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NDA 2025
Exam
NDA
Year
2025
Subject
Mathematics
Chapter
Matrices

NDA 2025 · Mathematics · Matrices

If A = pmatrix 1 & 2 & 2 2 & 1 & 2 2 & 2 & 1 pmatrix , then what is...

If A=(122212221)A = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix}, then what is A2−4AA^2 - 4A equal to?

Options

  1. A−5I3-5I_3
  2. B−I3-I_3
  3. CI3I_3
  4. D5I35I_3

Correct answer

D. 5I35I_3

Step-by-step solution

Given A=(122212221)A = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix}

First, calculating A2A^2:

A2=(122212221)(122212221)A^2 = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix}

A2=(1+4+42+2+42+4+22+2+44+1+44+2+22+4+24+2+24+4+1)=(988898889)A^2 = \begin{pmatrix} 1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1 \end{pmatrix} = \begin{pmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{pmatrix}

Now, calculating 4A4A:

4A=4(122212221)=(488848884)4A = 4 \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{pmatrix}

Subtracting 4A4A from A2A^2:

A2−4A=(988898889)−(488848884)A^2 - 4A = \begin{pmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{pmatrix} - \begin{pmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{pmatrix}

A2−4A=(500050005)A^2 - 4A = \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix}

A2−4A=5(100010001)=5I3A^2 - 4A = 5 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = 5I_3

Answer: 5I35I_3
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