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CBSE Class 12-commerce Answered

straight i space determine space the space value space of open vertical bar straight A close vertical bar space and space by space using space the space property space of space determinant space find space the space value space of space determinant table row 4 3 2 row 2 5 8 row 3 1 1 end table
Asked by rrajeshthapa1998 | 17 Mar, 2019, 12:36: PM
answered-by-expert Expert Answer
begin mathsize 12px style open vertical bar A close vertical bar space equals space open vertical bar table row 4 3 2 row 2 5 8 row 3 1 1 end table close vertical bar space equals space 4 open parentheses 5 space minus space 8 close parentheses space minus 3 open parentheses 2 space minus space 24 close parentheses space plus 2 open parentheses 2 space minus space 15 close parentheses space equals space 28 end style
 
Let us multiply column-2 by 4  and column-3 by 2, hence the determinant becomes 8 times of its initial value
 
begin mathsize 12px style 8 open vertical bar A close vertical bar space equals space open vertical bar table row 4 12 4 row 2 20 16 row 3 4 2 end table close vertical bar end style
Now, in the above determinant let us do, C2-3C1 ( column-1 is multiplied by 3 and subtracted from column-2)
and  C3-C1 (column-1 is subtracted from column-3)
 
begin mathsize 12px style 8 open vertical bar A close vertical bar space equals space open vertical bar table row 4 0 0 row 2 14 14 row 3 cell negative 5 end cell cell negative 1 end cell end table close vertical bar space equals space 4 open parentheses negative 14 plus 70 close parentheses space equals space 4 cross times 56 end style
hence |A| = (4×56)/8 = 28
Answered by Thiyagarajan K | 17 Mar, 2019, 02:06: PM

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CBSE 12-commerce - Maths
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