EXAMPLE 39 Using the properties of ...
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EXAMPLE 39 Using the properties of determinants, prove that la +b btc cra la b c b+c cta a+b=2 b c a. lcra a+b b+c le a b

11th - 12th Class
Maths
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sourtion We have [ begin{aligned} L H S &=left|begin{array}{lll} a+b & b+c & c+a b+c & c+a & a+b c+a & a+b & b+c end{array}right| &=left|begin{array}{ccc} -2 c & b+c & c+a -2 a & c+a & a+b -2 b & a+b & b+c end{array}right| text { [applying } C_{1} rightarrow C_{1}-left(C_{2}+C_{3}right) end{aligned} ] ( left.=(-2)left|begin{array}{ccc}c & b+c & c+a a & c+a & a+b b & a+b & b+cend{array}right| text { [taking out }(-2) text { common from } C_{1}right] ) ( left.=(-2) cdotleft|begin{array}{lll}c & b & a a & c & b b & a & cend{array}right| text { [applying } C_{2} rightarrow C_{2}-C_{1} text { and } C_{3} rightarrow C_{3}-C_{1}right] ) [ begin{aligned} &=2left|begin{array}{lll} a & b & c b & c & a c & a & b end{array}right| &left[text { applying } C_{1} leftrightarrow C_{3}right] =& text { RHS. } end{aligned} ] Hence, ( mathrm{LHS}=mathrm{RHS} )
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