InterviewSolution
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InterviewSolution
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By using properties of determinants. Show that:(i) `|[a-b-c,2a,2a],[2b,b-c-a,2b],[2c,2c,c-a-b]|=(a+b+c)^3` (ii) `|[x+y+2z, x, y],[ z, y+z+2x, y],[ z, x, z+x+2y]|=2(x+y+z)^3` |
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Answer» (i) `L.H.S. = |[a-b-c,2a,2a],[2b,b-c-a,2b],[2c,2c,c-a-b]|` Applying `R_1->R_1+R_2+R_3` `=|[a+b+c,a+b+c,a+b+c],[2b,b-c-a,2b],[2c,2c,c-a-b]|` `=(a+b+c)|[1,1,1],[2b,b-c-a,2b],[2c,2c,c-a-b]|` Applying `C_2->C_2-C_1 and C_3->C_3-C_1` `=(a+b+c)|[1,0,0],[2b,-(a+b+c),0],[2c,0,-(a+b+c)]|` `=(a+b+c)(-(a+b+c))(-(a+b+c))|[1,0,0],[2b,1,0],[2c,0,1]|` `=(a+b+c)^3[1(1-0)-0+0]` `= (a+b+c)^3 = R.H.S.` Following the same steps, we can prove the second part. |
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