Question

The maximum value of the determinant among all 2 x 2 real symmetric matrices
with trace 10 is
a)20
b)None
c) 25
d)36​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Concept:

A matrix is symmetric if and given that it's capable its transpose. All entries above the most diagonal of a symmetric matrix are reflected into equal entries below the diagonal.

Given:

Given that the $2\times 2$ real symmetric matrices with trace $10$.

Find:

We have to seek out the maximum value of the determinant.

Solution:

Let the symmetric matrix be $X=\left[\begin{array}{ll}b& a\\ a& c\end{array}\right]$

Now, we'll find the determinant of the symmetric matrix, we get

$|X|=bc-a^2$              .....(1)

Since the given matrix is real, to urgue the maximum determinant value $a$ should up to zero.

$b+c=10\\c=10-b$

now, substitute the worth of $a$ and $c$ in equation (1), we get

$|X|=b(10-b)\\ |X|=10b-b^2$

Further, we'll differentiate either side with regard to $b,$ we get

$|X|'=10-2b=0\\b=5$

Now, we'll again differentiate, we get

$|X|''=-2 < 0$

At $b=5$ it'll have maximum value

$c=10-5\\c=5$

Now, we'll substitute these values in equation (1), we get

$|X|=5\times 5-0\\|X|=25$

Hence, the maximum value is $25$ and option (C) is correct.

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