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Matrices and Determinants question

2021 · 22 Jul · Shift 2 · Q40
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Matrices and Determinants question

2021 · 22 Jul · Shift 2 · Q40

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1
Let A=[010100001]A = \left[ {\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{matrix} } \right]A=​010​100​001​​. Then the number of 3 ×\times× 3 matrices B with entries from the set {1, 2, 3, 4, 5} and satisfying AB = BA is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 3125

  1. We need to count all 3×33\times 33×3 matrices BBB with entries from {1,2,3,4,5}\{1,2,3,4,5\}{1,2,3,4,5} such that AB=BA,AB=BA,AB=BA, where A=[010100001].A=\begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix}.A=​010​100​001​​.

  2. Let B=[abcdefghi],B=\begin{bmatrix}a&b&c\\ d&e&f\\ g&h&i\end{bmatrix},B=​adg​beh​cfi​​, where each of a,b,c,d,e,f,g,h,i∈{1,2,3,4,5}a,b,c,d,e,f,g,h,i\in\{1,2,3,4,5\}a,b,c,d,e,f,g,h,i∈{1,2,3,4,5}.

  3. Compute ABABAB. Since left multiplication by AAA swaps the first two rows and keeps the third row unchanged, AB=[defabcghi].AB=\begin{bmatrix}d&e&f\\ a&b&c\\ g&h&i\end{bmatrix}.AB=​dag​ebh​fci​​.

  4. Compute BABABA. Since right multiplication by AAA swaps the first two columns and keeps the third column unchanged, BA=[bacedfhgi].BA=\begin{bmatrix}b&a&c\\ e&d&f\\ h&g&i\end{bmatrix}.BA=​beh​adg​cfi​​.

  5. Now use the condition AB=BAAB=BAAB=BA:

= \begin{bmatrix}b&a&c\\ e&d&f\\ h&g&i\end{bmatrix}.$$ Equating corresponding entries, - $d=b$ - $e=a$ - $f=c$ - $a=e$ - $b=d$ - $c=f$ - $g=h$ - $h=g$ - $i=i$ So the independent conditions are $$d=b,\quad e=a,\quad f=c,\quad g=h.$$ Thus $B$ must be of the form $$B=\begin{bmatrix}a&b&c\\ b&a&c\\ g&g&i\end{bmatrix}.$$ 6. Count the number of such matrices. The independent entries are: - $a$ : 5 choices - $b$ : 5 choices - $c$ : 5 choices - $g$ : 5 choices - $i$ : 5 choices Hence total number of matrices is $$5^5=3125.$$ 7. Therefore, the required number is $$\boxed{3125}.$$
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