{"id":19984,"date":"2024-04-15T05:46:50","date_gmt":"2024-04-15T05:46:50","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=19984"},"modified":"2024-04-15T05:46:50","modified_gmt":"2024-04-15T05:46:50","slug":"eigen-values-of-the-matrix-left-beginarray20c-3-1-1-13-1-1-13-endarray-right-are-a-1-1-1-b-1-1-2-c-1-4-4-d-1-2-4","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/eigen-values-of-the-matrix-left-beginarray20c-3-1-1-13-1-1-13-endarray-right-are-a-1-1-1-b-1-1-2-c-1-4-4-d-1-2-4\/","title":{"rendered":"Eigen values of the matrix \\[\\left[ {\\begin{array}{*{20}{c}} 3&{ – 1}&{ – 1} \\\\ { – 1}&3&{ – 1} \\\\ { – 1}&{ – 1}&3 \\end{array}} \\right]\\] are A. 1, 1, 1 B. 1, 1, 2 C. 1, 4, 4 D. 1, 2, 4"},"content":{"rendered":"

[amp_mcq option1=”1, 1, 1″ option2=”1, 1, 2″ option3=”1, 4, 4″ option4=”1, 2, 4″ correct=”option1″]<\/p>\n

The correct answer is $\\boxed{\\text{A) }1, 1, 1}$.<\/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

$$p(x) = |A – xI| = \\det \\left[ {\\begin{array}{*{20}{c}} 3&{ – 1}&{ – 1} \\ { – 1}&3&{ – 1} \\ { – 1}&{ – 1}&3 \\end{array}} \\right] = x^3 – 9x^2 + 6x – 3$$<\/p>\n

To find the eigenvalues of $A$, we need to solve the equation $p(x) = 0$.<\/p>\n

Solving $p(x) = 0$, we get the following eigenvalues:<\/p>\n

$$\\lambda = 1, 1, 1$$<\/p>\n

Therefore, the eigenvalues of the matrix $A$ are $\\boxed{\\text{A) }1, 1, 1}$.<\/p>\n

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