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Q : 8        If   \begin{vmatrix}x &2 \\18 &x \end{vmatrix}=\begin{vmatrix} 6 &2 \\ 18 &6 \end{vmatrix} ,  then x is equal to

                (A) 6        (B) \pm 6       (C) -6     (D) 0

Solving the L.H.S. determinant ; and solving R.H.S determinant; So equating both sides;         or             or      Hence answer is (B).  

Q : 7         Find values of x, if

                 (ii)  \begin{vmatrix}2 &3 \\ 4 &5 \end{vmatrix}=\begin{vmatrix}x &3 \\2x &5 \end{vmatrix}

Given  ; So, we here equate both sides after calculating each side's determinant values. L.H.S. determinant value; Similarly R.H.S. determinant value; So, we have then;     or    .

Q : 7         Find values of x, if

      (i)         \begin{vmatrix}2 &4 \\5 &1 \end{vmatrix} =\begin{vmatrix}2x &4 \\6 &x \end{vmatrix}

Given that  First, we solve the determinant value of L.H.S. and equate it to the determinant value of R.H.S.,       and    So, we have then,       or           or   

Q : 6     If   A=\begin{bmatrix}1 & 1 & -2\\ 2& 1 &-3 \\5 &4 &-9 \end{bmatrix} ,  then find  |A|.

Given the matrix  then, Finding the determinant value of A;

Q : 5     Evaluate the determinants. 

      (iv)    \begin{vmatrix}2 &-1 &2 \\0 &2 &-1 \\3 &-5 &0 \end{vmatrix}

Given determinant: , We now calculate determinant value:

Q : 5  Evaluate the determinants. 

        (iii)   \begin{vmatrix}0 & 1 & 2\\-1 &0 &-3 \\ -2 &3 &0 \end{vmatrix}

Given determinant ; Now calculating the determinant value;

Q : 5         Evaluate the determinants.

                (ii)  \begin{vmatrix}3 &-4 &5 \\1 &1 &-2 \\2 &3 &1 \end{vmatrix}

Given determinant ; Now calculating the determinant value; .

Q : 5         Evaluate the determinants.

                (i)    \begin{vmatrix}3 &-1 &-2 \\0 &0 &-1 \\3 &-5 & 0 \end{vmatrix}

GIven the determinant ; now, calculating its determinant value, .

Q : 4         If A =\begin{bmatrix} 1 &0 &1 \\ 0& 1& 2\\ 0& 0 &4 \end{bmatrix}  then show that  |3A|=27|A|

Given Matrix Calculating  So,   calculating , So,  Therefore . Hence proved.

Q : 3         If   A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix}  , then show that  | 2 A |=4|A|

Given determinant  then we have to show that , So,  then,  Hence we have  So, L.H.S. = |2A| = -24 then calculating R.H.S.  We have, hence R.H.S becomes  Therefore L.H.S. =R.H.S. Hence proved.

Q : 2      

(ii)        \begin{vmatrix}x^2-x+1 & x-1\\x+1 &x+1 \end{vmatrix}

We have determinant  So, 

Q : 2 

         (i)      \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta &\cos \theta \end{vmatrix}

The given two by two determinant is calculated as follows

 Q : 1      

             \begin{vmatrix} 2 & 4\\ -5 & -1\end{vmatrix}

The determinant is evaluated as follows  
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