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Matrices and Determinants question

2023 · 31 Jan · Shift 1 · Q36
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  5. /2023 · 31 Jan · Shift 1 · Q36

Matrices and Determinants question

2023 · 31 Jan · Shift 1 · Q36

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let A=(10004−1012−3)A = \left( {\begin{matrix} 1 & 0 & 0 \\ 0 & 4 & { - 1} \\ 0 & {12} & { - 3} \\ \end{matrix} } \right)A=​100​0412​0−1−3​​. Then the sum of the diagonal elements of the matrix (A+I)11{(A + I)^{11}}(A+I)11 is equal to :
  1. A
    4094
  2. B
    2050
  3. C
    6144
  4. D
    4097
View written solutionFree

Correct answer: D

  1. We need the sum of diagonal elements of (A+I)11(A+I)^{11}(A+I)11, i.e. the trace of (A+I)11(A+I)^{11}(A+I)11.

Given

A=(10004−1012−3)A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix}A=​100​0412​0−1−3​​

so

A+I=(20005−1012−2).A+I = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 5 & -1 \\ 0 & 12 & -2 \end{pmatrix}.A+I=​200​0512​0−1−2​​.
  1. Observe that A+IA+IA+I is block diagonal in the form
A+I=(200B),where B=(5−112−2).A+I = \begin{pmatrix} 2 & 0 \\ 0 & B \end{pmatrix}, \quad \text{where } B=\begin{pmatrix}5 & -1 \\ 12 & -2\end{pmatrix}.A+I=(20​0B​),where B=(512​−1−2​).

Hence,

(A+I)11=(21100B11).(A+I)^{11} = \begin{pmatrix} 2^{11} & 0 \\ 0 & B^{11} \end{pmatrix}.(A+I)11=(2110​0B11​).

Therefore,

tr⁡((A+I)11)=211+tr⁡(B11).\operatorname{tr}((A+I)^{11}) = 2^{11} + \operatorname{tr}(B^{11}).tr((A+I)11)=211+tr(B11).
  1. Now find eigenvalues of BBB.

Characteristic polynomial:

det⁡(B−λI)=∣5−λ−112−2−λ∣=(5−λ)(−2−λ)+12.\det(B-\lambda I)= \begin{vmatrix} 5-\lambda & -1 \\ 12 & -2-\lambda \end{vmatrix} =(5-\lambda)(-2-\lambda)+12.det(B−λI)=​5−λ12​−1−2−λ​​=(5−λ)(−2−λ)+12.

Expanding,

(5−λ)(−2−λ)+12=−10−5λ+2λ+λ2+12=λ2−3λ+2.(5-\lambda)(-2-\lambda)+12 = -10-5\lambda+2\lambda+\lambda^2+12 = \lambda^2-3\lambda+2.(5−λ)(−2−λ)+12=−10−5λ+2λ+λ2+12=λ2−3λ+2.

So,

λ2−3λ+2=(λ−1)(λ−2).\lambda^2-3\lambda+2=(\lambda-1)(\lambda-2).λ2−3λ+2=(λ−1)(λ−2).

Thus the eigenvalues of BBB are 111 and 222.

  1. Therefore the eigenvalues of B11B^{11}B11 are 1111^{11}111 and 2112^{11}211, so
tr⁡(B11)=111+211=1+2048=2049.\operatorname{tr}(B^{11})=1^{11}+2^{11}=1+2048=2049.tr(B11)=111+211=1+2048=2049.
  1. Hence,
tr⁡((A+I)11)=211+2049=2048+2049=4097.\operatorname{tr}((A+I)^{11})=2^{11}+2049=2048+2049=4097.tr((A+I)11)=211+2049=2048+2049=4097.
  1. So the required sum of diagonal elements is
4097.\boxed{4097}.4097​.
  1. Checking options:
  • A: 409440944094 ❌
  • B: 205020502050 ❌
  • C: 614461446144 ❌
  • D: 409740974097 ✅

Thus the correct option is D.

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