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Properties of Determinant
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Properties of Determinant
Last updated at May 29, 2023 by Teachoo
Question 11 By using properties of determinants, show that: (ii) |β 8(x+y+2z&x&[email protected]&y+z+2x&[email protected]&x&z+x+2y)| = 2(x + y + z)3 Solving L.H.S |β 8(x+y+2z&x&[email protected]&y+z+2x&[email protected]&x&z+x+2y)| Applying C1 β C1 + C2 + C3 = |β 8(π₯+π¦+2π§+π₯+π¦&π₯&π¦@π§+π¦+π§+2π₯+π¦&π¦+π§+2π₯&π¦@π§+π₯+π§+π₯+2π¦&π₯&π§+π₯+2π¦)| = |β 8(π(π+π+π)&π₯&π¦@π(π+π+π)&y+π§+2π₯&y@π(π+π+π)&x&z+x+2y)| Taking common 2(π₯+π¦+π§) from C1 = π(π±+π²+π³) |β 8(1&π₯&π¦@1&y+π§+2π₯&[email protected]&x&z+x+2y)| Applying R2 β R2 β R3 = 2(x+y+z)|β 8(1&π₯&π¦@πβπ&y+π§+2π₯βπ₯&yβ(π§+π₯+2π¦)@1&x&z+x+2y)| = 2(x+y+z)|β 8(1&π₯&π¦@π&π₯+π¦+π§&βπ₯βπ¦βπ§@1&x&z+x+2y)| = 2(x+y+z)|β 8(1&π₯&π¦@0&(π+π+π)&β(π+π+π)@1&x&z+x+2y)| Taking common (π₯+π¦+π§) from 2nd Row = 2(x+y+z)(x+y+z)|β 8(1&π₯&π¦@0&1&β[email protected]&x&z+x+2y)| Applying R3 β R3 β R1 = 2(x+y+z)2|β 8(1&π₯&π¦@0&1&β1@πβπ&xβπ₯&z+x+2yβy)| = 2(x+y+z)2|β 8(1&π₯&π¦@0&1&β1@π&0&x+y+z)| Taking common (π₯+π¦+π§) Common from 3rd Row = 2(x+y+z)2(x+y+z)|β 8(1&π₯&π¦@0&1&β[email protected]&0&1)| Expanding Determinant along C1 = 2(x+y+z)3 ( 1|β 8(1&β[email protected]&1)|β0|β 8(π₯&π¦@0&1)|+0|β 8(x&[email protected]&β1)|) = 2(x+y+z)3 ( 1|β 8(1&β[email protected]&1)|β0+0) = 2(x+y+z)3 (1(1β0)βπ₯(0)+π¦(0)) = 2(x+y+z)3 (1) = 2(x+y+z)3 = R.H.S Hence proved