1.

`(i+j+3k)x+(3i-3j+k)y+(-4i+5j)z=lambda(x i+yj+zk)` then lambda equal to

Answer» `(hati+hatj+3hatk)x+(3hati-3hatj+hatk)y +(-4hati+5hatj)z = lambda(xhati+yhatj+zhatk)`
Comparing coefficient of `hati, hatj and hatk`
`x+3y-4z = lambda x =>(1-lambda)x+3y-4z = 0->(1)`
`x-3y+5z = lambda y => x+(-3-lambda)y+5z = 0->(2)`
`3x+y = lambda z => 3x+y-lambdaz = 0->(3)`
Now, solving (1),(2),(3) using determinant,
`|[1-lambda,3,-4],[1,-3-lambda,5],[3,1,-lambda]| = 0`
`=>[(1-lambda)(3lambda+lambda^2-5)-3(-lambda-15)-4(1+9+3lambda)] = 0`
`=>lambda^2+3lambda-5-lambda^3-3lambda^2+5lambda+3lambda+45-40-12lambda = 0`
`=>-lambda^3-2lambda^2-lambda = 0`
`=>lambda(lambda^2+2lambda+1) = 0`
`=>lambda(lambda+1)^2 = 0`
`=>:. lambda = 0 and lambda = -1`


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