CHAPTER – 4: DETERMINANTS  
           
                                                                                MARKS WEIGHTAGE – 10 marks 
                                                                                                                         
          NCERT Important Questions & Answers 
           
          1.  If  x   2     6   2 , then find the value of x. 
                 18 x 18 6
              Ans: 
              Given that  x    2     6   2  
                          18 x 18 6
              On expanding both determinants, we get 
                                                     2
              x × x − 18 × 2 = 6 × 6 − 18 × 2  x  − 36 = 36 − 36 
                   2                2
                x  − 36 = 0  x  = 36  
               x = ± 6 
           
                            a     ab        abc
          2.  Prove that  2a     3a2b 4a3b2c a3 
                            3a 6a3b 10a6b3c
              Ans: 
              Applying operations R  → R  – 2R  and R  → R  – 3R  to the given determinant Δ, we have 
                                      2     2      1       3     3      1
                   a ab abc
                0      a      2ab  
                   0    3a     7a3b
              Now applying R  → R  – 3R  , we get 
                               3      3      2
                   a ab abc
                0      a      2ab  
                   0     0        a
              Expanding along C , we obtain 
                                  1
               aa 2ab00a(a20)a(a2)a3 
                     0     a
                           bc      a       a
          3.  Prove that     b     ca      b   4abc 
                             c       c    ab
              Ans: 
                        bc      a       a
               Let  b         ca      b    
                          c       c    ab
              Applying R  → R  – R  – R  to Δ, we get 
                          1      1    2    3
                   0    2c    2b
               b ca          b    
                   c     c     ab
              Expanding along R , we obtain 
                                  1
               0ca         b   (2c) b      b   (2b) b    ca  
                       c    ab            c  ab            c    c
          Prepared by: M. S. KumarSwamy, TGT(Maths)                                                Page - 1 - 
           
           
                             2                    2
              = 2 c (a b + b  – bc) – 2 b (b c – c  – ac) 
                             2       2      2         2
              = 2 abc + 2 cb  – 2 bc  – 2 b c + 2 bc  + 2abc 
              = 4 abc 
           
                                                x   x2   1x3
          4.  If x, y, z are different and   y    y2   1y3 0then show that 1 + xyz = 0 
                                                z   z2   1z3
              Ans: 
                             x   x2   1x3
              We have  y       y2   1y3  
                             z   z2   1z3
              Now, we know that If some or all elements of a row or column of a determinant are expressed as 
              sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) 
              determinants. 
                      x  x2   1    x  x2    x3
               y y2 1 y y2 y3  
                      z   z2  1    z  z2    z3
                      1 x x2           1 x x2
               (1)2 1   y   y2  xyz 1   y   y2                (Using C ↔C  and then C  ↔ C ) 
                                                                          3    2            1      2
                      1 z z2           1 z z2
                         1 x x2
               (1xyz)1 y y2  
                         1 z z2
                          1     x       x2
               (1xyz) 0     yx y2x2                  (Using R →R –R  and R  → R –R ) 
                                                                  2     2   1      3      3   1
                          0   z  x  z2  x2
              Taking out common factor (y – x) from R  and (z – x) from R , we get 
                                                         2                    3
                                                         1 x      x2
                              (1xyz)(yx)(zx) 0 1 yx  
                                                         0 1 zx
                              = (1 + xyz) (y – x) (z – x) (z – y) (on expanding along C ) 
                                                                                       1
              Since Δ = 0 and x, y, z are all different, i.e., x – y ≠ 0, y – z ≠ 0, z – x ≠ 0, we get 
              1 + xyz = 0 
                           1a     1      1              1   1   1
          5.  Show that      1    1b     1 abc1   abcbccaab 
                                                        a   b   c
                             1     1    1c                       
              Ans: 
                      1a      1      1
               LHS  1       1b      1  
                        1      1    1c
              Taking out factors a,b,c common from R , R  and R , we get 
                                                         1   2      3
          Prepared by: M. S. KumarSwamy, TGT(Maths)                                                Page - 2 - 
           
           
                       1 1     1      1
                       a        a      a
              abc 1         11      1    
                         b    b        b
                         1      1     11
                         c      c     c
              Applying R → R  + R  + R , we have 
                          1    1     2    3
                       1111 1111 1111
                          a   b   c      a   b   c      a   b   c
              abc          1            11              1        
                             b            b                b
                             1              1            11
                             c              c            c
               
              Now applying C  → C  – C , C  → C  – C , we get 
                               2     2    1   3      3    1
                                       1   0 0
                           1   1   1 1
              abc1                  1 0 
                          a   b   c b
                                    
                                       1   0 1
                                       c
                              1   1   1
              abc1   1(10)  
                                                
                             a   b   c
                                       
                         1   1   1
              abc1   abcbccaab= RHS 
                        a   b   c
                                  
          6.  Using the property of determinants and without expanding, prove that 
               bc qr yz              a   p   x
               ca rp zx 2b q y  
               ab pq xy              c   r   z
              Ans: 
                      bc qr yz
              LHS  ca r p zx  
                      ab pq xy
                 bc ca ab
               qr rp pq   (interchange row and column) 
                 yz zx xy
                 bc ca 2c
               qr rp 2r                      [usingC  → C  − (C  + C )] 
                                                        3      3     1    2
                 yz zx 2z
                      bc ca c
              (2) qr      r  p  r   (taking ‘–2’ common from C ) 
                                                                      3
                      yz zx z
          Prepared by: M. S. KumarSwamy, TGT(Maths)                                              Page - 3 - 
           
           
                   b   a  c
             (2) q   p r                            (using C  → C  – C  and C   → C  − C ) 
                                                             1     1    3     2      2    3
                   y   x  z
                 a  b c
             2 p q r                                                                             (using 
                 x  y   z
            C ↔ C ) 
              1     2
                a   p   x
             2b q y RHS                  (interchange row and column) 
                c   r   z
         7.  Using the property of determinants and without expanding, prove that 
             a2   ab    ac
              ba   b2   bc 4a2b2c2 
              ca   cb   c2
            Ans: 
                    a2   ab    ac       a   b    c
             LHS  ba    b2    bc abc a     b   c     [taking out factors a from R , b from R  and c from 
                                                                                  1         2
                    ca    cb   c2        a   b   c
            R] 
              3
                         1   1   1
             (abc)(abc) 1   1   1 (taking out factors a from C , b from C  and c from C ) 
                                                              1         2             3
                          1   1   1
                     0   0    2
             a2b2c2 0  2    2            (using R  → R  + R  and R  → R  − R ) 
                                                   1     1   2      2     2    3
                     1   1   1
            Expanding corresponding to first row R1, we get 
             a2b2c220 2  
                        1   1
             a2b2c22(02)4a2b2c2  RHS 
         8.  Using the property of determinants and without expanding, prove that 
             1 a a2
             1 b b2 (ab)(bc)(ca) 
             1 c c2
            Ans: 
                    1 a a2
             LHS  1 b b2  
                    1 c c2
            Applying R R R  and R R R , we get 
                       1    1   3        2    2   3
               0 ac a2c2        0 ac (ac)(ac)
              0 bc b2c2  0 bc (bc)(bc)  
               1    c      c2     1    c        c2
            Taking common factors (a − c) and (b − c) from R1 and R2 respectively, we get 
         Prepared by: M. S. KumarSwamy, TGT(Maths)                                      Page - 4 -