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Let a and b be real numbers.Show that there exists a unique 2×2 real symmetric matrix A with trace((A))=a and det((A))=b if and only if ((a^2))=4b.

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Let us consider a 2*2 real symmetric matrix of the form, A=

x  y

y  z

where x,y,z belongs to real line.

Given: trace(A)=x+z=a….1)

det(A)=xz-y^2=b……2)

From 1) we have z=a-x and putting it in 2) we get, x^2-ax+(b+y^2)=0(a quadratic in x). Solving by Sridhar Acharya’s Method for x we will obtain two values for x, which cannot take place since the matrix is unique, hence the discriminant D=a^2 – 4(b+y^2) = 0, i.e., y=+- square root(a^2-4b)/2.

Since A is to be an unique 2*2 Symmetric real matrix , this can happen iff y=0; i.e., iff a^2=4b. (PROVED).

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