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Matrices and Determinants question

2025 · 4 Apr · Shift 2 · Q31
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Matrices and Determinants question

2025 · 4 Apr · Shift 2 · Q31

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let the matrix A=[100101010]A=\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]A=​110​001​010​​ satisfy An=An−2+A2−IA^n=A^{n-2}+A^2-IAn=An−2+A2−I for n⩾3n \geqslant 3n⩾3. Then the sum of all the elements of A50\mathrm{A}^{50}A50 is :
  1. A
    44
  2. B
    39
  3. C
    52
  4. D
    53
View written solutionFree

Correct answer: D

  1. First compute small powers of AAA and identify a pattern.

Given

A=[100101010].A=\begin{bmatrix}1&0&0\\1&0&1\\0&1&0\end{bmatrix}.A=​110​001​010​​.

We are also told that for n≥3n\ge 3n≥3,

An=An−2+A2−I.A^n=A^{n-2}+A^2-I.An=An−2+A2−I.

We need the sum of all elements of A50A^{50}A50.

  1. Compute A2A^2A2.
A2=A⋅A=[100101010][100101010]=[100110101].A^2=A\cdot A =\begin{bmatrix}1&0&0\\1&0&1\\0&1&0\end{bmatrix} \begin{bmatrix}1&0&0\\1&0&1\\0&1&0\end{bmatrix} =\begin{bmatrix} 1&0&0\\ 1&1&0\\ 1&0&1 \end{bmatrix}.A2=A⋅A=​110​001​010​​​110​001​010​​=​111​010​001​​.

Hence

A2−I=[000100100].A^2-I= \begin{bmatrix} 0&0&0\\ 1&0&0\\ 1&0&0 \end{bmatrix}.A2−I=​011​000​000​​.

Let

B=A2−I.B=A^2-I.B=A2−I.

Then the recurrence becomes

An=An−2+B(n≥3).A^n=A^{n-2}+B \qquad (n\ge 3).An=An−2+B(n≥3).
  1. Use the recurrence separately for even powers.

For even powers,

A2k=A2k−2+B.A^{2k}=A^{2k-2}+B.A2k=A2k−2+B.

Starting from A2A^2A2, we get

A4=A2+B,A^4=A^2+B,A4=A2+B, A6=A4+B=A2+2B,A^6=A^4+B=A^2+2B,A6=A4+B=A2+2B,

and in general,

A2k=A2+(k−1)B.A^{2k}=A^2+(k-1)B.A2k=A2+(k−1)B.

Since B=A2−IB=A^2-IB=A2−I,

A2k=A2+(k−1)(A2−I)=kA2−(k−1)I.A^{2k}=A^2+(k-1)(A^2-I)=kA^2-(k-1)I.A2k=A2+(k−1)(A2−I)=kA2−(k−1)I.

For 50=2⋅2550=2\cdot 2550=2⋅25, we have k=25k=25k=25, so

A50=25A2−24I.A^{50}=25A^2-24I.A50=25A2−24I.
  1. Compute A50A^{50}A50 explicitly.

Since

25A2=25[100110101]=[25002525025025],25A^2=25\begin{bmatrix}1&0&0\\1&1&0\\1&0&1\end{bmatrix} =\begin{bmatrix}25&0&0\\25&25&0\\25&0&25\end{bmatrix},25A2=25​111​010​001​​=​252525​0250​0025​​,

and

24I=[240002400024],24I=\begin{bmatrix}24&0&0\\0&24&0\\0&0&24\end{bmatrix},24I=​2400​0240​0024​​,

we get

A50=25A2−24I=[10025102501].A^{50}=25A^2-24I =\begin{bmatrix}1&0&0\\25&1&0\\25&0&1\end{bmatrix}.A50=25A2−24I=​12525​010​001​​.
  1. Find the sum of all entries.

Sum of all elements:

1+0+0+25+1+0+25+0+1=53.1+0+0+25+1+0+25+0+1=53.1+0+0+25+1+0+25+0+1=53.
  1. Compare with options.

The required sum is

53.53.53.

So the correct option is D.

  1. Verification with stored answer.

Stored correct answer: D

Our derived answer: D

They match.

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