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Matrices and Determinants question

2010 · Shift 0 · Q38
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  5. /2010 · Shift 0 · Q38

Matrices and Determinants question

2010 · Shift 0 · Q38

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let AAA be a  2×2\,2 \times 22×2 matrix with non-zero entries and let A2=I,{A^2} = I,A2=I, where III is 2×22 \times 22×2 identity matrix. Define Tr(A)=Tr(A)=Tr(A)= sum of diagonal elements of AAA and ∣A∣=\left| A \right| =∣A∣= determinant of matrix AAA. Statement- 1: Tr(A)=0Tr(A)=0Tr(A)=0. Statement- 2: ∣A∣=1\left| A \right| = 1∣A∣=1 .
  1. A
    statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1.
  2. B
    statement - 1 is true, statement - 2 is false.
  3. C
    statement - 1 is false, statement -2 is true
  4. D
    statement -1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1.
View written solutionFree

Correct answer: B

  1. Let A=(abcd),A=\begin{pmatrix} a & b \\ c & d \end{pmatrix},A=(ac​bd​), where all entries are non-zero, and given A2=I.A^2=I.A2=I.

  2. Compute A2A^2A2:

    =\begin{pmatrix} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc \end{pmatrix}.$$ Since $A^2=I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$, we get: $$a^2+bc=1 \quad ...(1)$$ $$b(a+d)=0 \quad ...(2)$$ $$c(a+d)=0 \quad ...(3)$$ $$d^2+bc=1 \quad ...(4)$$

  3. Because all entries are non-zero, in particular b≠0b\neq 0b=0 and c≠0c\neq 0c=0. So from (2) and (3), a+d=0.a+d=0.a+d=0. Hence, Tr⁡(A)=a+d=0.\operatorname{Tr}(A)=a+d=0.Tr(A)=a+d=0.

    Therefore, Statement 1 is true.

  4. Now check the determinant. For any matrix, det⁡(A2)=det⁡(I)=1.\det(A^2)=\det(I)=1.det(A2)=det(I)=1. Thus, (det⁡A)2=1,\big(\det A\big)^2=1,(detA)2=1, so det⁡A=±1.\det A=\pm 1.detA=±1.

    This does not force det⁡A=1\det A=1detA=1.

  5. In fact, using d=−ad=-ad=−a from Step 3, det⁡(A)=ad−bc=−a2−bc.\det(A)=ad-bc=-a^2-bc.det(A)=ad−bc=−a2−bc. From (1), a2+bc=1  ⟹  −a2−bc=−1.a^2+bc=1 \implies -a^2-bc=-1.a2+bc=1⟹−a2−bc=−1. Hence, ∣A∣=−1.|A|=-1.∣A∣=−1.

    Therefore, Statement 2 is false.

  6. Final conclusion:

    • Statement 1: true
    • Statement 2: false

    So the correct option is B.\boxed{\text{B}}.B​.

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