefore, D=−2.D=-2.D=−2.

  1. Checking options:
  • A: −2-2−2 ✓
  • B: 000 ✗
  • C: (a+2)(a+3)(a+4)(a+2)(a+3)(a+4)(a+2)(a+3)(a+4) ✗
  • D: (a+1)(a+2)(a+3)(a+1)(a+2)(a+3)(a+1)(a+2)(a+3) ✗

So the correct option is A.

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