CBSE Important Questions Class 12 Mathematics Determinants

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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 myCBSEguide 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 and questions are framed as per the latest marking scheme and blue print issued by CBSE for class 12.

Class 12 Chapter 4 Maths Extra Questions

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Important Questions Class 12 Maths Chapter 4 Determinants

1. 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} are
   1. None of these
   2. 2 and 3
   3. 1, 2 and 3
   4. 1 and 3
2. If A’ is the transpose of a square matrix A, then
   1. |A| + |A’| = 0
   2. |A| = |A’|
   3. |A| {tex} \ne{/tex} |A’|
   4. None of these
3. 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} then, f ({tex}\frac{\pi }{3}{/tex}) =.
   1. 0
   2. 1
   3. -1
   4. 2
4. 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} are
   1. –1, –2
   2. –1, 2
   3. 1, –2
   4. 1, 2
5. If A and B are any {tex}{\text{2 }} \times {\text{ 2}}{/tex} matrices , then det. (A+B) = 0 implies
   1. det A + det B = 0
   2. det A = 0 or det B = 0
   3. None of these
   4. det A = 0 and det B = 0
6. If {tex}\begin{vmatrix}2x&5\\8&x\end{vmatrix}=\begin{vmatrix}6&5\\8&3\end{vmatrix}{/tex}, then x is ________.
7. Multiplying a determinant by k means multiplying the elements of only one row (or one column) by ________.
8. 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.
9. Find adj A for {tex}A = \left[ {\begin{array}{*{20}{c}} 2&3 \\ 1&4 \end{array}} \right].{/tex}
10. {tex}A = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 4&8 \end{array}} \right]{/tex}is singular or not.
11. Evaluate {tex}2 \left| \begin{array} { r r } { 7 } & { – 2 } \\ { – 10 } & { 5 } \end{array} \right|{/tex}.
12. 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}.
13. Find the area of {tex}\Delta {/tex}whose vertices are (3, 8) (-4, 2) and (5, 1).
14. 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} is 3 square unit.
15. If A = {tex} \left[ \begin{array} { c c c } { 1 } & { – 2 } & { 3 } \\ { 0 } & { – 1 } & { 4 } \\ { – 2 } & { 2 } & { 1 } \end{array} \right]{/tex}, then find (A’)^-1.
16. If {tex}A = \left[ {\begin{array}{*{20}{c}} 3&{ – 4} \\ { – 1}&2 \end{array}} \right],{/tex} find matrix B such that AB = I.
17. Using properties of determinants, prove that
    {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}.
18. Given {tex}A = \left[ {\begin{array}{*{20}{c}} 1&{ – 1}&1 \\ 1&{ – 2}&{ – 2} \\ 2&1&3 \end{array}} \right]{/tex} and {tex}B = \left[ {\begin{array}{*{20}{c}} { – 4}&4&4 \\ { – 7}&1&3 \\ 5&{ – 3}&{ – 1} \end{array}} \right]{/tex}. find AB and use this result in solving the following system of equation.
    x – y + z = 4, x – 2y – 2z = 9, 2x + y + 3z = 1

Chapter 4 Determinants

Solution

1. 3. 1 , 2 and 3
      Explanation: Expanding along C_{1
      {tex}\left| {\begin{array}{*{20}{c}} {1 – x}&2&3 \\ 0&{2 – x}&0 \\ 0&2&{3 – x} \end{array}} \right| = 0 \Rightarrow {/tex} (1 – x)(2 – x)(3 – x) = 0{tex}\Rightarrow{/tex} x = 1, 2 ,3.}
   2. |A| = |A’|
      Explanation: 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.
   3. –1
      Explanation: {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}
      Put x = {tex}\frac{\pi}{3}{/tex}, {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}
      {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}
      {tex}\Rightarrow \left| {\begin{array}{*{20}{c}} 1&1&0 \\ 1&1&1 \\ 0&1&1 \end{array}} \right| {/tex}{tex} \Rightarrow 1(0) – 1(1) = – 1 \\{/tex}
   2. –1 , 2
      Explanation: {tex}\left| {\begin{array}{*{20}{c}} 1&4&{20} \\ 1&{ – 2}&5 \\ 1&{2x}&{5{x^2}} \end{array}} \right| = 0{/tex}
      Apply, R₃{tex}\rightarrow{/tex}R₃ – R₁, R₂{tex}\rightarrow{/tex}R₂ – R₁,
      {tex}\Rightarrow{/tex} {tex}\left| {\begin{array}{*{20}{c}} 1&4&{20} \\ 0&{ – 6}&{ – 15} \\ 0&{2x – 4}&{5{x^2} – 20} \end{array}} \right| = 0{/tex}
      {tex}\Rightarrow{/tex} -6(5x² – 20) + 15(2x – 4) =0
      {tex}\Rightarrow{/tex} (x – 2)(x + 1) = 0 {tex}\Rightarrow{/tex} x= 2 , -1.
   3. None of these
      Explanation: If det (A+B)=0 implies that A+B a Singular matrix.
2. x = {tex}\pm{/tex}3
3. k
4. sum
5. {tex}adjA = \left[ {\begin{array}{*{20}{c}} 4&{ – 3} \\ { – 1}&2 \end{array}} \right]{/tex}
   
6. {tex}\left| A \right| = \left| {\begin{array}{*{20}{c}} 1&2 \\ 4&8 \end{array}} \right|{/tex}
   = 8 – 8
   = 0
   Hence A is singular
7. According to the question, we have to evaluate {tex}2 \left| \begin{array} { r r } { 7 } & { – 2 } \\ { – 10 } & { 5 } \end{array} \right|{/tex}.
   Now, {tex}2 \left| \begin{array} { r r } { 7 } & { – 2 } \\ { – 10 } & { 5 } \end{array} \right| = 2 [ 35 – ( 20 ) ]{/tex}
   {tex}= 2 \times 15 = 30{/tex}
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}
   Expanding along first row,
   {tex}= \cos \alpha \cos \beta \left( {\cos \alpha \cos \beta – 0} \right) {/tex} {tex}- \cos \alpha \sin \beta \left( { – \cos \alpha \sin \beta – 0} \right) {/tex} {tex}- \sin \alpha \left( { – \sin \alpha {{\sin }^2}\beta – \sin \alpha {{\cos }^2}\beta } \right){/tex}
   {tex}= {\cos ^2}\alpha {\cos ^2}\beta + {\cos ^2}\alpha {\sin ^2}\beta {/tex} {tex}+ {\sin ^2}\alpha \left( {{{\sin }^2}\beta + {{\cos }^2}\beta } \right){/tex}
   {tex}= {\cos ^2}\alpha \left( {{{\cos }^2}\beta + {{\sin }^2}\beta } \right){/tex} {tex} + {\sin ^2}\alpha \left( {{{\sin }^2}\beta + {{\cos }^2}\beta } \right){/tex}
   {tex}= {\cos ^2}\alpha + {\sin ^2}\alpha{/tex}
   = 1
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}
   {tex}= \frac{1}{2}\left| {\begin{array}{*{20}{c}} 3&8&1 \\ { – 4}&2&1 \\ 5&1&1 \end{array}} \right|{/tex}
   {tex} = \frac{1}{2}\left[ {3\left( {2 – 1} \right) – 8\left( { – 4 – 5} \right) + 1\left( { – 4 + 10} \right)} \right]{/tex}
   {tex} = \frac{1}{2}\left[ {3 + 72 – 14} \right] = \frac{{61}}{2}{/tex}
10. Let P (x, y) be any point on AB. Then the equation of line AB is,
    {tex}\frac{1}{2}\left| {\begin{array}{*{20}{c}} 0&0&1 \\ 1&3&1 \\ x&y&1 \end{array}} \right| = 0{/tex}
    y = 3x
    Area {tex}\Delta ABD = 3{/tex} square unit
    {tex}\frac{1}{2}\left| {\begin{array}{*{20}{c}} 1&3&1 \\ 0&0&1 \\ K&0&1 \end{array}} \right| = \pm 3{/tex}
    {tex}k = \pm 2{/tex}
11. If A = {tex} \left[ \begin{array} { c c c } { 1 } & { – 2 } & { 3 } \\ { 0 } & { – 1 } & { 4 } \\ { – 2 } & { 2 } & { 1 } \end{array} \right]{/tex}, then we have to find (A’)^-1.
    Now, A = {tex} \left[ \begin{array} { c c c } { 1 } & { – 2 } & { 3 } \\ { 0 } & { – 1 } & { 4 } \\ { – 2 } & { 2 } & { 1 } \end{array} \right]{/tex} Therefore, we have, {tex} |A|=\left| \begin{array} { c c c } { 1 } & { – 2 } & { 3 } \\ { 0 } & { – 1 } & { 4 } \\ { – 2 } & { 2 } & { 1 } \end{array} \right|{/tex}
    = 1 (-1 – 8) + 2 (0 + 8) + 3 (0 – 2)
    [expanding along R₁] =-9+16-6=1{tex} \neq{/tex}0
    Therefore, A is non-singular matrix and hence its inverse exists.
    Cofactors of an element of |A| are given by
    {tex} A _ { 11 } = ( – 1 ) ^ { 1 + 1 } \left| \begin{array} { c c } { – 1 } & { 4 } \\ { 2 } & { 1 } \end{array} \right| = ( – 1 – 8 ) = – 9{/tex}
    {tex} A _ { 12 } = ( – 1 ) ^ { 1 } + 2 \left| \begin{array} { c c } { 0 } & { 4 } \\ { – 2 } & { 1 } \end{array} \right| = – ( 0 + 8 ) = – 8{/tex}
    {tex} A _ { 13 } = ( – 1 ) ^ { 1 + 3 } \left| \begin{array} { c c } { 0 } & { -1 } \\ { – 2 } & { 2 } \end{array} \right| = ( 0 – 2 ) = – 2{/tex}
    {tex} A _ { 21 } = ( – 1 ) ^ { 2 + 1 } \left| \begin{array} { c c } { – 2 } & { 3 } \\ { 2 } & { 1 } \end{array} \right| = – ( – 2 – 6 ) = 8{/tex}
    {tex} A _ { 22 } = ( – 1 ) ^ { 2 + 2 } \left| \begin{array} { c c } { 1 } & { 3 } \\ { – 2 } & { 1 } \end{array} \right| = ( 1 + 6 ) = 7{/tex}
    {tex} A _ { 23 } = ( – 1 ) ^ { 2 + 3 } \left| \begin{array} { c c } { 1 } & { – 2 } \\ { – 2 } & { 2 } \end{array} \right| = – ( 2 – 4 ) = 2{/tex}
    {tex} A _ { 31 } = ( – 1 ) ^ { 3 + 1 } \left| \begin{array} { c c } { – 2 } & { 3 } \\ { – 1 } & { 4 } \end{array} \right| = ( – 8 + 3 ) = – 5{/tex}
    {tex} A _ { 32 } = ( – 1 ) ^ { 3 + 2 } \left| \begin{array} { l l } { 1 } & { 3 } \\ { 0 } & { 4 } \end{array} \right| = – ( 4 – 0 ) = – 4{/tex}
    {tex} A _ { 33 } = ( – 1 ) ^ { 3 +3 } \left| \begin{array} { c c } { 1 } & { – 2 } \\ { 0 } & { – 1 } \end{array} \right| = ( – 1 – 0 ) = – 1{/tex}
    Thus, 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}
    Hence, {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}
    Now, (A’)^-1 = (A^-1)’ = {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}
12. {tex}\left| A \right| = 2 \ne 0{/tex}
    Therefore A^-1 exists
    AB = I
    A^-1 AB = A^-1I
    B = A^-1
    {tex}adjA = \left[ {\begin{array}{*{20}{c}} 2&4 \\ 1&3 \end{array}} \right]{/tex}
    {tex}{A^{ – 1}} = \frac{1}{{\left| A \right|}}\left( {adjA} \right){/tex}
    {tex}= \frac{1}{2}\left[ {\begin{array}{*{20}{c}} 2&4 \\ 1&3 \end{array}} \right]{/tex}
    {tex}= \left[ {\begin{array}{*{20}{c}} 1&2 \\ {\frac{1}{2}}&{\frac{3}{2}} \end{array}} \right]{/tex}
    Hence {tex}B = \left[ {\begin{array}{*{20}{c}} 1&2 \\ {\frac{1}{2}}&{\frac{3}{2}} \end{array}} \right]{/tex}
13. According to the question,we have to use properties of determinants to prove that,
    {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}
    Let 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}
    Therefore,on applying C₁{tex}\rightarrow{/tex} C₁ + C₂ + C₃ we get,
    {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}
    on taking 2 common from {tex}C_1{/tex},we get,
    {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}
    On applying C₂{tex}\rightarrow{/tex} C₂ – C₁ and C₃ {tex}\rightarrow{/tex} C₃ – C12,
    we get
    {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}
    on applying {tex}C_1\rightarrow C_1+C_2+C_3,we\ get,{/tex}
    {tex}\Delta = 2 \left| \begin{array} { c c c } { a } & { – b } & { – c } \\ { p } & { – q } & { – r } \\ { x } & { – y } & { – z } \end{array} \right|{/tex}
    {tex}\therefore{/tex} {tex}\Delta = 2 \left| \begin{array} { l l l } { a } & { b } & { c } \\ { p } & { q } & { r } \\ { x } & { y } & { z } \end{array} \right|{/tex} [taking (-1) common from both C₂ and C₃] = RHS
14. x – y + z = 4
    x – 2y – 2z = 9
    2x + y + 3z = 1
    Let {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}
    AX = C
    {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} {tex} = \left[ {\begin{array}{*{20}{c}} 8&0&0 \\ 0&8&0 \\ 0&0&8 \end{array}} \right]{/tex}
    AB = 8I
    {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}
    {tex} = \frac{1}{8}\left[ {\begin{array}{*{20}{c}} { – 4}&4&4 \\ { – 7}&1&3 \\ 5&{ – 3}&{ – 1} \end{array}} \right]{/tex}
    X = A^-1C
    {tex}\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3 \\ { – 2} \\ { – 1} \end{array}} \right]{/tex}
    x = 3, y = -2, z = -1

Chapter Wise Important Questions Class 12 Maths Part I and Part II

1. Relations and Functions
2. Inverse Trigonometric Functions
3. Matrices
4. Determinants
5. Continuity and Differentiability
6. Application of Derivatives
7. Integrals
8. Application of Integrals
9. Differential Equations
10. Vector Algebra
11. Three Dimensional Geometry
12. Linear Programming
13. Probability

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