Question 1 :
Without expanding the determinant, prove that
Solution :
To prove the given determinant as zero, we may try to show any two rows or columns as identical.
For that we use elementary operations.
Now we may factor out "s" and "(a^{2} + b^{2} + c^{2})" from the 1^{st} and and 3^{rd} column respectively.
So, we get
If any two columns are identical, then the value of determinant will become zero.
= s(a^{2} + b^{2} + c^{2}) (0)
= 0
Hence proved.
Question 2 :
Show that
Solution :
In order to show any two rows or columns are same, let us multiply "a", "b" and "c" by the 1^{st}, 2^{nd} and 3^{rd} row respectively.
Now we may factor abc from 2^{nd} and 3^{rd} column respectively.
Since column 1 and 2 are identical, the value of determinant will become 0.
So, we get (abc)^{2} (ab + bc + ca) (0).
Hence the answer is 0.
Question 3 :
Prove that
Solution :
First let us factor out a from a^{2} + ab, b from b^{2} + bc and c from c^{2} + ac respectively.
Factor out a, b and c from column 1, 2 and 3 respectively.
So, we get
Now we may factor out 2 from the first row.So, we get
Now let us subtract column 3 from column 2. So we get
= 2abc [ c(ab + b^{2} - b(b - a)) ]
= 2abc [ c(ab + b^{2} - b^{2} + ab) ]
= 2abc [ 2abc ]
= 4a^{2}b^{2}c^{2}
Hence proved.
Question 4 :
Prove that
Solution :
First let us subtract second row from 1^{st} row.
Since we cannot simplify this hereafter, let us find the expansion.
= a [(1+b)(1+c) - 1] + b[(1+c) - 1]
= a [(1 + c + b + bc) - 1] + b[1+c-1]
= a [c + b + bc] + b[c]
= ac + ab + abc + bc
= abc +ab + bc + ca
= abc (1 + 1/a + 1/b + 1/c)
Hence proved.
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