{"id":19991,"date":"2024-04-15T05:46:56","date_gmt":"2024-04-15T05:46:56","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=19991"},"modified":"2024-04-15T05:46:56","modified_gmt":"2024-04-15T05:46:56","slug":"consider-the-5-a%c2%97-5-matrix-texta-left-beginarray20c-12345-51234-45123-34512-23451-endarray-right-it-is-given-that-a-ha","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/consider-the-5-a%c2%97-5-matrix-texta-left-beginarray20c-12345-51234-45123-34512-23451-endarray-right-it-is-given-that-a-ha\/","title":{"rendered":"Consider the 5 \u00c3\u0097 5 matrix \\[{\\text{A}} = \\left[ {\\begin{array}{*{20}{c}} 1&2&3&4&5 \\\\ 5&1&2&3&4 \\\\ 4&5&1&2&3 \\\\ 3&4&5&1&2 \\\\ 2&3&4&5&1 \\end{array}} \\right]\\] It is given that A has only one real eigen value. Then the real eigen value of A is A. -2.5 B. 0 C. 15 D. 25"},"content":{"rendered":"

[amp_mcq option1=”-2.5″ option2=”0″ option3=”15″ option4=”25″ correct=”option1″]<\/p>\n

The correct answer is $\\boxed{0}$.<\/p>\n

To find the eigenvalues of a matrix, we can use the following formula:<\/p>\n

$$\\lambda = \\frac{-b \\pm \\sqrt{b^2 – 4ac}}{2a}$$<\/p>\n

where $a$, $b$, and $c$ are the coefficients of the characteristic polynomial of the matrix.<\/p>\n

The characteristic polynomial of the matrix $A$ is given by:<\/p>\n

$$p(x) = |xI – A| = x^5 – 15x^4 + 75x^3 – 125x^2 + 75x – 15$$<\/p>\n

We can factor the characteristic polynomial as follows:<\/p>\n

$$p(x) = (x – 1)(x – 5)(x – 3)(x – 2)(x – 1)$$<\/p>\n

This means that the eigenvalues of $A$ are $1$, $5$, $3$, $2$, and $1$.<\/p>\n

Since $A$ has only one real eigenvalue, the real eigenvalue of $A$ must be $1$.<\/p>\n

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