{"id":20002,"date":"2024-04-15T05:47:04","date_gmt":"2024-04-15T05:47:04","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20002"},"modified":"2024-04-15T05:47:04","modified_gmt":"2024-04-15T05:47:04","slug":"the-value-of-x-for-which-all-the-eigen-values-of-the-matrix-given-below-are-real-is-left-beginarray20c-105-textj4-textx202-42-10-endarray","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-value-of-x-for-which-all-the-eigen-values-of-the-matrix-given-below-are-real-is-left-beginarray20c-105-textj4-textx202-42-10-endarray\/","title":{"rendered":"The value of x for which all the eigen-values of the matrix given below are real is \\[\\left[ {\\begin{array}{*{20}{c}} {10}&{5 + {\\text{j}}}&4 \\\\ {\\text{x}}&{20}&2 \\\\ 4&2&{ – 10} \\end{array}} \\right]\\] A. 5 + j B. 5 – j C. 1 – 5j D. 1 + 5j"},"content":{"rendered":"

[amp_mcq option1=”5 + j” option2=”5 – j” option3=”1 – 5j” option4=”1 + 5j” correct=”option3″]<\/p>\n

The correct answer is $\\boxed{5 + j}$.<\/p>\n

To find the eigenvalues of a matrix, we can use the characteristic equation. The characteristic equation for a 3×3 matrix is given by:<\/p>\n

$$|A – \\lambda I| = 0$$<\/p>\n

where $A$ is the matrix and $\\lambda$ is the eigenvalue.<\/p>\n

In this case, we have:<\/p>\n

$$\\begin{vmatrix} {10}&{5 + {\\text{j}}}&4 \\ {\\text{x}}&{20}&2 \\ 4&2&{ – 10} \\end{vmatrix} – \\lambda \\begin{vmatrix} {1}&{0}&{0} \\ {0}&{1}&{0} \\ {0}&{0}&{1} \\end{vmatrix} = 0$$<\/p>\n

Expanding the determinant, we get:<\/p>\n

$$-\\lambda^3 + 35 \\lambda^2 – 250 \\lambda + 1200 = 0$$<\/p>\n

Solving for $\\lambda$, we get the following eigenvalues:<\/p>\n

$$\\lambda = 5 + j, \\lambda = 5 – j, \\lambda = -10$$<\/p>\n

Since the eigenvalues are all real, the value of $x$ for which all the eigenvalues of the matrix are real is $\\boxed{5 + j}$.<\/p>\n

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