and 
are collinear then the value of 
is 
Solution: Option(C) To find: Area of $A B C$ Given: $A(3,-2), B(k, 2)$ and $C(8,8)$ The formula used: $\Delta=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} &...
are 
and 
The area of a is 
units 
unitsSolution: Option(B) To find: Area of $A B C$ Given: $A(-2,4), B(2,-6)$ and $C(5,4)$ Formula used: $\Delta=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\...
is 
Solution: Option(B) To find: Value of $x$ We have, $\left|\begin{array}{ccc}3 x-8 & 3 & 3 \\ 3 & 3 x-8 & 3 \\ 3 & 3 & 3 x-8\end{array}\right|=0$ Applying $\mathrm{R}_{1}...
is 
Solution: Option(B) To find: Value of $x$ We have, $\left|\begin{array}{ccc}a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x\end{array}\right|=0$ Applying...
is 
Solution: Option(A) To find: Value of $x$ We have, $\left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \\ x-4 & 2 x-9 & 3 x-16 \\ x-8 & 2 x-27 & 3 x-64\end{array}\right|=0$ Applying...
is 
Solution: Option(C) To find: Value of $x$ We have, $\left|\begin{array}{lll}\mathrm{x} & 3 & 7 \\ 2 & \mathrm{X} & 2 \\ 7 & 6 & \mathrm{x}\end{array}\right|=0$ Applying...
then 
Solution: Option(C) To find: Value of $\mathrm{x}$ We have, $\left|\begin{array}{ccc}5 & 3 & -1 \\ -7 & x & 2 \\ 9 & 6 & -2\end{array}\right|=0$ Applying $R_{1} \rightarrow 2...
Solution: Option(A) To find: Value of $\left|\begin{array}{ccc}\mathrm{x}+1 & \mathrm{x}+2 & \mathrm{x}+4 \\ \mathrm{x}+3 & \mathrm{x}+5 & \mathrm{x}+8 \\ \mathrm{x}+7 &...
Solution: Option(C) To find: Value of $\left|\begin{array}{lll}a & 1 & b+c \\ b & 1 & c+a \\ c & 1 & a+b\end{array}\right|$ We have, $\left|\begin{array}{lll}a & 1 &...
Solution: Option(A) To find: Value of $\left|\begin{array}{ccc}b+c & a & a \\ b & c+a & b \\ c & c & a+b\end{array}\right|$ We have, $\left|\begin{array}{ccc}b+c & a...
Solution: Option(A) To find: Value of $\left|\begin{array}{lll}\mathrm{bc} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{ca} & \mathrm{a}+\mathrm{c} & 1 \\ \mathrm{ab} &...
Solution: Option(C) To find: Value of $\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|$ We have, $\left|\begin{array}{ccc}1 & 1 & 1...
Solution: Option(C) To find: Value of $\left|\begin{array}{lll}b+c & a & b \\ c+a & c & a \\ a+b & b & c\end{array}\right|$ We have,...
Solution: Option(B) To find: Value of $\left|\begin{array}{ccc}a & a+2 b & a+2 b+3 c \\ 3 a & 4 a+6 b & 5 a+7 b+9 c \\ 6 a & 9 a+12 b & 11 a+15 b+18 c\end{array}\right|$ We...
Solution: Option(C) To find: Value of $\left|\begin{array}{ccc}a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|$ We have,...
Solution: Option( B) To find: Value of $\left|\begin{array}{ccc}x+y & x & x \\ 5 x+4 y & 4 x & 2 x \\ 10 x+8 y & 8 x & 3 x\end{array}\right|$ We have,...
be distinct positive real numbers then the value of 
is Solution: Option(B) To find: Nature of $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ We have, $\left|\begin{array}{lll}a & b & c \\...
Solution: Option(A) To find: Value of $\left|\begin{array}{lll}\operatorname{sind} & \cos \alpha & \sin (\alpha+\bar{\delta}) \\ \sin \beta & \cos \beta & \sin (\beta+\bar{\delta})...
Solution: Option(C) To find: Value of $\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3}\end{array}\right|$ We have, $\left|\begin{array}{ccc}1 & 1...
Solution: Option(B) To find: Value of $\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 1+3 p+2 q \\ 3 & 6+3 p & 1+6 p+3 q\end{array}\right|$ We have,...
Solution: Option(D) To find: Value of $\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$ We have, $\left|\begin{array}{lll}a-b...
Solution: Option(24) To find: Value of $\left|\begin{array}{lll}1 ! & 2 ! & 3 ! \\ 2 ! & 3 ! & 4 ! \\ 3 ! & 4 ! & 5 !\end{array}\right|$ We have,...
Solution: Option(A) To find: Value of $\left|\begin{array}{lll}1^{2} & 2^{2} & 3^{2} \\ 2^{2} & 3^{2} & 4^{2} \\ 3^{2} & 4^{2} & 5^{2}\end{array}\right|$ We have,...
is a complex cube root of unity then the value of 
is Solution: Option(0) To find: Value of $\left|\begin{array}{ccc}1 & \omega & 1+\omega \\ 1+\omega & 1 & \omega \\ \omega & 1+\omega & 1\end{array}\right|$ Formula used: (i)...
is a complex root of unity then 
Solution: Option(C) To find: Value of $\left|\begin{array}{ccc}1 & \omega & \omega^{2} \\ \omega & \omega^{2} & 1 \\ \omega^{2} & 1 & \omega\end{array}\right|$ We have,...
Solution: Option(C) To find: Value of $\left|\begin{array}{cc}a+i b & c+i d \\ -c+i d & a-i b\end{array}\right|$ Formula used: $\mathrm{i}^{2}=-1$ We have, $\left|\begin{array}{cc}a+i b...
Solution: Option(B) To find: Value of $\left|\begin{array}{cc}\sin 23^{\circ} & -\sin 7^{\circ} \\ \cos 23^{\circ} & \cos 7^{\circ}\end{array}\right|$ Formula used: (i) $\sin (A+B)=\sin A...
Solution: Option(C) To find: Value of $\left|\begin{array}{ll}\cos 15^{\circ} & \sin 15^{\circ} \\ \sin 15^{\circ} & \cos 15^{\circ}\end{array}\right|$ Formula used: (i) $\cos (A+B)=\cos A...
Solution: Option(B) To find: Value of $\left|\begin{array}{cc}\cos 70^{\circ} & \sin 20^{\circ} \\ \sin 70^{\circ} & \cos 20^{\circ}\end{array}\right|$ Formula used: (i) $\cos \theta=\sin...
and 
are collinear, prove that 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
be three points such that area of a is 4 sq units, find the value of .Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the area of a ABC having vertices 
and 
is 35 sq units.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the points 
and 
are collinear.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the points 
and 
are collinear.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the points and are collinear.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
![\[{{e}^{x}}~dy\text{ }+\text{ }\left( y\text{ }{{e}^{x}}~+\text{ }2x \right)\text{ }dx\text{ }=\text{ }0\text{ }isA.\text{ }x\text{ }ey\text{ }+\text{ }{{x}^{2}}~=\text{ }C\text{ }B.\text{ }x\text{ }ey\text{ }+\text{ }{{y}^{2}}~=\text{ }C\text{ }C.\text{ }y\text{ }ex\text{ }+\text{ }{{x}^{2}}~=\text{ }C\text{ }D.\text{ }y\text{ }ey\text{ }+\text{ }{{x}^{2}}~=\text{ }C\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a5083fc3f37724ded34eb973899cb501_l3.png "Rendered by QuickLaTeX.com")
Therefore, the correct option is option(c).
SOLUTION: Therefore, yje correct option is option(c).
![\[\mathbf{A}.\text{ }\mathbf{xy}\text{ }=\text{ }\mathbf{C}\text{ }\mathbf{B}.\text{ }\mathbf{x}\text{ }=\text{ }\mathbf{C}{{\mathbf{y}}^{\mathbf{2}}}~\mathbf{C}.\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{Cx}\text{ }\mathbf{D}.\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{C}{{\mathbf{x}}^{\mathbf{2}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1d706555279c605d673ea5651ada04da_l3.png "Rendered by QuickLaTeX.com")
Given question is Therefore, the correct option is OPTION(C)
![\[\left( \mathbf{1}\text{ }+\text{ }{{\mathbf{e}}^{\mathbf{2x}}} \right)\text{ }\mathbf{dy}\text{ }+\text{ }\left( \mathbf{1}\text{ }+\text{ }{{\mathbf{y}}^{\mathbf{2}}} \right)\text{ }{{\mathbf{e}}^{\mathbf{x}}}~\mathbf{dx}\text{ }=\text{ }\mathbf{0}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a741e3dfd2b281497b24b3de70542d10_l3.png "Rendered by QuickLaTeX.com")
On integrating, we get, ⇒ sin-1x + sin-1y = C
(x -a)2 + (y –a)2 = a2 …………1 differentiating eq 1 with respect to x, we get, \[\begin{array}{*{35}{l}} 2\left( x-a \right)\text{ }+\text{ }2\left( y-a \right)\text{ }dy/dx~=\text{ }0 \\ \Rightarrow...
![\[{{\mathbf{x}}^{\mathbf{2}}}~-\text{ }{{\mathbf{y}}^{\mathbf{2}}}~=\text{ }\mathbf{c}\text{ }{{\left( {{\mathbf{x}}^{\mathbf{2}}}~+\text{ }{{\mathbf{y}}^{\mathbf{2}}} \right)}^{\mathbf{2}}}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3aba527b09857b2a939cd844ff559f93_l3.png "Rendered by QuickLaTeX.com")
![\[({{\mathbf{x}}^{\mathbf{3}}}-\mathbf{3x}{{\mathbf{y}}^{\mathbf{2}}})\text{ }\mathbf{dx}\text{ }=\text{ }\left( {{\mathbf{y}}^{\mathbf{3}}}-\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{y} \right)\text{ }\mathbf{dy},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f17e91ba3477eee6c20ed07cc8909f13_l3.png "Rendered by QuickLaTeX.com")
(iv) 
(ii) 
(ii) 
Therefore, its degree is three.
the functions in exercise 
Solution: Now let's consider $y=\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}}$ Derivating the above function: $\frac{d y}{d x}=\frac{\sqrt{2 x+7} \frac{d}{d x} \cos ^{-1} \frac{x}{2}-\cos ^{-1}...
Then:Fig 1= (A) Det (A) = 0 (B) Det (A) (C) Det (A) (D) Det (A) SOLUTION: Given: Matrix A = ……….(i) Since [ cannot be negative] Therefore, The correct option is option...
(A) (B) (C) (D) SOLUTION: Given: Matrix A = exists and unique solution is ……….(i) Now and and adj. A = = And = = = Therefore, option (A) is...
fig 1: (A) 0 (B) 1 (C) (D) solution: According to question, ……….(i) Let = = [From eq. (i)] = 0 [ R2 and R3 have become identical] Therefore, option (A) is...
SOLUTION: Putting and in the given equations, the matrix form of given equations is [AX= B] Here, A = X = and B = = = exists and unique solution is ……….(i) Now and and adj....
=0 SOLUTION: L.H.S. = = = = = = [ C2 and C3 have become identical] = 0 = R.H.S.
SOLUTION: L.H.S. = = = = = = 1 = R.H.S.
SOLUTION: L.H.S. = = = = = = = = = = R.H.S.
SOLUTION: L.H.S. = = = (say) ……….(i) Now = = = From eq. (i), L.H.S. = ……….(ii) Now = Expanding along third column, = = = = = From eq. (i), L.H.S. = = =...
SOLUTION: L.H.S. = = = = Expanding along third column, = = = = = = = R.H.S.
SOLUTION: Let = = = =
SOLUTION: Let = = = = = = = = =
Given: Matrix A = = Therefore, exists. and and adj. A = = B (say) = ………(i) = = Therefore, exists. and and adj. B = = = = ….(ii) Now to find (say), where C...
fig 1: , B= solution: Given: and B = Since, [Reversal law] ……….(i) Now = = Therefore, exists. and and adj. B = = From eq. (i), ...
fig 1: solution: L.H.S. = = = = = = = = = R.H.S. Proved.
fig1: Either ……….(i) Or But this is contrary as given that . Therefore, from eq. (i), is only the solution.
fig 1= Given: Here, Either ……….(i) Or [Expanding along first row] and and and and ……….(ii) Therefore, from eq. (i) and (ii),...
fig 1: Expanding along first row, = = = = = 1
LHS: $\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|$ Multiplying R 1 by a $\mathrm{R} 2$ by $\mathrm{b}$ and $\mathrm{R} 3$...
is independent of 
.Let $\Delta=\left[\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right]$ $\Delta=x\left|\begin{array}{cc}-x &...
Let $x, y$ and $z$ be the per $k g$. prices of onion, wheat and rice respectively. According to given statement, we have following equations, $4 x+3 y+2 z=60$ $2 x+4 y+6 z=90$ $6 x+2 y+3 z=70$ The...
![A=\left[\begin{array}{llr}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e30d4eac9a4d4350e1fb200d7e82f2e4_l3.png "Rendered by QuickLaTeX.com")
find A^-1. using A^-1 solve 
$A=\left[\begin{array}{llr}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right]$ $|A|=\left|\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 &...
Given set of equations is : $x-y+2 z=7$ $3 x+4 y-5 z=-5$ $2 x-y+3 z=12$ This set of equation in the form of matrix is $A X=B$ $\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 4 & -5 \\ 2...
Given set of equations is : $2 x+3 y+3 z=5$ $x-2 y+z=-4$ $3 x-y-2 z=3$ This set of equation in the form of matrix is $A X=B$ $\left[\begin{array}{ccc}2 & 3 & 3 \\ 1 & -2 & 1 \\ 3...
and 
and 
Given set of equations is $: x-y+z=4$ and $2 x+y-3 z=0$ and $x+y+z=2$ This set of equation can be written in the form of matrix as $A X=B$ $\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1...
and 
and 
Given set of equations is $: 2 x+y+z=1$ and $x-2 y-z=3 / 2$ and $3 y-5 z=9$ This set of equation in the form of matrix is $A X=B$ $\left[\begin{array}{ccc}2 & 1 & 1 \\ 1 & -2 & -1 \\...
and 
Given set of equations is : $5 x+2 y=3$ and $3 x+2 y=5$ This set of equation in the form of matrix is $A X=B$ $\left[\begin{array}{ll}5 & 2 \\ 3 & 2\end{array}\right]\left[\begin{array}{l}x...
and 
Given set of equations is : $4 x-3 y=3$ and $3 x-5 y=7$ This set of equation in the form of matrix is $A X=B$ $\left[\begin{array}{ll}4 & -3 \\ 3 &...
and 
Let $A=\left[ {{a}_{ij}} \right]$ be a square matrix of order n. The adjoint of a matrix A is the transporse of the cofactor matrix of A. It is denoted by adj A. Given set of equations is : $2...
and 
Let $A=\left[ {{a}_{ij}} \right]$ be a square matrix of order n. The adjoint of a matrix A is the transporse of the cofactor matrix of A. It is denoted by adj A. Given set of equations is : $5 x+2...
& 
& 
Given set of equations is: $5 x-y+4 z=5 ; 2 x+3 y+5 z=2 ; 5 x-2 y+6 z=-1$ This set of equation in the form of matrix is $A X=B$ $=140-13-76$ $=140-89$ $=51$ $\neq 0$ System of equations is...
and 
Given set of equations is : $3 x-y-2 z=2 ; 2 y-z=-1$ and $3 x-5 y=3$ This set of equation in the form of matrix is $A X=B$ $|A|=\left|\begin{array}{ccc}3 & -1 & -2 \\ 0 & 2 & -1 \\ 3...
and 
Given set of equations is $: x+y+z=1 ; 2 x+3 y+2 z=2$ and ax $+a y+2 a z=4$ This set of equationin the form of matrix is $A X=B$ $\left[\begin{array}{lll}1 & 1 & 1 \\ 2 & 3 & 2 \\ a...
and 
Given set of equations is : $x+3 y$ and $2 x+6 y=8$ This set of equation in the form of matrix is $A X=B$. $\left[\begin{array}{ll}1 & 3 \\ 2 & 6\end{array}\right]\left[\begin{array}{l}x \\...
and 
Given set of equations is $: 2 x-y=5$ and $x+y=4$ This set of equation in the form of matrix is $A X=B$, where $\mathrm{A}=\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right]$ and...
and 
Given set of equations is : $x+2 y=2$ : and $2 x+3 y=3$ The set of equation in the form of matrix is $A X=B$, where $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]$ and...
is an invertible matrix of order 2 , then 
is equal to: (A) det A (B) 
det A (C) 1 (D) 0The correct answer is Option (B). Explanation: $A A^{-1}=\mid$ $\operatorname{det}\left(A A^{-1}=I\right)$ $\operatorname{det}(A) \operatorname{det}\left(A^{-1}\right)=1$...
be a non-singular matrix of order 
. Then |adj. 
is equal to: (A) 
(B) 
(C) 
(D) 
The correct answer is Option (B). Explanation: $\mid$ adj. $\left.A|=| A\right|^{n-1}=|A|^{2} \quad($ for $n=3)$
and hence find A inverse.fig(1) A = = Now = = L.H.S. = = = = = = = R.H.S. Now, to find , multiplying by =
=0 Hence find A inverse.fig(1) A = = $A^{3}=A^{2} A=\left[\begin{array}{ccc}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right]\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2...
![A=\left[\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-866d9f588aa87affb3349473fc083592_l3.png "Rendered by QuickLaTeX.com")
, find the numbers 
and 
such that 
.$A^{2}=A A=\left[\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right]=\left[\begin{array}{ll}11 & 8 \\ 4 &...
![A=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5709cd602ca367d80c27d0813b877fa5_l3.png "Rendered by QuickLaTeX.com")
, show that 
. Hence find 
.$A^{2}=A A$ $A^{2}=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]=\left[\begin{array}{rr}8 & 5 \\ -5 &...
![A=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1c124b99bf1a57d3fb3721bea26a7e1e_l3.png "Rendered by QuickLaTeX.com")
and ![B=\left[\begin{array}{cc}6 & 8 \\ 7 & 9\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e3a0eb5107384b32c24edef968000318_l3.png "Rendered by QuickLaTeX.com")
verify that 
$A=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]$ $|\mathrm{A}|=\left|\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right|=1 \neq 0$ $\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|}$...
![A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-04334529244081d6181667faa6b1c6cf_l3.png "Rendered by QuickLaTeX.com")
Let A = = exists. A11 = , A12 = , A13 = , A21 = , A22 = , A23 = , A31 = , A32 = , A33 = adj. A =
![\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e042c29b7aac5879ad60e2d53ba39a82_l3.png "Rendered by QuickLaTeX.com")
Let A = = exists. A11 = , A12 = , A13 = , A21 = , A22 = , A23 = , A31 = , A32 = , A33 = adj. A =
![\left[\begin{array}{ccc}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ef9ad04080fa72421925cb52af2f44d9_l3.png "Rendered by QuickLaTeX.com")
Let $\mathrm{A}=\left[\begin{array}{ccc}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{array}\right]$ Therefore, $A^{-1}$ exists A11 = , A12 = , A13 = , A21 = , A22 = , A23 = ,...
![\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9a6d041e3d25f78d49c2d03a26066b9f_l3.png "Rendered by QuickLaTeX.com")
Let $A=\left[\begin{array}{lll}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$ Therefore, $A^{-1}$ exists Find adj A: 212. \text { 212. }...
![\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7c2504929a778bb0735e27cd0c3c76e5_l3.png "Rendered by QuickLaTeX.com")
$\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]$ Solution: Let $\mathrm{A}=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0...
![\left[\begin{array}{ll}-1 & 5 \\ -3 & 2\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e37f391dc108d81264eceb67aeb2556f_l3.png "Rendered by QuickLaTeX.com")
Let $A=\left[ {{a}_{ij}} \right]$ be a square matrix of order n. The adjoint of a matrix A is the transporse of the cofactor matrix of A. It is denoted by adj A. Given...
![\left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e53bb962fa8ed0816e1952010b40c21f_l3.png "Rendered by QuickLaTeX.com")
Let $A=\left[\begin{array}{rr}2 & -2 \\ 4 & 3\end{array}\right]$ $|A|=\left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right]=14 \neq 0$ Since determinant of the matrix is not zero,...
in ![\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a1a867b9b309d524c23c55036905efd7_l3.png "Rendered by QuickLaTeX.com")
. Let A = = adj. A = A. (adj. A) = = = ……….(i) Again (adj. A). A = = = ……….(ii) And = Also = ……….(iii) From eq. (i), (ii) and (iii) A. (adj. A) = (adj. A). A...
in ![\left[\begin{array}{cc}2 & 3 \\ -4 & -6\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3d3a43d3aa5a6911398ef00d207d8087_l3.png "Rendered by QuickLaTeX.com")
Let A = adj. A = A.(adj. A) = = = …..(i) Again (adj. A). A = = = …..(ii) And = Again …..(iii) From eq. (i), (ii) and (iii) A. (adj. A) = (adj. A). A...
![\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1c97d8dc1896c9f6190641181d8c86d5_l3.png "Rendered by QuickLaTeX.com")
Let $\mathrm{A}=\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{array}\right]$ Cofactors of the above matrix are $\mathrm{A}_{11}=+\left|\begin{array}{ll}3...
![\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d3ff67a5276113f580775455111647e6_l3.png "Rendered by QuickLaTeX.com")
Let $A=$ $\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$ Cofactors of the above matrix are 11 $A_{11}=4$ $A_{12}=-3$ $\mathrm{A}_{21}=-2$ $A_{22}=1$ adj....
. 
Elements of third column of are A13 = Cofactor of = A23 = Cofactor of = A33 = Cofactor of = = = = = = = = = =
Find Cofactors of elements of second row: $A_{21}=$ Cofactor of element $a_{21}=(-1)^{2+1}\left|\begin{array}{cc}3 & 8 \\ 2 & 3\end{array}\right|-(-1)^{3}(9-16)=7$ $A_{22}=C$ ofactor of...
(ii) 
(i) $\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$ Find Minors and cofactors of elements: Say, $M_{i j}$ is minor of element $a_{i j}$ and...
(ii) 
Find Minors of elements: Say, $M_{i j}$ is minor of element $a_{i i}$ $\mathrm{M}_{11}=$ Minor of element $\mathrm{a}_{11}=3$ $\mathrm{M}_{12}=$ Minor of element $\mathrm{a}_{12}=0$...
and 
Then is (A) 12 (B) (C) 
(D) 
The correct option is OPTION(D) Given: Area of triangle = Modulus of = 35 Modulus of = 35 Taking positive sign, Taking negative sign,
and 
using determinants. (ii) Find equation of line joining 
and 
using determinants.Let $A(x, y)$ be any vertex of a triangle. All points are on one line if area of triangle is zero. $\frac{1}{2}[x(2-6)-y(1-3)+1(6-6)]=0$ $-4 x+2 y=0$ $y=2 x$ Which is equation of line. (ii) Let...
if area of triangle is 4 sq. units and vertices are (i) 
(ii) 
(i) $\frac{1}{2}[k(0-2)-0+1(8-0)]=4$ $1 / 2(-2 k+4)=4$ $-k+4=4$ Now: $-k+4=\pm 4$ $-k+4=4$ and $-k+4=-4$ $\mathrm{k}=0$ and $\mathrm{k}=8$ $\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} &...
are collinear.Points are collinear if area of triangle is equal to zero. i.e. Area of triangle $=0$ Area of Triangle $=\frac{1}{2}\left|\begin{array}{ccc}a & b+c & 1 \\ b & c+a & 1 \\ c & a+b...
Area $=\frac{1}{2}\left|\begin{array}{ccc}-2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1\end{array}\right|$ $=\frac{1}{2}[-2(10)+3(4)-22]$ $=15$ sq. Units
(ii) 
Formula for Area of triangle: $\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|$ (i)...
The correct option is Option (C) Since, Determinant is a number associated to a square matrix.
be a square matrix of order , then 
is equal to (A) 
(C) 
(D) 
The correct option is Option (C). Let A = be a square matrix of order 3 x 3. ……….(i) = [From eq. (i)]
LHS $\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|$ Multiplying by respectively and then dividing the...
LHS $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-b \mathrm{C}_{3}$ and $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}+a \mathrm{C}_{3}$ $=\left|\begin{array}{ccc}1+a^{2}+b^{2} & 0 & -2 b \\ 0...
![\[\left|\begin{array}{ccc} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right|=\left(1-x^{3}\right)^{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-31bd0220c4c348d80de219a56446e48b_l3.png "Rendered by QuickLaTeX.com")
L.H.S. = = = = = = = = = = = = = R.H.S. Proved.
(ii) 
LHS $\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$ $\mathrm{R}_{1} \rightarrow...
![\[\text { (i) }\left|\begin{array}{ccc} x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4 \end{array}\right|=(5 x+4)(4-x)^{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-761ecebc01ac7ac73d668008d102850f_l3.png "Rendered by QuickLaTeX.com")
![\[\text { (ii) }\left|\begin{array}{ccc} y+k & y & y \\ y & y+k & y \\ y & y & y+k \end{array}\right|=k^{2}(3 y+k)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-96eaf73e0a5c4da52bd9f5777af34135_l3.png "Rendered by QuickLaTeX.com")
(i) LHS = [operating and ] = = = R.H.S. (ii) L.H.S. = = = = [operating and ] = = = = R.H.S. Proved.
LHS= $\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|$ Mulitiplying $R_{1}, R_{2}, R_{3}$ by $x, y, z$ respectively...
(ii) 
(i) LHS: $\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|$ $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}$ and...
Solving L.H.S $\left| \begin{matrix} -{{a}^{2}} & ab & ac \\ ba & -{{b}^{2}} & bc \\ ca & cb & -{{c}^{2}} \\ \end{matrix} \right|$ Taking a common from${{R}_{1}}$,b common...
Let $\Delta=\left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|$ Taking (-1) common from all the 3 rows. Again, interchanging rows and columns,...
LHS: Applying: $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ $\left|\begin{array}{ccc}b+c+c+a+a+b & q+r+r+p+p+q & y+z+z+x+x+y \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right|$...
$\left|\begin{array}{ccc}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|$ Applying: $\mathrm{C}_{3}->\mathrm{C}_{3}+\mathrm{C}_{2}$...
$\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|$ Applying: $\mathrm{C}_{3}->\mathrm{C}_{3}-\mathrm{C}_{1}$ $\left|\begin{array}{lll}2...
$\left|\begin{array}{ccc}a-b & b-c & a-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$ Applying: $\mathrm{C}_{1} \rightarrow...
$\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=0$ L.H.S. $\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z &...
then 
is equal to (A) 6 (B) 
(C) (D) 0The correct option is Option (B). Reason : $x^2{\prime} =36$ $x=\pm 6$
, if (i) 
(ii) 
(i) $\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$ $2-20=2 \times^{-2}-24$ $2 \times^{\prime} 2=6$ $x^{-2}=3$ or...
find |A|$A=\left|\begin{array}{ccc}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right|$ $|A|=\left|\begin{array}{ccc}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 &...
(iv) 
(iii) = 0 + 6 – 6 = 0 (iv) =5
(ii) 
(i) $\left|\begin{array}{rrr}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|=3\left|\begin{array}{rr}0 & -1 \\ -5 &...
then show that | 3 A | = 27 | A |$3 A=\left[\begin{array}{ccc}3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12\end{array}\right]$ LHS: $|3 A|=\left|\begin{array}{lll}3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0...
![A=\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9e263b57f5b17d5ce7c731eb0cc6b8e0_l3.png "Rendered by QuickLaTeX.com")
then show that 
.$A=\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right]$ $2 \mathrm{~A}=\left[\begin{array}{ll}2 & 4 \\ 8 & 4\end{array}\right]$ L.H.S. $=|2 A|=\left|\begin{array}{ll}2 & 4 \\...
& 
(i) $\left| \begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{matrix} \right|$ = $\cos \theta X\cos \theta -(-\sin \theta )Xsin\theta =1$ (ii) $\left|...
$\left|\begin{array}{cc}2 & 4 \\ -5 & -1\end{array}\right|=2(-1)-4(-5)=18$