{"id":27897,"date":"2019-10-18T15:59:04","date_gmt":"2019-10-18T10:29:04","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/?p=27897"},"modified":"2019-10-25T11:38:15","modified_gmt":"2019-10-25T06:08:15","slug":"cbse-important-questions-class-12-mathematics-determinants","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/cbse-important-questions-class-12-mathematics-determinants\/","title":{"rendered":"CBSE Important Questions Class 12 Mathematics Determinants"},"content":{"rendered":"

Important Questions Class 12 Mathematics Determinants. <\/strong>myCBSEguide has just released Chapter Wise Question Answers for class 12 Maths. There chapter wise Practice Questions with complete solutions are available for download in\u00a0myCBSEguide<\/a>\u00a0<\/strong>website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools for years. There are around 4-5 set of solved Chapter 4 Mathematics Extra Questions from each and every chapter. The students will not miss any concept in these Chapter wise question that are specially designed to tackle Board Exam. We have taken care of every single concept given in CBSE Class 12 Mathematics syllabus<\/a><\/strong>\u00a0and questions are framed as per the latest marking scheme and blue print issued by CBSE for class 12.<\/p>\n

Class 12 Chapter 4 Maths Extra Questions<\/strong><\/p>\n

Download as PDF<\/a><\/strong><\/p>\n

Important Questions Class 12 Maths Chapter 4 Determinants<\/h2>\n
    \n
  1. \n
    \n
    The roots of the equation det. {tex}\\left| {\\begin{array}{*{20}{c}} {1 – x}&2&3 \\\\ 0&{2 – x}&0 \\\\ 0&2&{3 – x} \\end{array}} \\right| = 0{\/tex}<\/span> are<\/div>\n<\/div>\n
    \n
    \n
    \n
      \n
    1. None of these<\/li>\n
    2. 2 and 3<\/li>\n
    3. 1, 2 and 3<\/li>\n
    4. 1 and 3<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
    5. \n
      \n
      If A\u2019 is the transpose of a square matrix A, then<\/div>\n<\/div>\n
      \n
      \n
      \n
        \n
      1. |A| + |A’| = 0<\/li>\n
      2. |A| = |A’|<\/li>\n
      3. |A| {tex} \\ne{\/tex}<\/span> |A’|<\/li>\n
      4. None of th<\/span>ese<\/span><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
      5. \n
        \n
        If f(x) = {tex}\\left| {\\begin{array}{*{20}{c}} {2\\cos x}&1&0 \\\\ 1&{2\\cos x}&1 \\\\ 0&1&{2\\cos x} \\end{array}} \\right|{\/tex}<\/span> then, f ({tex}\\frac{\\pi }{3}{\/tex}<\/span>) =.<\/div>\n<\/div>\n
        \n
        \n
        \n
          \n
        1. 0<\/li>\n
        2. 1<\/li>\n
        3. -1<\/li>\n
        4. 2<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
        5. \n
          \n
          The roots of the equation {tex}\\left| {\\begin{array}{*{20}{c}} 1&4&{20} \\\\ 1&{ – 2}&5 \\\\ 1&{2x}&{5{x^2}} \\end{array}} \\right| = 0{\/tex}<\/span> are<\/div>\n<\/div>\n
          \n
          \n
          \n
            \n
          1. \u20131, \u20132<\/li>\n
          2. \u20131, 2<\/li>\n
          3. 1, \u20132<\/li>\n
          4. 1, 2<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
          5. \n
            \n
            If A and B are any {tex}{\\text{2 }} \\times {\\text{ 2}}{\/tex}<\/span> matrices , then det. (A+B) = 0 implies<\/div>\n<\/div>\n
            \n
            \n
            \n
              \n
            1. det A + det B = 0<\/li>\n
            2. det A = 0 or det B = 0<\/li>\n
            3. None of these<\/li>\n
            4. det A = 0 and det B = 0<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
            5. If {tex}\\begin{vmatrix}2x&5\\\\8&x\\end{vmatrix}=\\begin{vmatrix}6&5\\\\8&3\\end{vmatrix}{\/tex}<\/span>, then x is ________.<\/li>\n
            6. Multiplying\u00a0a determinant by k means multiplying the elements of only one row (or one column) by ________.<\/li>\n
            7. If elements of a row (or a column) in a determinant can be expressed as the sum of two or more elements, then the given determinant can be expressed as the ________ of two or more determinants.<\/li>\n
            8. \n
              \n
              Find adj A for {tex}A = \\left[ {\\begin{array}{*{20}{c}} 2&3 \\\\ 1&4 \\end{array}} \\right].{\/tex}<\/span><\/div>\n<\/div>\n<\/li>\n
            9. \n
              \n
              {tex}A = \\left[ {\\begin{array}{*{20}{c}} 1&2 \\\\ 4&8 \\end{array}} \\right]{\/tex}<\/span>is singular or not.<\/div>\n<\/div>\n<\/li>\n
            10. \n
              \n
              Evaluate {tex}2 \\left| \\begin{array} { r r } { 7 } & { – 2 } \\\\ { – 10 } & { 5 } \\end{array} \\right|{\/tex}<\/span>.<\/div>\n<\/div>\n<\/li>\n
            11. \n
              \n
              Evaluate: {tex}\\left| {\\begin{array}{*{20}{c}} {\\cos \\alpha \\cos \\beta }&{\\cos \\alpha \\sin \\beta }&{ – \\sin \\alpha } \\\\ { – \\sin \\beta }&{\\cos \\beta }&0 \\\\ {\\sin \\alpha \\cos \\beta }&{\\sin \\alpha \\sin \\beta }&{\\cos \\alpha } \\end{array}} \\right|{\/tex}<\/span>.<\/div>\n<\/div>\n<\/li>\n
            12. \n
              \n
              Find the area of {tex}\\Delta {\/tex}<\/span>whose vertices are (3, 8) (-4, 2) and (5, 1).<\/div>\n<\/div>\n<\/li>\n
            13. \n
              \n
              Find the equation of the line joining A (1, 3) and B (0, 0) using det. Find K if D (K, 0) is a point such that area of {tex}\\Delta ABD{\/tex}<\/span> is 3 square unit.<\/div>\n<\/div>\n<\/li>\n
            14. \n
              \n
              If A = {tex} \\left[ \\begin{array} { c c c } { 1 } & { – 2 } & { 3 } \\\\ { 0 } & { – 1 } & { 4 } \\\\ { – 2 } & { 2 } & { 1 } \\end{array} \\right]{\/tex}<\/span>, then find (A’)-1<\/sup>.<\/div>\n<\/div>\n<\/li>\n
            15. \n
              \n
              If {tex}A = \\left[ {\\begin{array}{*{20}{c}} 3&{ – 4} \\\\ { – 1}&2 \\end{array}} \\right],{\/tex}<\/span> find matrix B such that AB = I.<\/div>\n<\/div>\n<\/li>\n
            16. \n
              \n
              Using properties of determinants, prove that
              \n{tex} \\left| \\begin{array} { c c c } { b + c } & { c + a } & { a + b } \\\\ { q + r } & { r + p } & { p + q } \\\\ { y + z } & { z + x } & { x + y } \\end{array} \\right| = 2 \\left| \\begin{array} { c c c } { a } & { b } & { c } \\\\ { p } & { q } & { r } \\\\ { x } & { y } & { z } \\end{array} \\right|{\/tex}<\/span>.<\/div>\n<\/div>\n<\/li>\n
            17. \n
              \n
              Given {tex}A = \\left[ {\\begin{array}{*{20}{c}} 1&{ – 1}&1 \\\\ 1&{ – 2}&{ – 2} \\\\ 2&1&3 \\end{array}} \\right]{\/tex}<\/span>\u00a0and {tex}B = \\left[ {\\begin{array}{*{20}{c}} { – 4}&4&4 \\\\ { – 7}&1&3 \\\\ 5&{ – 3}&{ – 1} \\end{array}} \\right]{\/tex}<\/span>. find AB and use this result in solving the following system of equation.
              \nx – y + z = 4, x – 2y – 2z = 9, 2x + y + 3z = 1<\/div>\n<\/div>\n<\/li>\n<\/ol>\n
              \n

              Chapter 4 Determinants<\/strong><\/p>\n

              Solution<\/b><\/p>\n

                \n
              1. \n
                  \n
                1. 1 , 2 and 3
                  \nExplanation: <\/strong>Expanding along C1<\/sub>
                  \n{tex}\\left| {\\begin{array}{*{20}{c}} {1 – x}&2&3 \\\\ 0&{2 – x}&0 \\\\ 0&2&{3 – x} \\end{array}} \\right| = 0 \\Rightarrow {\/tex}<\/span> (1 – x)(2 – x)(3 – x) = 0{tex}\\Rightarrow{\/tex}<\/span> x = 1, 2 ,3.<\/sub><\/li>\n<\/ol>\n
                    \n
                  1. |A| = |A’|
                    \nExplanation: <\/strong>The determinant of a matrix A and its transpose always same. Because if we interchange the rows into column in a determinant the value of determinant remains unaltered.<\/li>\n<\/ol>\n
                      \n
                    1. \u20131
                      \nExplanation: <\/strong> {tex} \\left| {\\begin{array}{*{20}{c}} {2\\cos x}&1&0 \\\\ 1&{2\\cos x}&1 \\\\ 0&1&{2\\cos x} \\end{array}} \\right|{\/tex}<\/span>
                      \nPut x = {tex}\\frac{\\pi}{3}{\/tex}<\/span>, {tex}\\left| {\\begin{array}{*{20}{c}} {2\\cos \\frac{\\pi }{3}}&1&0 \\\\ 1&{2\\cos \\frac{\\pi }{3}}&1 \\\\ 0&1&{2\\cos \\frac{\\pi }{3}} \\end{array}} \\right| \\\\ {\/tex}<\/span>
                      \n{tex}\\Rightarrow \\left| {\\begin{array}{*{20}{c}} {2.\\frac{1}{2}}&1&0 \\\\ 1&{2.\\frac{1}{2}}&1 \\\\ 0&1&{2.\\frac{1}{2}} \\end{array}} \\right| \\\\{\/tex}<\/span>
                      \n{tex}\\Rightarrow \\left| {\\begin{array}{*{20}{c}} 1&1&0 \\\\ 1&1&1 \\\\ 0&1&1 \\end{array}} \\right| {\/tex}<\/span>{tex} \\Rightarrow 1(0) – 1(1) = – 1 \\\\{\/tex}<\/span><\/li>\n<\/ol>\n
                        \n
                      1. \u20131 , 2
                        \nExplanation:<\/strong> {tex}\\left| {\\begin{array}{*{20}{c}} 1&4&{20} \\\\ 1&{ – 2}&5 \\\\ 1&{2x}&{5{x^2}} \\end{array}} \\right| = 0{\/tex}<\/span>
                        \nApply, R3<\/sub>{tex}\\rightarrow{\/tex}<\/span>R3<\/sub> – R1<\/sub>, R2<\/sub>{tex}\\rightarrow{\/tex}<\/span>R2 <\/sub>– R1<\/sub>,
                        \n{tex}\\Rightarrow{\/tex}<\/span> {tex}\\left| {\\begin{array}{*{20}{c}} 1&4&{20} \\\\ 0&{ – 6}&{ – 15} \\\\ 0&{2x – 4}&{5{x^2} – 20} \\end{array}} \\right| = 0{\/tex}<\/span>
                        \n{tex}\\Rightarrow{\/tex}<\/span> -6(5x2<\/sup> – 20) + 15(2x – 4) =0
                        \n{tex}\\Rightarrow{\/tex}<\/span> (x – 2)(x + 1) = 0 {tex}\\Rightarrow{\/tex}<\/span> x= 2 , -1.<\/li>\n<\/ol>\n
                          \n
                        1. None of these
                          \nExplanation: <\/strong>If det (A+B)=0 implies that A+B a Singular matrix.<\/li>\n<\/ol>\n<\/li>\n
                        2. x = {tex}\\pm{\/tex}<\/span>3<\/li>\n
                        3. k<\/li>\n
                        4. sum<\/li>\n
                        5. {tex}adjA = \\left[ {\\begin{array}{*{20}{c}} 4&{ – 3} \\\\ { – 1}&2 \\end{array}} \\right]{\/tex}<\/span>
                          \n\"CBSE<\/li>\n
                        6. {tex}\\left| A \\right| = \\left| {\\begin{array}{*{20}{c}} 1&2 \\\\ 4&8 \\end{array}} \\right|{\/tex}<\/span>
                          \n= 8 – 8
                          \n= 0
                          \nHence A is singular<\/li>\n
                        7. According to the question, we have to evaluate {tex}2 \\left| \\begin{array} { r r } { 7 } & { – 2 } \\\\ { – 10 } & { 5 } \\end{array} \\right|{\/tex}<\/span>.
                          \nNow, {tex}2 \\left| \\begin{array} { r r } { 7 } & { – 2 } \\\\ { – 10 } & { 5 } \\end{array} \\right| = 2 [ 35 – ( 20 ) ]{\/tex}<\/span>
                          \n{tex}= 2 \\times 15 = 30{\/tex}<\/span><\/li>\n
                        8. Let {tex}\\Delta = \\left| {\\begin{array}{*{20}{c}} {\\cos \\alpha \\cos \\beta }&{\\cos \\alpha \\sin \\beta }&{ – \\sin \\alpha } \\\\ { – \\sin \\beta }&{\\cos \\beta }&0 \\\\ {\\sin \\alpha \\cos \\beta }&{\\sin \\alpha \\sin \\beta }&{\\cos \\alpha } \\end{array}} \\right|{\/tex}<\/span>
                          \nExpanding along first row,
                          \n{tex}= \\cos \\alpha \\cos \\beta \\left( {\\cos \\alpha \\cos \\beta – 0} \\right) {\/tex}<\/span> {tex}- \\cos \\alpha \\sin \\beta \\left( { – \\cos \\alpha \\sin \\beta – 0} \\right) {\/tex}<\/span> {tex}- \\sin \\alpha \\left( { – \\sin \\alpha {{\\sin }^2}\\beta – \\sin \\alpha {{\\cos }^2}\\beta } \\right){\/tex}<\/span>
                          \n{tex}= {\\cos ^2}\\alpha {\\cos ^2}\\beta + {\\cos ^2}\\alpha {\\sin ^2}\\beta {\/tex}<\/span> {tex}+ {\\sin ^2}\\alpha \\left( {{{\\sin }^2}\\beta + {{\\cos }^2}\\beta } \\right){\/tex}<\/span>
                          \n{tex}= {\\cos ^2}\\alpha \\left( {{{\\cos }^2}\\beta + {{\\sin }^2}\\beta } \\right){\/tex}<\/span> {tex} + {\\sin ^2}\\alpha \\left( {{{\\sin }^2}\\beta + {{\\cos }^2}\\beta } \\right){\/tex}<\/span>
                          \n{tex}= {\\cos ^2}\\alpha + {\\sin ^2}\\alpha{\/tex}<\/span>
                          \n= 1<\/li>\n
                        9. {tex}\\Delta = \\frac{1}{2}\\left| {\\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\\\ {{x_2}}&{{y_2}}&1 \\\\ {{x_3}}&{{y_3}}&1 \\end{array}} \\right|{\/tex}<\/span>
                          \n{tex}= \\frac{1}{2}\\left| {\\begin{array}{*{20}{c}} 3&8&1 \\\\ { – 4}&2&1 \\\\ 5&1&1 \\end{array}} \\right|{\/tex}<\/span>
                          \n{tex} = \\frac{1}{2}\\left[ {3\\left( {2 – 1} \\right) – 8\\left( { – 4 – 5} \\right) + 1\\left( { – 4 + 10} \\right)} \\right]{\/tex}<\/span>
                          \n{tex} = \\frac{1}{2}\\left[ {3 + 72 – 14} \\right] = \\frac{{61}}{2}{\/tex}<\/span><\/li>\n
                        10. Let P (x, y) be any point on AB. Then the equation of line AB is,
                          \n{tex}\\frac{1}{2}\\left| {\\begin{array}{*{20}{c}} 0&0&1 \\\\ 1&3&1 \\\\ x&y&1 \\end{array}} \\right| = 0{\/tex}<\/span>
                          \ny = 3x
                          \nArea {tex}\\Delta ABD = 3{\/tex}<\/span> square unit
                          \n{tex}\\frac{1}{2}\\left| {\\begin{array}{*{20}{c}} 1&3&1 \\\\ 0&0&1 \\\\ K&0&1 \\end{array}} \\right| = \\pm 3{\/tex}<\/span>
                          \n{tex}k = \\pm 2{\/tex}<\/span><\/li>\n
                        11. If A = {tex} \\left[ \\begin{array} { c c c } { 1 } & { – 2 } & { 3 } \\\\ { 0 } & { – 1 } & { 4 } \\\\ { – 2 } & { 2 } & { 1 } \\end{array} \\right]{\/tex}<\/span>, then we have to find (A’)-1<\/sup>.
                          \nNow, A = {tex} \\left[ \\begin{array} { c c c } { 1 } & { – 2 } & { 3 } \\\\ { 0 } & { – 1 } & { 4 } \\\\ { – 2 } & { 2 } & { 1 } \\end{array} \\right]{\/tex}<\/span>\u00a0Therefore, we have, {tex} |A|=\\left| \\begin{array} { c c c } { 1 } & { – 2 } & { 3 } \\\\ { 0 } & { – 1 } & { 4 } \\\\ { – 2 } & { 2 } & { 1 } \\end{array} \\right|{\/tex}<\/span>
                          \n= 1 (-1 – 8) + 2 (0 + 8) + 3 (0 – 2)
                          \n[expanding along R1<\/sub>]\n=-9+16-6=1{tex} \\neq{\/tex}<\/span>0
                          \nTherefore, A is non-singular matrix and hence its inverse exists.
                          \nCofactors of an element of |A| are given by
                          \n{tex} A _ { 11 } = ( – 1 ) ^ { 1 + 1 } \\left| \\begin{array} { c c } { – 1 } & { 4 } \\\\ { 2 } & { 1 } \\end{array} \\right| = ( – 1 – 8 ) = – 9{\/tex}<\/span>
                          \n{tex} A _ { 12 } = ( – 1 ) ^ { 1 } + 2 \\left| \\begin{array} { c c } { 0 } & { 4 } \\\\ { – 2 } & { 1 } \\end{array} \\right| = – ( 0 + 8 ) = – 8{\/tex}<\/span>
                          \n{tex} A _ { 13 } = ( – 1 ) ^ { 1 + 3 } \\left| \\begin{array} { c c } { 0 } & { -1 } \\\\ { – 2 } & { 2 } \\end{array} \\right| = ( 0 – 2 ) = – 2{\/tex}<\/span>
                          \n{tex} A _ { 21 } = ( – 1 ) ^ { 2 + 1 } \\left| \\begin{array} { c c } { – 2 } & { 3 } \\\\ { 2 } & { 1 } \\end{array} \\right| = – ( – 2 – 6 ) = 8{\/tex}<\/span>
                          \n{tex} A _ { 22 } = ( – 1 ) ^ { 2 + 2 } \\left| \\begin{array} { c c } { 1 } & { 3 } \\\\ { – 2 } & { 1 } \\end{array} \\right| = ( 1 + 6 ) = 7{\/tex}<\/span>
                          \n{tex} A _ { 23 } = ( – 1 ) ^ { 2 + 3 } \\left| \\begin{array} { c c } { 1 } & { – 2 } \\\\ { – 2 } & { 2 } \\end{array} \\right| = – ( 2 – 4 ) = 2{\/tex}<\/span>
                          \n{tex} A _ { 31 } = ( – 1 ) ^ { 3 + 1 } \\left| \\begin{array} { c c } { – 2 } & { 3 } \\\\ { – 1 } & { 4 } \\end{array} \\right| = ( – 8 + 3 ) = – 5{\/tex}<\/span>
                          \n{tex} A _ { 32 } = ( – 1 ) ^ { 3 + 2 } \\left| \\begin{array} { l l } { 1 } & { 3 } \\\\ { 0 } & { 4 } \\end{array} \\right| = – ( 4 – 0 ) = – 4{\/tex}<\/span>
                          \n{tex} A _ { 33 } = ( – 1 ) ^ { 3 +3 } \\left| \\begin{array} { c c } { 1 } & { – 2 } \\\\ { 0 } & { – 1 } \\end{array} \\right| = ( – 1 – 0 ) = – 1{\/tex}<\/span>
                          \nThus, adj A = {tex} \\left[ \\begin{array} { l l l } { A _ { 11 } } & { A _ { 21 } } & { A _ { 31 } } \\\\ { A _ { 12 } } & { A _ { 22 } } & { A _ { 32 } } \\\\ { A _ { 13 } } & { A _ { 23 } } & { A _ { 33 } } \\end{array} \\right] = \\left[ \\begin{array} { c c c } { – 9 } & { 8 } & { – 5 } \\\\ { – 8 } & { 7 } & { – 4 } \\\\ { – 2 } & { 2 } & { – 1 } \\end{array} \\right]{\/tex}<\/span>
                          \nHence, {tex} A ^ { -1 } = \\frac { 1 } { | A | } \\text { adj } A = \\frac { 1 } { 1 } \\left[ \\begin{array} { c c c } { – 9 } & { 8 } & { – 5 } \\\\ { – 8 } & { 7 } & { – 4 } \\\\ { – 2 } & { 2 } & { – 1 } \\end{array} \\right]{\/tex}<\/span>
                          \nNow, (A’)-1<\/sup> = (A-1<\/sup>)’ =\u00a0{tex} \\left[ \\begin{array} { c c c } { – 9 } & { 8 } & { – 5 } \\\\ { – 8 } & { 7 } & { – 4 } \\\\ { – 2 } & { 2 } & { – 1 } \\end{array} \\right]’ = \\left[ \\begin{array} { c c c } { – 9 } & { – 8 } & { – 2 } \\\\ { 8 } & { 7 } & { 2 } \\\\ { – 5 } & { – 4 } & { – 1 } \\end{array} \\right]{\/tex}<\/span><\/li>\n
                        12. {tex}\\left| A \\right| = 2 \\ne 0{\/tex}<\/span>
                          \nTherefore A-1<\/sup> exists
                          \nAB = I
                          \nA-1 <\/sup>AB = A-1<\/sup>I
                          \nB = A-1<\/sup>
                          \n{tex}adjA = \\left[ {\\begin{array}{*{20}{c}} 2&4 \\\\ 1&3 \\end{array}} \\right]{\/tex}<\/span>
                          \n{tex}{A^{ – 1}} = \\frac{1}{{\\left| A \\right|}}\\left( {adjA} \\right){\/tex}<\/span>
                          \n{tex}= \\frac{1}{2}\\left[ {\\begin{array}{*{20}{c}} 2&4 \\\\ 1&3 \\end{array}} \\right]{\/tex}<\/span>
                          \n{tex}= \\left[ {\\begin{array}{*{20}{c}} 1&2 \\\\ {\\frac{1}{2}}&{\\frac{3}{2}} \\end{array}} \\right]{\/tex}<\/span>
                          \nHence {tex}B = \\left[ {\\begin{array}{*{20}{c}} 1&2 \\\\ {\\frac{1}{2}}&{\\frac{3}{2}} \\end{array}} \\right]{\/tex}<\/span><\/li>\n
                        13. According to the question,we have to use properties of determinants to prove that,
                          \n{tex} \\left| \\begin{array} { c c c c } { b + c } & { c + a } & { a + b } \\\\ { q + r } & { r + p } & { p + q } \\\\ { y + z } & { z + x } & { x + y } \\end{array} \\right| = 2 \\left| \\begin{array} { l l l } { a } & { b } & { c } \\\\ { p } & { q } & { r } \\\\ { x } & { y } & { z } \\end{array} \\right|{\/tex}<\/span>
                          \nLet LHS = {tex}\\left| \\begin{array} { c c c } { b + c } & { c + a } & { a + b } \\\\ { q + r } & { r + p } & { p + q } \\\\ { y + z } & { z + x } & { x + y } \\end{array} \\right|{\/tex}<\/span>
                          \nTherefore,on applying C1<\/sub>{tex}\\rightarrow{\/tex}<\/span> C1<\/sub> + C2 <\/sub>+ C3<\/sub> we get,
                          \n{tex}\\Delta = \\left| \\begin{array} { c c c } { 2 ( a + b + c ) } & { c + a } & { a + b } \\\\ { 2 ( p + q + r ) } & { r + p } & { p + q } \\\\ { 2 ( x + y + z ) } & { z + x } & { x + y } \\end{array} \\right|{\/tex}<\/span>
                          \non taking 2 common from {tex}C_1{\/tex}<\/span>,we get,
                          \n{tex}\\Delta = 2\\left| {\\begin{array}{*{20}{c}} {a + b + c}&{c + a}&{a + b} \\\\ {p + q + r}&{r + p}&{p + q} \\\\ {x + y + z}&{z + x}&{x + y} \\end{array}} \\right|{\/tex}<\/span>
                          \nOn applying C2<\/sub>{tex}\\rightarrow{\/tex}<\/span> C2<\/sub> – C1<\/sub> and C3<\/sub> {tex}\\rightarrow{\/tex}<\/span> C3 <\/sub>– C12<\/span>,
                          \nwe get
                          \n{tex}\\Delta = 2 \\left| \\begin{array} { c c c } { a + b + c } & { – b } & { – c } \\\\ { p + q + r } & { – q } & { – r } \\\\ { x + y + z } & { – y } & { – z } \\end{array} \\right|{\/tex}<\/span>
                          \non applying {tex}C_1\\rightarrow C_1+C_2+C_3,we\\ get,{\/tex}<\/span>
                          \n{tex}\\Delta = 2 \\left| \\begin{array} { c c c } { a } & { – b } & { – c } \\\\ { p } & { – q } & { – r } \\\\ { x } & { – y } & { – z } \\end{array} \\right|{\/tex}<\/span>
                          \n{tex}\\therefore{\/tex}<\/span>\u00a0{tex}\\Delta = 2 \\left| \\begin{array} { l l l } { a } & { b } & { c } \\\\ { p } & { q } & { r } \\\\ { x } & { y } & { z } \\end{array} \\right|{\/tex}<\/span> [taking (-1) common from both C2<\/sub> and C3<\/sub>]\n= RHS<\/li>\n
                        14. x – y + z = 4
                          \nx – 2y – 2z = 9
                          \n2x + y + 3z = 1
                          \nLet {tex}A = \\left[ {\\begin{array}{*{20}{c}} 1&{ – 1}&1 \\\\ 1&{ – 2}&{ – 2} \\\\ 2&1&3 \\end{array}} \\right]X = \\left[ {\\begin{array}{*{20}{c}} x \\\\ y \\\\ z \\end{array}} \\right]C = \\left[ {\\begin{array}{*{20}{c}} 4 \\\\ 9 \\\\ 1 \\end{array}} \\right]{\/tex}<\/span>
                          \nAX = C
                          \n{tex}AB = \\left[ {\\begin{array}{*{20}{c}} 1&{ – 1}&1 \\\\ 1&{ – 2}&{ – 2} \\\\ 2&1&3 \\end{array}} \\right]\\left[ {\\begin{array}{*{20}{c}} { – 4}&4&4 \\\\ { – 7}&1&3 \\\\ 5&{ – 3}&{ – 1} \\end{array}} \\right]{\/tex}<\/span>\u00a0{tex} = \\left[ {\\begin{array}{*{20}{c}} 8&0&0 \\\\ 0&8&0 \\\\ 0&0&8 \\end{array}} \\right]{\/tex}<\/span>
                          \nAB = 8I
                          \n{tex}{A^{ – 1}} = \\frac{1}{8}B\\left[ \\begin{gathered} \\because {A^{ – 1}}AB = 8{A^{ – 1}}I \\hfill \\\\ B = 8{A^{ – 1}} \\hfill \\\\ \\end{gathered} \\right]{\/tex}<\/span>
                          \n{tex} = \\frac{1}{8}\\left[ {\\begin{array}{*{20}{c}} { – 4}&4&4 \\\\ { – 7}&1&3 \\\\ 5&{ – 3}&{ – 1} \\end{array}} \\right]{\/tex}<\/span>
                          \nX = A-1<\/sup>C
                          \n{tex}\\left[ {\\begin{array}{*{20}{c}} x \\\\ y \\\\ z \\end{array}} \\right] = \\left[ {\\begin{array}{*{20}{c}} 3 \\\\ { – 2} \\\\ { – 1} \\end{array}} \\right]{\/tex}<\/span>
                          \nx = 3, y = -2, z = -1<\/li>\n<\/ol>\n

                          Chapter Wise Important Questions Class 12 Maths Part I and Part II<\/h2>\n
                            \n
                          1. Relations and Functions<\/a><\/li>\n
                          2. Inverse Trigonometric Functions<\/a><\/li>\n
                          3. Matrices<\/a><\/li>\n
                          4. Determinants<\/a><\/li>\n
                          5. Continuity and Differentiability<\/a><\/li>\n
                          6. Application of Derivatives<\/a><\/li>\n
                          7. Integrals<\/a><\/li>\n
                          8. Application of Integrals<\/a><\/li>\n
                          9. Differential Equations<\/a><\/li>\n
                          10. Vector Algebra<\/a><\/li>\n
                          11. Three Dimensional Geometry<\/a><\/li>\n
                          12. Linear Programming<\/a><\/li>\n
                          13. Probability<\/a><\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

                            Important Questions Class 12 Mathematics Determinants. myCBSEguide has just released Chapter Wise Question Answers for class 12 Maths. There chapter wise Practice Questions with complete solutions are available for download in\u00a0myCBSEguide\u00a0website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools for years. … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1346,1432],"tags":[1867,1839,1838,1833,1832,1854],"class_list":["post-27897","post","type-post","status-publish","format-standard","hentry","category-cbse","category-mathematics-cbse-class-12","tag-cbse-class-12-mathematics","tag-extra-questions","tag-important-questions","tag-latest-exam-questions","tag-practice-questions","tag-practice-test"],"yoast_head":"\nCBSE Important Questions Class 12 Mathematics Determinants<\/title>\n<meta name=\"description\" content=\"CBSE Important Questions Class 12 Mathematics Determinants 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