Kerala KEAM 5 June 2024 Solved Paper

We are given that $$A = \begin{pmatrix} x & 22 & x \end{pmatrix}$$and $$\det(A^2) = 25.$$We are asked to find the value of $x$.Step 1: Recall that the determinant of a matrix product is the product of the determinants:$$\det(A^2) = \det(A) \cdot \det(A).$$Thus, we need to first calculate $\det(A)$.Step 2: The determinant of $A$ is given by: $$\det(A) = \det \begin{pmatrix} x & 22 & x \end{pmatrix} = (x)(x) - (2)(2) = x^2 - 4.$$Step 3: Now, we can calculate $\det(A^2)$: $$\det(A^2) = (x^2 - 4)^2.$$We are given that $\det(A^2) = 25$, so: $$(x^2 - 4)^2 = 25.$$Step 4: Taking the square root of both sides: $$x^2 - 4 = \pm 5.$$Step 5: Solving for $x^2$:- If $x^2 - 4 = 5$, then $x^2 = 9$, so $x = \pm 3$.- If $x^2 - 4 = -5$, then $x^2 = -1$, which has no real solutions.Thus, the only possible solution is $x = \pm 3$.Therefore, the correct answer is option (A).Quick Tip: When dealing with determinants, remember that $\det(A^2) = (\det(A))^2$. This simplifies the problem of finding values for $x$ by equating the determinant expression to the given value.