Extra questions on Determinants notes for Class 12

Multiple Choice Questions

Question 1
If $x,y \in R$, then the determinant $\Delta=\begin{vmatrix} cos x & -sin x & 1 \\ sinx & cos x & 1 \\ cos(x+y) & -sin(x+y) & 0 \\ \end{vmatrix}$ lies in
(a) $[-\sqrt 2 , \sqrt 2]$
(b) $[-1 ,1]$
(c) $[-\sqrt 2 , 1]$
(d) $[-1,-\sqrt 2]$

Answer

The correct choice is A. Indeed applying $R_3 -> R_3 - cosy R_1 + siny R_2$, we get
$\Delta=\begin{vmatrix} cos x & -sin x & 1 \\ sinx & cos x & 1 \\ cos(x+y) & -sin(x+y) & 0 \\ \end{vmatrix}$
$\Delta=\begin{vmatrix} cos x & -sin x & 1 \\ sinx & cos x & 1 \\ 0 & 0 & sin y - cos y \\ \end{vmatrix}$
Now expanding
$\Delta= sin y - cos y = \sqrt 2 sin( y - \frac {\pi}{4})$
Hence $\Delta$ lies in $[-\sqrt 2 , \sqrt 2]$

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Question 2
If A,B,C are the angles of the triangle, then the determinant $\begin{vmatrix} -1 & cos C & cos B \\ cos C & -1 & Cos A \\ cos B & cos A & -1 \\ \end{vmatrix}$ is equal to
(a) 1
(b) -1
(c) 0
(d) Nones of these

Answer

Doing $C_1 -> aC_1 + bC_2 + cC_3$
$\begin{vmatrix} -a + bcos C + c cos B & cos C & cos B \\ a cos C -b + ccos A & -1 & Cos A \\ a cos B + bcos A - c & cos A & -1 \\ \end{vmatrix}$
Now we know that in a triangle
a=bcos C + c cos A
b= acos C + c cos A
c= aCos B + b cos A
hence
$\begin{vmatrix} -a + a & cos C & cos B \\ b -b & -1 & Cos A \\ c - c & cos A & -1 \\ \end{vmatrix}$
$\begin{vmatrix} 0 & cos C & cos B \\ 0 & -1 & Cos A \\ 0 & cos A & -1 \\ \end{vmatrix} =0$

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Question 3
if $f(x) = \begin{vmatrix} (1+x)^{17} & (1+x)^{19} & (1+x)^{23} \\ (1+x)^{23} & (1+x)^{29} & (1+x)^{34} \\ (1+x)^{41} & (1+x)^{43} & (1+x)^{47} \\ \end{vmatrix} = A + Bx + Cx^2 +$...
(a)0
(b)1
(c)-1
(d)Nones of these

Answer

$f(x) = \begin{vmatrix} (1+x)^{17} & (1+x)^{19} & (1+x)^{23} \\ (1+x)^{23} & (1+x)^{29} & (1+x)^{34} \\ (1+x)^{41} & (1+x)^{43} & (1+x)^{47} \\ \end{vmatrix}$
$=(1+x)^{59} \begin{vmatrix} 1 & 1 & 1 \\ (1+x)^6 & (1+x)^{10} & (1+x)^{11} \\ (1+x)^{24} & (1+x)^{24} & (1+x)^{24} \\ \end{vmatrix}$
$=(1+x)^83 \begin{vmatrix} 1 & 1 & 1 \\ (1+x)^6 & (1+x)^{10} & (1+x)^{11} \\ 1 & 1 & 1 \\ \end{vmatrix} =0$

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Question 4
if $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ ^nC_1 & ^{n+1}C_1 & ^{n+2}C_1 \\ ^nC_2 & ^{n+2}C_2 & ^{n+4}C_2 \\ \end{vmatrix}$,then what is the vale of it
(a) 4
(b) 8
(c) dependent on n
(d) None of these

Answer

$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ ^nC_1 & ^{n+1}C_1 & ^{n+2}C_1 \\ ^nC_2 & ^{n+2}C_2 & ^{n+4}C_2 \\ \end{vmatrix}$
$=\begin{vmatrix} 1 & 1 & 1 \\ n & n+1 & n+2 \\ \frac {n(n-1)}{2} & \frac {(n+2)(n+1)}{2} & \frac {(n+4)(n+3)}{2} \\ \end{vmatrix}$
Applying $C_2 > C_2 -C_1$ and $C_3 > C_3 -C_1$
$=\begin{vmatrix} 1 & 0 & 0 \\ n & 1 & 2 \\ \frac {n(n-1)}{2} & 2n+ 1 & 4n+6 \\ \end{vmatrix}$
$=8$

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Short questions

Question 5
if A is a square matrix ,such that |A|=5 , then find the value of $|AA^T|$

Question 6
$A=\begin{bmatrix} 1 & 2 \\ 3 & -1 \\ \end{bmatrix}$
and $B=\begin{bmatrix} 1 & -4 \\ 3 & -2 \\ \end{bmatrix}$
Find the value |AB|

Question 7
A is a square matrix with |A| = 4. Then find the value of |A . (adj A)|.

Long questions

Question 8
$\begin{vmatrix} (b+c)^2 & a^2 & bc \\ (a+c)^2 & b^2 & ac \\ (a+b)^2 & c^2 & ab \\ \end{vmatrix} = (a-b)(b-c)(c-a)(a+b+c)(a^2 + b^2 + c^2)$

Question 8
$\begin{vmatrix} 1 & 1+p & 1+p +q \\ 3 & 4+3p & 2+ 4p + 3q \\ 4 & 7 + 4p & 2 + 7p + 4q\\ \end{vmatrix} = 1$

Question 9
Using properties of determent ,show that triangle ABC is isoceles if
$\begin{vmatrix} 1 & 1 & 1 \\ 1+ cos A & 1 + cos B & 1 + cos C \\ cos^2A + cos A & cos^2B + cos B & cos^2C + cos C \\ \end{vmatrix} = 0$

Question 10
Solve for x
$\begin{vmatrix} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \\ \end{vmatrix} = 0$

Question 11
Prove using properties of determinants
$\begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \\ \end{vmatrix} = 4abc$

Question 12
Prove using properties of determinants
$\begin{vmatrix} a^2 & bc & ac + c^2 \\ a^2 +ab & b^2 & ac \\ ab & b^2 + bc & c^2 \\ \end{vmatrix} = 4a^2b^2c^2$

Question 13
Prove using properties of determinants
$\begin{vmatrix} 3a & -a+b & -a+c \\ -b +a & 3b & -b+c \\ -c+a & -c+b & 3c \\ \end{vmatrix} = 3(a+b+c)(ab+ bc+ac)$

Question 14
Prove using properties of determinants
$\begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \\ \end{vmatrix} = a^3 + b^3 + c^3 -3abc$
Also Read

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• Notes
  • What is Determinants
  • Properties of Determinants
  • Minor and Cofactors of Determinants
  • Adjoint and Inverse of Matrix
  NCERT Solutions & Assignments
  • Determinants Extra Questions for Class 12

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