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Mathematics · Matrices

JEE Advanced 2019 — Paper 2 — Question 22

Let x∈Rx \in \mathbb{R} and let P=[111022003],Q=[2xx040xx6]and R=PQP−1P = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{bmatrix}, \quad Q = \begin{bmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \end{bmatrix} \quad \text{and } R = PQP^{-1} Then which of the following options is/are correct?

  1. Option A:
    For x=0, if R[1ab]=6[1ab], then a+b=5\text{For } x = 0, \text{ if } R \begin{bmatrix} 1 \\ a \\ b \end{bmatrix} = 6 \begin{bmatrix} 1 \\ a \\ b \end{bmatrix}, \text{ then } a + b = 5
    Correct
  2. Option B:
    For x=1, there exists a unit vector αi^+βj^+γk^ for which R[αβγ]=[000]\text{For } x = 1, \text{ there exists a unit vector } \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k} \text{ for which } R \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}
  3. Option C:
    det⁡R=det⁡[2xx040xx5]+8, for all x∈R\det R = \det \begin{bmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 5 \end{bmatrix} + 8, \text{ for all } x \in \mathbb{R}
    Correct
  4. Option D:

    There exists a real number x such that PQ=QP\mathrm{PQ}=\mathrm{QP}

Answer: A, C

Step-by-step solution

R=PQP−1\mathbf{R} = \mathbf{PQP}^{-1} det⁡(R)=det⁡(PQP−1)\det(\mathbf{R}) = \det(\mathbf{PQP}^{-1}) =det⁡(P)det⁡(Q)det⁡(P−1)= \det(\mathbf{P}) \det(\mathbf{Q}) \det(\mathbf{P}^{-1}) =det⁡(P)det⁡(Q)1det⁡(P)= \det(\mathbf{P}) \det(\mathbf{Q}) \frac{1}{\det(\mathbf{P})} =det⁡(Q)= \det(\mathbf{Q}) det⁡(Q)=∣2x2044x06∣=2(24−0)−x(0−4x)+2(0−4x)\det(\mathbf{Q}) = \begin{vmatrix} 2 & x & 2 \\ 0 & 4 & 4 \\ x & 0 & 6 \end{vmatrix} = 2(24 - 0) - x(0 - 4x) + 2(0 - 4x) =48+4x2−8x= 48 + 4x^2 - 8x Given det⁡(R)=det⁡(Q)=48−8x+4x2\text{Given } \det(\mathbf{R}) = \det(\mathbf{Q}) = 48 - 8x + 4x^2 (A) For x=0\text{(A) For } x = 0 P=[112022004],Q=[202044006]\mathbf{P} = \begin{bmatrix} 1 & 1 & 2 \\ 0 & 2 & 2 \\ 0 & 0 & 4 \end{bmatrix}, \mathbf{Q} = \begin{bmatrix} 2 & 0 & 2 \\ 0 & 4 & 4 \\ 0 & 0 & 6 \end{bmatrix} PQ=[241608200024]\mathbf{PQ} = \begin{bmatrix} 2 & 4 & 16 \\ 0 & 8 & 20 \\ 0 & 0 & 24 \end{bmatrix} det⁡(P)=8\det(\mathbf{P}) = 8 P−1=1∣P∣(Adj P)′=18[800−440−4−22]′=18[8−4−404−2002]\mathbf{P}^{-1} = \frac{1}{|\mathbf{P}|} (\text{Adj } \mathbf{P})' = \frac{1}{8} \begin{bmatrix} 8 & 0 & 0 \\ -4 & 4 & 0 \\ -4 & -2 & 2 \end{bmatrix}' = \frac{1}{8} \begin{bmatrix} 8 & -4 & -4 \\ 0 & 4 & -2 \\ 0 & 0 & 2 \end{bmatrix} R=PQP−1=[241608200024]18[8−4−404−2002]=18[160003200048]=[200040006]\mathbf{R} = \mathbf{PQP}^{-1} = \begin{bmatrix} 2 & 4 & 16 \\ 0 & 8 & 20 \\ 0 & 0 & 24 \end{bmatrix} \frac{1}{8} \begin{bmatrix} 8 & -4 & -4 \\ 0 & 4 & -2 \\ 0 & 0 & 2 \end{bmatrix} = \frac{1}{8} \begin{bmatrix} 16 & 0 & 0 \\ 0 & 32 & 0 \\ 0 & 0 & 48 \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \end{bmatrix} Given, R[ab1]=6[ab1]⇒[2a4b6]=[6a6b6]\text{Given, } \mathbf{R} \begin{bmatrix} a \\ b \\ 1 \end{bmatrix} = 6 \begin{bmatrix} a \\ b \\ 1 \end{bmatrix} \Rightarrow \begin{bmatrix} 2a \\ 4b \\ 6 \end{bmatrix} = \begin{bmatrix} 6a \\ 6b \\ 6 \end{bmatrix} ⇒4a=0⇒a=0\Rightarrow 4a = 0 \Rightarrow a = 0 2b=0⇒b=02b = 0 \Rightarrow b = 0 (B) det⁡(R)=48−8x+4x2=40\text{(B) } \det(\mathbf{R}) = 48 - 8x + 4x^2 = 40 4x2−8x+8=04x^2 - 8x + 8 = 0 x2−2x+2=0x^2 - 2x + 2 = 0 x=2±4−82=1±ix = \frac{2 \pm \sqrt{4 - 8}}{2} = 1 \pm i

R=αi+βj+γk\mathbf{R} = \alpha \mathbf{i} + \beta \mathbf{j} + \gamma \mathbf{k} cannot a unit vector.

(C) det⁡(Q)=∣202044008∣=64\text{(C) } \det(\mathbf{Q}) = \begin{vmatrix} 2 & 0 & 2 \\ 0 & 4 & 4 \\ 0 & 0 & 8 \end{vmatrix} = 64 det⁡(R)=64\det(\mathbf{R}) = 64 48−8x+4x2=6448 - 8x + 4x^2 = 64 4x2−8x−16=04x^2 - 8x - 16 = 0 x2−2x−4=0x^2 - 2x - 4 = 0 x=2±4+162=1±5x = \frac{2 \pm \sqrt{4 + 16}}{2} = 1 \pm \sqrt{5} (D) PQ=QP\text{(D) } \mathbf{PQ} = \mathbf{QP} [111022003][2xx040xx6]=[2xx040xx6][111022003]\begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{bmatrix} \begin{bmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \end{bmatrix} = \begin{bmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{bmatrix}

If we equate a12a_{12} from both

x+4+x=2+2xx + 4 + x = 2 + 2x 4+2x=2+2x4 + 2x = 2 + 2x 4=24 = 2

⇒x∈ϕ,\Rightarrow x \in \phi, no value exists.

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2019
Paper
Paper 2
Subject
Mathematics
Chapter
Matrices
Topic
Inverse of a Matrix

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