{"id":20042,"date":"2024-04-15T05:47:35","date_gmt":"2024-04-15T05:47:35","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20042"},"modified":"2024-04-15T05:47:35","modified_gmt":"2024-04-15T05:47:35","slug":"consider-the-following-matrix-texta-left-beginarray20c-23-textxtexty-endarray-right-if-the-eigen-values-of-a-are-4-and-8-then-a-x-4-y-10","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/consider-the-following-matrix-texta-left-beginarray20c-23-textxtexty-endarray-right-if-the-eigen-values-of-a-are-4-and-8-then-a-x-4-y-10\/","title":{"rendered":"Consider the following matrix. \\[{\\text{A}} = \\left[ {\\begin{array}{*{20}{c}} 2&3 \\\\ {\\text{x}}&{\\text{y}} \\end{array}} \\right]\\] If the eigen values of A are 4 and 8, then A. x = 4, y = 10 B. x = 5, y = 8 C. x = -3, y = 9 D. x = -4, y = 10"},"content":{"rendered":"

[amp_mcq option1=”x = 4, y = 10″ option2=”x = 5, y = 8″ option3=”x = -3, y = 9″ option4=”x = -4, y = 10″ correct=”option3″]<\/p>\n

The correct answer is $\\boxed{\\text{C}}$.<\/p>\n

The eigenvalues of a matrix are the roots of its characteristic polynomial. The characteristic polynomial of a matrix $A$ is given by $$p(x) = \\det(xI – A)$$ where $I$ is the identity matrix.<\/p>\n

In this case, we have $$p(x) = \\det \\left( \\begin{array}{cc} x – 2 & -3 \\\\ x & x – y \\end{array} \\right) = x^2 – (2+y)x + 2y – 6$$<\/p>\n

We are given that the eigenvalues of $A$ are 4 and 8. Therefore, we must have $$4^2 – (2+y)4 + 2y – 6 = 0$$ and $$8^2 – (2+y)8 + 2y – 6 = 0$$<\/p>\n

Solving these equations, we find that $y = -3$ and $x = 9$.<\/p>\n

Therefore, the correct answer is $\\boxed{\\text{C}}$.<\/p>\n

Here is a brief explanation of each option:<\/p>\n