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GUJCET 2025 Math Paper

Question : 17 of 40

Marks: +1, -0

If the matrix

\begin{bmatrix} x & x^2+3x & 5 \\ -2x-6 & x^2 & -4x-2 \\ 5 & x^2+2 & x^3 \end{bmatrix}

x

Solution:

Concept:

A matrix is symmetric if

A^T=A

, so every element satisfies

A_{ij}=A_{ji}

Explanation:

Compare the corresponding off-diagonal elements.

A_{12}=A_{21}

gives:

x^2+3x=-2x-6

x^2+5x+6=0

(x+2)(x+3)=0

So,

x=-2

x=-3

Now compare

A_{23}=A_{32}

-4x-2=x^2+2

x^2+4x+4=0

(x+2)^2=0

This gives

x=-2

only.

Both conditions must be true together, so the common value is

x=-2

Answer:

The value of

x

-2

, which matches option A.

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