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We are given that | A | = 4 ...(1)

Since A is of order 3 × 3

∴ | 3 A | = 3³ |A| = 27 |A|

= 27 × 4 [∵ of (1)]

= 108

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32. Which of the following is correct

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33.

Solve space for space straight x space colon open square brackets table row cell 2 straight x end cell cell space 0 end cell cell space space 1 end cell row 0 cell space 1 end cell cell space space 2 end cell row cell negative 1 end cell cell space 2 end cell cell space space 0 end cell end table close square brackets space equals space open vertical bar table row 1 cell space space space space 0 end cell cell space space space 0 end cell row 2 cell space space space space 3 end cell cell space minus 4 end cell row 3 cell negative 3 end cell cell space space 5 end cell end table close vertical bar

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34.

If space open square brackets table row 1 cell space space space 1 end cell cell space space minus 2 end cell row 2 cell space space 1 end cell cell space space minus 3 end cell row 5 cell space space 4 end cell cell space space minus 9 end cell end table close square brackets space space Find space open vertical bar space straight A space close vertical bar

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35.

If space space straight A equals open square brackets table row cell 1 space space end cell cell space space 0 end cell cell space space space space space 1 end cell row 0 cell space space 1 end cell cell space space space space space 2 end cell row 0 cell space space 0 end cell cell space space space space 4 end cell end table close square brackets

then show that | 3A | = 27 | A |.

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36.

Syntax error from line 1 column 244 to line 1 column 251.

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37.

Let space space space space space straight A equals space open square brackets table row 1 cell sin space straight theta end cell 1 row cell negative sin space straight theta end cell 1 cell sin space straight theta end cell row cell negative 1 end cell cell negative sin space straight theta end cell 1 end table close square brackets space comma space w h e r e space 0 space element of space ө space 2 space pi. space T h e n

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38. The value of the determinant remains unchanged if its rows and columns are interchanged.

Verify space the space property space for space space increment space equals space open vertical bar table row 2 cell space space minus 3 end cell cell space space 5 end cell row 6 cell space space space 0 end cell cell space space 4 end cell row 1 cell space space space 5 end cell cell space 7 end cell end table close vertical bar

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39. If any two rows of a determinant are interchanged, then sign of determinant

verify space this space property space for space increment space equals space open vertical bar table row 2 cell space space minus 3 end cell cell space space space space space space 5 end cell row 6 cell space space space space space 0 end cell cell space space space space space space 4 end cell row 1 cell space space space space 5 end cell cell space space minus 7 end cell end table close vertical bar

changes.

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40. Without expanding, prove that

open vertical bar table row 9 cell space space space space space space 9 end cell cell space space 12 end cell row 1 cell space space minus 3 end cell cell space space minus 4 end cell row 1 cell space space space space space space 9 end cell cell space 12 end cell end table close vertical bar space equals 0

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