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Matrices and Determinants question

2020 · 3 Sep · Shift 1 · Q25
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  5. /2020 · 3 Sep · Shift 1 · Q25

Matrices and Determinants question

2020 · 3 Sep · Shift 1 · Q25

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If Δ\DeltaΔ=∣x−22x−33x−42x−33x−44x−53x−55x−810x−17∣\left| {\begin{matrix} {x - 2} & {2x - 3} & {3x - 4} \\ {2x - 3} & {3x - 4} & {4x - 5} \\ {3x - 5} & {5x - 8} & {10x - 17} \\ \end{matrix} } \right|​x−22x−33x−5​2x−33x−45x−8​3x−44x−510x−17​​ = Ax3 + Bx2 + Cx + D, then B + C is equal to :
  1. A
    -1
  2. B
    -3
  3. C
    9
  4. D
    1
View written solutionFree

Correct answer: B

  1. Let
Δ=∣x−22x−33x−42x−33x−44x−53x−55x−810x−17∣\Delta=\begin{vmatrix} x-2 & 2x-3 & 3x-4\\ 2x-3 & 3x-4 & 4x-5\\ 3x-5 & 5x-8 & 10x-17 \end{vmatrix}Δ=​x−22x−33x−5​2x−33x−45x−8​3x−44x−510x−17​​

and suppose

Δ=Ax3+Bx2+Cx+D.\Delta=Ax^3+Bx^2+Cx+D.Δ=Ax3+Bx2+Cx+D.

We need B+CB+CB+C.

  1. Compute the determinant by expanding along the first row:
Δ=(x−2)∣3x−44x−55x−810x−17∣−(2x−3)∣2x−34x−53x−510x−17∣+(3x−4)∣2x−33x−43x−55x−8∣.\Delta=(x-2)\begin{vmatrix}3x-4 & 4x-5\\ 5x-8 & 10x-17\end{vmatrix} -(2x-3)\begin{vmatrix}2x-3 & 4x-5\\ 3x-5 & 10x-17\end{vmatrix} +(3x-4)\begin{vmatrix}2x-3 & 3x-4\\ 3x-5 & 5x-8\end{vmatrix}.Δ=(x−2)​3x−45x−8​4x−510x−17​​−(2x−3)​2x−33x−5​4x−510x−17​​+(3x−4)​2x−33x−5​3x−45x−8​​.
  1. Evaluate each 2×22\times 22×2 minor.

First minor:

(3x−4)(10x−17)−(4x−5)(5x−8).(3x-4)(10x-17)-(4x-5)(5x-8).(3x−4)(10x−17)−(4x−5)(5x−8).

Now,

(3x−4)(10x−17)=30x2−91x+68,(3x-4)(10x-17)=30x^2-91x+68,(3x−4)(10x−17)=30x2−91x+68, (4x−5)(5x−8)=20x2−57x+40.(4x-5)(5x-8)=20x^2-57x+40.(4x−5)(5x−8)=20x2−57x+40.

So,

M1=10x2−34x+28.M_1=10x^2-34x+28.M1​=10x2−34x+28.

Second minor:

(2x−3)(10x−17)−(4x−5)(3x−5).(2x-3)(10x-17)-(4x-5)(3x-5).(2x−3)(10x−17)−(4x−5)(3x−5).

Now,

(2x−3)(10x−17)=20x2−64x+51,(2x-3)(10x-17)=20x^2-64x+51,(2x−3)(10x−17)=20x2−64x+51, (4x−5)(3x−5)=12x2−35x+25.(4x-5)(3x-5)=12x^2-35x+25.(4x−5)(3x−5)=12x2−35x+25.

So,

M2=8x2−29x+26.M_2=8x^2-29x+26.M2​=8x2−29x+26.

Third minor:

(2x−3)(5x−8)−(3x−4)(3x−5).(2x-3)(5x-8)-(3x-4)(3x-5).(2x−3)(5x−8)−(3x−4)(3x−5).

Now,

(2x−3)(5x−8)=10x2−31x+24,(2x-3)(5x-8)=10x^2-31x+24,(2x−3)(5x−8)=10x2−31x+24, (3x−4)(3x−5)=9x2−27x+20.(3x-4)(3x-5)=9x^2-27x+20.(3x−4)(3x−5)=9x2−27x+20.

So,

M3=x2−4x+4=(x−2)2.M_3=x^2-4x+4=(x-2)^2.M3​=x2−4x+4=(x−2)2.
  1. Substitute back:
Δ=(x−2)(10x2−34x+28)−(2x−3)(8x2−29x+26)+(3x−4)(x2−4x+4).\Delta=(x-2)(10x^2-34x+28)-(2x-3)(8x^2-29x+26)+(3x-4)(x^2-4x+4).Δ=(x−2)(10x2−34x+28)−(2x−3)(8x2−29x+26)+(3x−4)(x2−4x+4).

Expand each term.

First term:

(x−2)(10x2−34x+28)=10x3−54x2+96x−56.(x-2)(10x^2-34x+28)=10x^3-54x^2+96x-56.(x−2)(10x2−34x+28)=10x3−54x2+96x−56.

Second term:

(2x−3)(8x2−29x+26)=16x3−82x2+139x−78.(2x-3)(8x^2-29x+26)=16x^3-82x^2+139x-78.(2x−3)(8x2−29x+26)=16x3−82x2+139x−78.

Hence,

−(2x−3)(8x2−29x+26)=−16x3+82x2−139x+78.-(2x-3)(8x^2-29x+26)=-16x^3+82x^2-139x+78.−(2x−3)(8x2−29x+26)=−16x3+82x2−139x+78.

Third term:

(3x−4)(x2−4x+4)=3x3−16x2+28x−16.(3x-4)(x^2-4x+4)=3x^3-16x^2+28x-16.(3x−4)(x2−4x+4)=3x3−16x2+28x−16.
  1. Add all terms:
Δ=(10x3−54x2+96x−56)+(−16x3+82x2−139x+78)+(3x3−16x2+28x−16).\Delta=(10x^3-54x^2+96x-56)+(-16x^3+82x^2-139x+78)+(3x^3-16x^2+28x-16).Δ=(10x3−54x2+96x−56)+(−16x3+82x2−139x+78)+(3x3−16x2+28x−16).

So,

Δ=−3x3+12x2−15x+6.\Delta=-3x^3+12x^2-15x+6.Δ=−3x3+12x2−15x+6.

Thus,

A=−3,B=12,C=−15,D=6.A=-3,\quad B=12,\quad C=-15,\quad D=6.A=−3,B=12,C=−15,D=6.

Therefore,

B+C=12+(−15)=−3.B+C=12+(-15)=-3.B+C=12+(−15)=−3.
  1. Checking options:
  • A: −1-1−1
  • B: −3-3−3 ✅
  • C: 999
  • D: 111

Hence the correct option is B.

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