## Filters

Sort by :
Clear All
Q

Q : 19        Let $A=\begin{vmatrix} 1 &\sin \theta &1 \\ -\sin \theta & 1 & \sin \theta \\ -1&-\sin \theta &1 \end{vmatrix},$     where $0\leq \theta \leq 2\pi$.  Then

(A)$Det(A)=0$                         (B) $Det(A)\in (2,\infty)$

(C) $Det(A)\in (2,4)$                 (D)$Det(A)\in [2,4]$

Given determinant  Now, given the range of  from  Therefore the  . Hence the correct answer is D.

Q : 18         If  x, y, z are nonzero real numbers, then the inverse of matrix $A=\begin{bmatrix} x &0 &0 \\ 0 &y &0 \\ 0 & 0 & z \end{bmatrix}$    is

$(A)\begin{bmatrix} x^-^1 &0 &0 \\ 0 &y^-^1 &0 \\ 0 & 0 & z^-^1 \end{bmatrix}$                                   $(B)xyz\begin{bmatrix} x^-^1 &0 &0 \\ 0 &y^-^1 &0 \\ 0 & 0 & z^-^1 \end{bmatrix}$

$(C)\frac{1}{xyz}\begin{bmatrix} x &0 &0 \\ 0 &y &0 \\ 0 & 0 & z \end{bmatrix}$                                        $(D)\frac{1}{xyz}\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

Given Matrix , As we know,  So, we  will find the , Determining its cofactor first,                Hence  Therefore the correct answer is (A).

Q : 17     If  $\dpi{100} a,b,c,$  are in A.P, then the determinant
$\dpi{100} \begin{vmatrix} x+2 &x+3 &x+2a \\ x+3 & x+4 & x+2b\\ x+4 & x+5 &x+2c \end{vmatrix}$   is

(A) $\dpi{100} 0$            (B) $\dpi{100} 1$           (C) $\dpi{100} x$        (D) $\dpi{100} 2x$

Given determinant  and given that a, b, c are in A.P. That means , 2b =a+c Applying the row transformations,    and then  we have; Now, applying another row transformation, , we have Clearly we have the determinant value equal to zero; Hence the option (A) is correct.

Q : 16         Solve the system of equations

$\dpi{100} \frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$

$\dpi{100} \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$

$\dpi{100} \frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$

We have a system of equations;           So, we will convert the given system of equations in a simple form to solve the problem by the matrix method; Let us take, ,  Then we have the equations;          We can write it in the matrix form as  , where Now, Finding the determinant value of A;                 Hence we can say that A is non-singular  its invers exists; Finding cofactors of...

Q : 15        Using properties  of determinants, prove that

$\dpi{100} \begin{vmatrix} \sin \alpha &\cos \alpha &\cos (\alpha +\delta ) \\ \sin \beta & \cos \beta & \cos (\beta +\delta )\\ \sin \gamma &\cos \gamma & \cos (\gamma +\delta ) \end{vmatrix}=0$

Given determinant  Multiplying the first column by  and the second column by , and expanding the third column, we get Applying column transformation,  we have then; Here we can see that two columns  are identical. The determinant value is equal to zero.  Hence proved.

Q :14        Using properties  of determinants, prove that

$\dpi{100} \begin{vmatrix} 1 &1+p &1+p+q \\ 2&3+2p &4+3p+2q \\ 3&6+3p & 10+6p+3q \end{vmatrix}=1$

Given determinant  Applying the row transformation;   and  we have then; Now, applying another row transformation  we have; We can expand the remaining determinant along , we have; Hence the result is proved.

Q : 13        Using properties  of determinants, prove that

$\dpi{100} \begin{vmatrix} 3a &-a+b &-a+c \\ -b+c &3b &-b+c \\ -c+a &-c+b &3c \end{vmatrix}=3(a+b+c)(ab+bc+ca)$

Given determinant  Applying the column transformation,  we have then; Taking common factor (a+b+c) out from the column first; Applying   and  , we have then; Now we can expand the remaining determinant along  we have;   Hence proved.

Q : 12        Using properties  of determinants, prove that

$\dpi{100} \begin{vmatrix} x & x^2&1+px^3 \\ y& y^2& 1+py^3\\ z&z^2 & 1+pz^3 \end{vmatrix}=(1+pxyz)(x-y)(y-z)(z-x),$  where p is any scalar.

Given the determinant  Applying the row transformations;   and  then we have; Applying row transformation  we have then; Now we can expand the remaining determinant to get the result; hence the given result is proved.

Q : 11        Using properties  of determinants, prove that

$\dpi{100} \begin{vmatrix} \alpha & \alpha ^2 &\beta +\gamma \\ \beta & \beta ^2 &\gamma +\alpha \\ \gamma &\gamma ^2 &\alpha +\beta \end{vmatrix}=(\beta -\gamma )(\gamma -\alpha )(\alpha -\beta )(\alpha +\beta +\gamma )$

Given determinant   Applying Row transformations; and  , then we have; Expanding the remaining determinant; hence the given result is proved.

Q : 9        Evaluate  $\dpi{100} \begin{vmatrix} x & y &x+y \\ y & x+y &x \\ x+y & x & y \end{vmatrix}$

We have determinant   Applying row transformations;   , we have then; Taking out the common factor 2(x+y) from the row first. Now, applying the column transformation;  and   we have ; Expanding the remaining determinant; .

Q : 10        Evaluate  $\dpi{100} \begin{vmatrix} 1 & x &y \\ 1 &x+y &y \\ 1 &x &x+y \end{vmatrix}$

We have determinant   Applying row transformations;    and  then  we have then; Taking out the common factor -y from the row first. Expanding the remaining determinant;

Q : 8        Let $\dpi{100} A=\begin{bmatrix} 1 &2 &1 \\ 2 & 3 &1 \\ 1 & 1 & 5 \end{bmatrix}$, Verify that

(ii) $\dpi{100} (A^-^1)^-^1=A$

Given that ; So, let us assume that   Hence its inverse exists;     or  ; so, we now calculate the value of  Cofactors of A;                                               Finding the inverse of B ; Hence its inverse exists; Now, finding the ;                                                                                           Hence proved L.H.S. =R.H.S..

Q : 8        Let $\dpi{100} A=\begin{bmatrix} 1 &2 &1 \\ 2 &3 &1 \\ 1 & 1 & 5 \end{bmatrix}$. Verify that,

(i) $\dpi{100} [adj A]^-^1 = adj (A^-^1)$

Given that ; So, let us assume that  matrix and  then; Hence its inverse exists;     or  ; so, we now calculate the value of  Cofactors of A;                                              ??????? ??????? Finding the inverse of C; Hence its inverse exists; Now, finding the ;                                                                                            or  Now, finding the...

Q : 7        If     $\dpi{100} A^-^1=\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 5 &-2 &2 \end{bmatrix}$  and $\dpi{100} B=\begin{bmatrix} 1 &2 &-2 \\ -1&3 &0 \\ 0 &-2 &1 \end{bmatrix}$, find $\dpi{100} (AB)^-^1$.

We know from the identity that; . Then we can find easily,  Given   and   Then we have to basically find the  matrix.   So, Given matrix  Hence its inverse  exists; Now, as we know that So, calculating cofactors of B,                                                                                   Now, We have both  as well as  ; Putting in the relation we know;

Q : 6        Prove that    $\dpi{100} \begin{vmatrix} a^2 &bc &ac+a^2 \\ a^2+ab & b^2 & ac\\ ab &b^2+bc &c^2 \end{vmatrix}=4a^2b^2c^2$.

Given matrix  Taking common factors a,b and c from the column  respectively. we have;  Applying , we have; Then applying  , we get; Applying , we have; Now, applying column transformation; , we have So we can now expand the remaining determinant along  we have; Hence proved.

Q : 5        Solve the equation

$\dpi{100} \begin{vmatrix} x+a & x &x \\ x &x+a &x\\ x & x & x+a \end{vmatrix}=0; a\neq 0$

Given determinant  Applying the row transformation;  we have; Taking common factor (3x+a) out from first row. Now applying the column transformations;  and . we get;            as    , or   or

Q : 4        If $\dpi{100} a,b$ and $\dpi{100} c$ are real numbers, and

$\dpi{100} \Delta =\begin{vmatrix} b+c & c+a &a+b \\ c+a &a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}=0$

Show that either $\dpi{100} a+b+c=0$ or $\dpi{100} a=b=c$

We have given  Applying the row transformations;  we have; Taking out common factor 2(a+b+c) from the first row; Now, applying the column transformations;  we have; and given that the determinant is equal to zero. i.e., ;  So, either  or . we can write  as;  are non-negative. Hence . we get then  Therefore, if given  = 0 then either  or .

Q : 3        Evaluate    $\dpi{100} \begin{vmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta &-\sin \alpha \\ -\sin \beta & \cos \beta &0 \\ \sin \alpha \cos \beta &\sin \alpha \sin \beta & \cos \alpha \end{vmatrix}$.

Given determinant ; .

Q : 2         Without expanding the determinant, prove that
$\dpi{100} \begin{vmatrix} a &a^2 &bc \\ b& b^2 &ca \\ c & c^2 &ab \end{vmatrix}= \begin{vmatrix} 1 &a^2 &a^3 \\ 1 &b^2 &b^3 \\ 1 & c^2 &c^3 \end{vmatrix}$

We have the   Multiplying rows with a, b, and c respectively. we get;                                   = R.H.S. Hence proved. L.H.S. =R.H.S.

Q : 1         Prove that the determinant  $\dpi{100} \begin{vmatrix} x & \sin \theta &\cos \theta \\ -\sin \theta &-x & 1\\ \cos \theta &1 &x \end{vmatrix}$  is independent of $\dpi{100} \theta$.

Calculating the determinant value of ; Clearly, the determinant is independent of .
Exams
Articles
Questions