Q&A - Ask Doubts and Get Answers

Sort by :
Clear All
Q

Choose the correct answer. 

Q : 16         Which of the following is correct


                  (A) Determinant is a square matrix.
                  (B) Determinant is a number associated to a matrix.
                  (C) Determinant is a number associated to a square matrix.
                  (D) None of these

The answer is (C) Determinant is a number associated to a square matrix. As we know that To every square matrix of order n, we can associate a number (real or complex) called determinant of the square matrix A, where element of A.

Choose the correct answer. 

Q : 15        Let A be a square matrix of order 3\times 3 , then |kA| is equal to

                  (A) k|A|          (B) k^2|A|        (C) k^3|A|        (D)  3k|A|

Assume a square matrix A of order of . Then we have; (Taking the common factors k from each row.)   Therefore correct option is (C).    

Q : 14            By using properties of determinants, show that:

                       \begin{vmatrix} a^2+1 &ab &ac \\ ab &b^2+1 &bc \\ ca & cb &c^2+1 \end{vmatrix}=1+a^2+b^2+c^2
    

Given determinant:                               Let  Then we can clearly see that each column can be reduced by taking common factors like a,b, and c respectively from C1,C2,and C3. We then get; Now, applying column transformations:   and   then we have; Now, expanding the remaining determinant: . Hence proved.  

Q : 13        By using properties of determinants, show that:

                \begin{vmatrix} 1+a^2-b^2 &2ab &-2b \\ 2ab &1-a^2+b^2 &2a \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}=(1+a^2+b^2)^3

We have determinant:                              Applying row transformations,     and    then we have; taking common factor out of the determinant; Now expanding the remaining determinant we get; Hence proved.

Q : 12        By using properties of determinants, show that:

                \begin{vmatrix} 1 &x &x^2 \\ x^2 &1 &x \\ x &x^2 &1 \end{vmatrix}=(1-x^3)^2

Give determinant   Applying column transformation  we get;       [after taking the (1+x+x2 ) factor common out.] Now, applying row transformations,     and then . we have now, As we know  Hence proved.

Q : 11        By using properties of determinants, show that:

                  (ii) \begin{vmatrix} x+y+2z &x &y \\ z & y+z+2x & y\\ z & x &z+x+2y \end{vmatrix}=2(x+y+z)^3

Given determinant    Applying    we get; Taking 2(x+y+z) factor out, we get; Now, applying row transformations,   and then . we get; Hence proved.

Q : 11         By using properties of determinants, show that:

                   (i) \begin{vmatrix} a-b-c &2a &2a \\ 2b &b-c-a &2b \\ 2c &2c &c-a-b \end{vmatrix}=(a+b+c)^3

Given determinant:                                 We apply row transformation:  we have; Taking common factor (a+b+c) out. Now, applying column tranformation     and then   We have; Hence Proved.        

Q : 10          By using properties of determinants, show that:

 

                (ii)\begin{vmatrix} y+k & y & y\\ y & y+k &y \\ y & y & y+k \end{vmatrix}=k^2(3y+k)    

Given determinant:                                        Applying row transformation   we get;                             [taking common (3y + k) factor] Now, applying column transformation   and  Hence proved.

Q : 10        By using properties of determinants, show that:

                (i) \begin{vmatrix} x+4 &2x &2x \\ 2x & x+4 & 2x\\ 2x & 2x & x+4 \end{vmatrix}=(5x+4)(4-x)

Given determinant:                                        Applying row transformation:   then we have; Taking a common factor: 5x+4 Now, applying column transformations   and   

Q : 9        By using properties of determinants, show that:

                \begin{vmatrix} x & x^2 & yz\\ y & y^2 &zx \\ z & z^2 & xy \end{vmatrix}=(x-y)(y-z)(z-x)(xy+yz+zx)

We have the determinant:                                             Applying the row transformations  and then  , we have; Now, applying ; we have Now, expanding the remaining determinant; Hence proved.    

Q : 8        By using properties of determinants, show that:

              (ii) \begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ a^3 &b^3 &c^3 \end{vmatrix}=(a-b)(b-c)(c-a)(a+b+c)

Given determinant :                                           , Applying column transformation  and then  We get, Now, applying column transformation , we have: Hence proved.

Q : 8        By using properties of determinants, show that:

                (i)      \begin{vmatrix} 1 &a &a^2 \\ 1 &b &b^2 \\ 1 &c &c^2 \end{vmatrix}=(a-b)(b-c)(c-a)
 

We have the determinant  Applying the row transformations   and then   we have: Now, applying  we have:     or     Hence proved.

Q : 7          Using the property of determinants and without expanding, prove that 

                  \begin{vmatrix} -a^2 &ab &ac \\ ba &-b^2 &bc \\ ca & cb & -c^2 \end{vmatrix}=4a^2b^2c^2

Given determinant :  As we can easily take out the common factors a,b,c from rows  respectively. So, get then: Now, taking common factors a,b,c from the columns  respectively. Now, applying rows transformations    and then  we have; Expanding to get R.H.S.

Q : 6          Using the property of determinants and without expanding, prove that     

                  \begin{vmatrix} 0 &a &-b \\-a &0 & -c\\b &c &0 \end{vmatrix}=0

We have given determinant   Applying transformation,  we have then, We can make the first row identical to the third row so, Taking another row transformation:  we have, So, determinant has two rows  identical. Hence .

Q : 5           Using the property of determinants and without expanding, prove that 

                   \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a+b &p+q & x+y \end{vmatrix}=2\begin{vmatrix} a &p &x \\ b &q &y \\ c &r & z \end{vmatrix}

Given determinant :    Splitting the third row; we get, . Then we have, On Applying row transformation    and then  ; we get,  Applying Rows exchange transformation    and   , we have: also  On applying rows transformation,  and then    and then   Then applying rows exchange transformation;    and then . we have then; So, we now calculate the sum =  Hence proved.  

Q : 4           Using the property of determinants and without expanding, prove that 

                    \begin{vmatrix}1 &bc &a(b+c) \\1 &ca &b(c+a) \\1 &ab & c(a+b) \end{vmatrix}=0

We have determinant:    Applying  we have then; So, here column 3 and column 1 are proportional. Therefore, .

Q : 3          Using the property of determinants and without expanding, prove that 

                    \begin{vmatrix}2 & 7 &65 \\3 &8 &75 \\5 &9 &86 \end{vmatrix}=0

Given determinant  So, we can split it in two addition determinants:      [ Here two columns are identical ] and     [ Here two columns are identical ]                   Therefore we have the value of determinant = 0.     

Q : 2      Using the property of determinants and without expanding, prove that 

              \begin{vmatrix}a-b &b-c &c-a \\b-c &c-a &a-b \\c-a &a-b &b-c \end{vmatrix}=0

   

Given determinant  Applying the rows addition    then we have; So, we have two rows  and  identical hence we can say that the value of determinant = 0  Therefore .  

Q : 1        Using the property of determinants and without expanding, prove that 

                  \begin{vmatrix}x &a &x+a \\y &b &y+b \\z &c &z+c \end{vmatrix}=0

We can split it in manner like;   So, we know the identity that If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero. Clearly, expanded determinants have identical columns. Hence the sum is zero.
Exams
Articles
Questions