{"id":20105,"date":"2024-04-15T05:48:28","date_gmt":"2024-04-15T05:48:28","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20105"},"modified":"2024-04-15T05:48:28","modified_gmt":"2024-04-15T05:48:28","slug":"for-the-matrix-texta-left-beginarray20c-53-13-endarray-right-one-of-the-normalized-eigen-vectors-is-given-as-a-left-beginarray20c-fra","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/for-the-matrix-texta-left-beginarray20c-53-13-endarray-right-one-of-the-normalized-eigen-vectors-is-given-as-a-left-beginarray20c-fra\/","title":{"rendered":"For the matrix \\[{\\text{A}} = \\left[ {\\begin{array}{*{20}{c}} 5&#038;3 \\\\ 1&#038;3 \\end{array}} \\right],\\] ONE of the normalized eigen vectors is given as A. \\[\\left( {\\begin{array}{*{20}{c}} {\\frac{1}{2}} \\\\ {\\frac{{\\sqrt 3 }}{2}} \\end{array}} \\right)\\] B. \\[\\left( {\\begin{array}{*{20}{c}} {\\frac{1}{{\\sqrt 2 }}} \\\\ {\\frac{{ &#8211; 1}}{{\\sqrt 2 }}} \\end{array}} \\right)\\] C. \\[\\left( {\\begin{array}{*{20}{c}} {\\frac{3}{{\\sqrt {10} }}} \\\\ {\\frac{{ &#8211; 1}}{{\\sqrt {10} }}} \\end{array}} \\right)\\] D. \\[\\left( {\\begin{array}{*{20}{c}} {\\frac{1}{{\\sqrt 5 }}} \\\\ {\\frac{2}{{\\sqrt 5 }}} \\end{array}} \\right)\\]"},"content":{"rendered":"<p>\r\n    <!-- Check if it's an AMP page -->\r\n            <!-- Non-AMP version -->\r\n        <div class=\"mcq-container\" data-quiz-id=\"quizState_6a98709521f84\">\r\n                                            <div class=\"option\" data-option-key=\"option1\" data-is-correct=\"false\">\r\n                    &#8221;[left(                <\/div>\r\n                                                                                                                                            \r\n            <!-- Feedback messages for non-AMP -->\r\n            <div class=\"feedback\" data-feedback=\"wrong\">Answer is Right!<\/div>\r\n            <div class=\"feedback\" data-feedback=\"right\">Answer is Wrong!<\/div>\r\n        <\/div>\r\n\r\n        <script>\r\n        document.addEventListener('DOMContentLoaded', function () {\r\n            var containers = document.querySelectorAll('.mcq-container');\r\n\r\n            containers.forEach(function(container) {\r\n                var options = container.querySelectorAll('.option');\r\n                var feedbackSelect = container.querySelector('[data-feedback=\"select\"]');\r\n                var feedbackWrong = container.querySelector('[data-feedback=\"wrong\"]');\r\n                var feedbackRight = container.querySelector('[data-feedback=\"right\"]');\r\n\r\n                options.forEach(function(option) {\r\n                    option.addEventListener('click', function() {\r\n                        var selectedOption = option.getAttribute('data-option-key');\r\n                        var isCorrect = option.getAttribute('data-is-correct') === 'true';\r\n\r\n                        \/\/ Remove previous selections\r\n                        options.forEach(function(opt) {\r\n                            opt.classList.remove('correct', 'incorrect');\r\n                        });\r\n\r\n                        \/\/ Add the correct\/incorrect class\r\n                        if (isCorrect) {\r\n                            option.classList.add('correct');\r\n                            feedbackRight.hidden = false;\r\n                            feedbackWrong.hidden = true;\r\n                        } else {\r\n                            option.classList.add('incorrect');\r\n                            feedbackRight.hidden = true;\r\n                            feedbackWrong.hidden = false;\r\n                        }\r\n\r\n                        \/\/ Hide select feedback\r\n                        feedbackSelect.hidden = true;\r\n                    });\r\n                });\r\n            });\r\n        });\r\n        <\/script>\r\n    \r\n    &#8221; option2=&#8221;\\[\\left( {\\begin{array}{*{20}{c}} {\\frac{1}{{\\sqrt 2 }}} \\\\ {\\frac{{ &#8211; 1}}{{\\sqrt 2 }}} \\end{array}} \\right)\\]&#8221; option3=&#8221;\\[\\left( {\\begin{array}{*{20}{c}} {\\frac{3}{{\\sqrt {10} }}} \\\\ {\\frac{{ &#8211; 1}}{{\\sqrt {10} }}} \\end{array}} \\right)\\]&#8221; option4=&#8221;\\[\\left( {\\begin{array}{*{20}{c}} {\\frac{1}{{\\sqrt 5 }}} \\\\ {\\frac{2}{{\\sqrt 5 }}} \\end{array}} \\right)\\]&#8221; correct=&#8221;option3&#8243;]<!--more--><\/p>\n<p>The correct answer is $\\boxed{\\left( {\\begin{array}{*{20}{c}} {\\frac{1}{{\\sqrt 2 }}} \\ {\\frac{{ &#8211; 1}}{{\\sqrt 2 }}} \\end{array}} \\right)}$.<\/p>\n<p>To find the eigenvalues and eigenvectors of a matrix, we can use the following formula:<\/p>\n<p>$$\\lambda v = A v$$<\/p>\n<p>where $\\lambda$ is the eigenvalue, $v$ is the eigenvector, and $A$ is the matrix.<\/p>\n<p>In this case, we have the following matrix:<\/p>\n<p>$$A = \\left[ {\\begin{array}{*{20}{c}} 5&amp;3 \\ 1&amp;3 \\end{array}} \\right]$$<\/p>\n<p>To find the eigenvalues, we can solve the following equation:<\/p>\n<p>$$| A &#8211; \\lambda I | = 0$$<\/p>\n<p>where $I$ is the identity matrix.<\/p>\n<p>Solving this equation, we find that <div class=\"youtube-subscribe-container\">\r\n        <a href=\"https:\/\/www.youtube.com\/channel\/UCNHT8lW-JmLC68rjBfZhdkg?sub_confirmation=1\" target=\"_blank\" class=\"youtube-subscribe-button\">\r\n            <span class=\"youtube-icon\">\r\n                <svg xmlns=\"http:\/\/www.w3.org\/2000\/svg\" viewBox=\"0 0 576 512\">\r\n                    <path d=\"M549.7 124.1c-6.3-23.7-24.8-42.3-48.3-48.6C458.8 64 288 64 288 64S117.2 64 74.6 75.5c-23.5 6.3-42 24.9-48.3 48.6-11.4 42.9-11.4 132.3-11.4 132.3s0 89.4 11.4 132.3c6.3 23.7 24.8 41.5 48.3 47.8C117.2 448 288 448 288 448s170.8 0 213.4-11.5c23.5-6.3 42-24.2 48.3-47.8 11.4-42.9 11.4-132.3 11.4-132.3s0-89.4-11.4-132.3zm-317.5 213.5V175.2l142.7 81.2-142.7 81.2z\"\/>\r\n                <\/svg>\r\n            <\/span>\r\n            <div class=\"telegram-channel-container\">\r\n        <a href=\"https:\/\/t.me\/pscnotes2025\" target=\"_blank\" class=\"telegram-channel-button\">\r\n            <span class=\"telegram-icon\">\r\n                <svg xmlns=\"http:\/\/www.w3.org\/2000\/svg\" viewBox=\"0 0 496 512\">\r\n                    <path fill=\"white\" d=\"M248,8C111,8,0,119,0,256s111,248,248,248s248-111,248-248S385,8,248,8z M362,177L320,367c-3,14-10,18-20,14l-56-41l-27,26 c-3,3-5,5-10,5l4-63L323,196c5-5-1-7-8-3l-98,62l-42-13c-9-3-10-9,2-14l162-63C351,160,365,164,362,177z\"\/>\r\n                <\/svg>\r\n            <\/span>\r\n            Join Our Telegram Channel\r\n        <\/a>\r\n    <\/div> Subscribe on YouTube\r\n        <\/a>\r\n    <\/div> the eigenvalues are $\\lambda = 5$ and $\\lambda = 3$.<\/p>\n<p>To find the eigenvectors corresponding to each eigenvalue, we can substitute each eigenvalue into the following equation:<\/p>\n<p>$$(A &#8211; \\lambda I) v = 0$$<\/p>\n<p>In this case, we have the following equations:<\/p>\n<p>$$\\left[ {\\begin{array}{*{20}{c}} 5 &#8211; \\lambda &amp;3 \\ 1&amp;3 &#8211; \\lambda \\end{array}} \\right] v = 0$$<\/p>\n<p>For the eigenvalue $\\lambda = 5$, we have the following equation:<\/p>\n<p>$$\\left[ {\\begin{array}{*{20}{c}} 5 &#8211; 5 &amp;3 \\ 1&amp;3 &#8211; 5 \\end{array}} \\right] v = 0$$<\/p>\n<p>Solving this equation, we find that the eigenvector corresponding to $\\lambda = 5$ is $v = \\left( {\\begin{array}{*{20}{c}} 1 \\ 0 \\end{array}} \\right)$.<\/p>\n<p>For the eigenvalue $\\lambda = 3$, we have the following equation:<\/p>\n<p>$$\\left[ {\\begin{array}{*{20}{c}} 3 &#8211; 3 &amp;3 \\ 1&amp;3 &#8211; 3 \\end{array}} \\right] v = 0$$<\/p>\n<p>Solving this equation, we find that the eigenvector corresponding to $\\lambda = 3$ is $v = \\left( {\\begin{array}{*{20}{c}} 0 \\ 1 \\end{array}} \\right)$.<\/p>\n<p>To normalize the eigenvectors, we divide them by their norm. The norm of a vector is given by the following formula:<\/p>\n<p>$$\\| v \\| = \\sqrt{v^T v}$$<\/p>\n<p>In this case, we have the following eigenvectors:<\/p>\n<p>$$v_1 = \\left( {\\begin{array}{*{20}{c}} 1 \\ 0 \\end{array}} \\right)$$<\/p>\n<p>$$v_2 = \\left( {\\begin{array}{*{20}{c}} 0 \\ 1 \\end{array}} \\right)$$<\/p>\n<p>The norms of these vectors are $\\| v_1 \\| = 1$ and $\\| v_2 \\| = 1$. Therefore, the normalized eigenvectors are given by:<\/p>\n<p>$$v_1 = \\left( {\\begin{array}{*{20}{c}} {\\frac{1}{{\\sqrt 2 }}} \\ {\\frac{{ &#8211; 1}}{{\\sqrt 2 }}} \\end{array}} \\right)$$<\/p>\n<p>$$v_2 = \\left( {\\begin{array}{*{20}{c}} {\\frac{1}{{\\sqrt 2 }}} \\ {\\frac{{ 1}}{{\\sqrt 2 }}} \\end{array}} \\right)$$<\/p>\n<p>Therefore, one of the normalized eigenvectors of the matrix $A$ is $\\boxed{\\left( {\\begin{array}{*{20}{c}} {\\frac{1}{{\\sqrt 2 }}} \\ {\\frac{{ &#8211; 1}}{{\\sqrt 2 }}} \\end{array}} \\right)}$.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>&#8221; option2=&#8221;\\[\\left( {\\begin{array}{*{20}{c}} {\\frac{1}{{\\sqrt 2 }}} \\\\ {\\frac{{ &#8211; 1}}{{\\sqrt 2 }}} \\end{array}} \\right)\\]&#8221; option3=&#8221;\\[\\left( {\\begin{array}{*{20}{c}} {\\frac{3}{{\\sqrt {10} }}} \\\\ Join Our Telegram Channel Subscribe on YouTube {\\frac{{ &#8211; 1}}{{\\sqrt {10} }}} \\end{array}} \\right)\\]&#8221; option4=&#8221;\\[\\left( {\\begin{array}{*{20}{c}} {\\frac{1}{{\\sqrt 5 }}} \\\\ {\\frac{2}{{\\sqrt 5 }}} \\end{array}} \\right)\\]&#8221; correct=&#8221;option3&#8243;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[489],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.2 (Yoast SEO v23.3) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>For the matrix \\[{\\text{A}} = \\left[ {\\begin{array}{*{20}{c}} 5&amp;3 \\\\ 1&amp;3 \\end{array}} \\right],\\] ONE of the normalized eigen vectors is given as A. \\[\\left( {\\begin{array}{*{20}{c}} {\\frac{1}{2}} \\\\ {\\frac{{\\sqrt 3 }}{2}} \\end{array}} \\right)\\] B. \\[\\left( {\\begin{array}{*{20}{c}} {\\frac{1}{{\\sqrt 2 }}} \\\\ {\\frac{{ - 1}}{{\\sqrt 2 }}} \\end{array}} \\right)\\] C. \\[\\left( {\\begin{array}{*{20}{c}} {\\frac{3}{{\\sqrt {10} }}} \\\\ {\\frac{{ - 1}}{{\\sqrt {10} }}} \\end{array}} \\right)\\] D. \\[\\left( {\\begin{array}{*{20}{c}} {\\frac{1}{{\\sqrt 5 }}} \\\\ {\\frac{2}{{\\sqrt 5 }}} \\end{array}} \\right)\\]<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/for-the-matrix-texta-left-beginarray20c-53-13-endarray-right-one-of-the-normalized-eigen-vectors-is-given-as-a-left-beginarray20c-fra\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"For the matrix \\[{\\text{A}} = \\left[ {\\begin{array}{*{20}{c}} 5&amp;3 \\\\ 1&amp;3 \\end{array}} \\right],\\] ONE of the normalized eigen vectors is given as A. \\[\\left( {\\begin{array}{*{20}{c}} {\\frac{1}{2}} \\\\ {\\frac{{\\sqrt 3 }}{2}} \\end{array}} \\right)\\] B. \\[\\left( {\\begin{array}{*{20}{c}} {\\frac{1}{{\\sqrt 2 }}} \\\\ {\\frac{{ - 1}}{{\\sqrt 2 }}} \\end{array}} \\right)\\] C. \\[\\left( {\\begin{array}{*{20}{c}} {\\frac{3}{{\\sqrt {10} }}} \\\\ {\\frac{{ - 1}}{{\\sqrt {10} }}} \\end{array}} \\right)\\] D. \\[\\left( {\\begin{array}{*{20}{c}} {\\frac{1}{{\\sqrt 5 }}} \\\\ {\\frac{2}{{\\sqrt 5 }}} \\end{array}} \\right)\\]\" \/>\n<meta property=\"og:description\" content=\"&#8221; 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