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} &...

### Mark the tick against the correct answer in the following: If the points and are collinear then the value of is

### Mark the tick against the correct answer in the following: The vertices of a a are and The area of a is

A. units

B. 35 sq units

C. 32 sq units

D. units

Solution: 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 \\...

### Mark the tick against the correct answer in the following: The solution set of the equation is

A.

B.

C.

D. None of these

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}...

### Mark the tick against the correct answer in the following: The solution set of the equation is

A.

B.

C.

D. None of these

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...

### Mark the tick against the correct answer in the following: The solution set of the equation is

A.

B.

C.

D.

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...

### Mark the tick against the correct answer in the following: The solution set of the equation is

A.

B.

C.

D. none of these

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...

### Mark the tick against the correct answer in the following: If then

A. 0

B. 6

C.

D. 9

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...

### Mark the tick against the correct answer in the following:

A.

B. 2

C.

D.

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 &...

### Mark the tick against the correct answer in the following:

A.

B.

C.

D.

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 &...

### Mark the tick against the correct answer in the following:

A.

B.

C.

D. none of these

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...

### Mark the tick against the correct answer in the following:

A.

B.

C.

D. None of these

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} &...

### Mark the tick against the correct answer in the following:

A.

B.

C.

D. none of these

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...

### Mark the tick against the correct answer in the following:

A.

B.

C.

D.

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,...

### Mark the tick against the correct answer in the following:

A.

B.

C. 0

D. none of these

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...

### Mark the tick against the correct answer in the following:

A.

B.

C.

D. none of these

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,...

### Mark the tick against the correct answer in the following:

A. 0

B.

C.

D. none of these

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,...

### Mark the tick against the correct answer in the following: If be distinct positive real numbers then the value of is

A. positive

B. negative

C. a perfect square

D. 0

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 \\...

### Mark the tick against the correct answer in the following:

A. 0

B. 1

C.

D. none of these

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})...

### Mark the tick against the correct answer in the following:

A.

B.

C.

D.

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...

### Mark the tick against the correct answer in the following:

A. 0

B. 1

C.

D. none of these

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,...

### Mark the tick against the correct answer in the following:

A.

B.

C.

D. 0

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...

### Mark the tick against the correct answer in the following:

A. 2

B. 6

C. 24

D. 120

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,...

### Mark the tick against the correct answer in the following:

A. 8

B.

C. 16

D. 142

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,...

### Mark the tick against the correct answer in the following: If is a complex cube root of unity then the value of is

A. 2

B. 4

C. 0

D.

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)...

### Mark the tick against the correct answer in the following: If is a complex root of unity then

A. 1

B.

C. 0

D. none of these

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,...

### Mark the tick against the correct answer in the following:

A.

B.

C.

D. none of these

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...

### Mark the tick against the correct answer in the following:

A.

B.

C.

D.

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...

### Mark the tick against the correct answer in the following:

A. 1

B.

C.

D. none of these

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...

### Mark the tick against the correct answer in the following:

A. 1

B. 0

C.

8D.

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...

### If the points 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}...

### If 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}...

### Find the value of 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}...

### Find the value of for which the points and are collinear.

### Find the value of for which the points and are collinear.

### Find the value of for which the points and are collinear.

### Use determinants to show that the following points are collinear. and

### Use determinants to show that the following points are collinear. and

### Use determinants to show that the following points are collinear. and

### Find the area of the triangle whose vertices are: and

### Find the area of the triangle whose vertices are: and

### Find the area of the triangle whose vertices are: and

### Find the area of the triangle whose vertices are: and

### Find the area of the triangle whose vertices are: and

### The general solution of the differential equation

Therefore, the correct option is option(c).

### The general solution of a differential equation of the type is

SOLUTION: Therefore, yje correct option is option(c).

### The general solution of the differential equation is

Given question is Therefore, the correct option is OPTION(C)

### The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

### Find a particular solution of the differential equation, given that y = 0 when x = 0.

### Find a particular solution of the differential equation (x ≠ 0), given that y = 0 when x = π/2

### Solve the differential equation

### Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0. (Hint: put x – y = t)

### Solve the differential equation

### Find the particular solution of the differential equation

, given that y = 1 when x = 0.

### Find the equation of the curve passing through the point (0, π/4) whose differential equation is sin x cos y dx + cos x sin y dy = 0.

### Show that the general solution of the differential equation is given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter.

### Find the general solution of the differential equation

On integrating, we get, ⇒ sin-1x + sin-1y = C

### Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

(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...

### Prove that

is the general solution of differential equation

where c is a parameter.

### Form the differential equation representing the family of curves given by (x – a)^2 + 2y^2 = a^2, where a is an arbitrary constant.

### verify that the given function (implicit or explicit) is a solution of the corresponding differential equation (iii) (iv)

### verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.(i) (ii)

### indicate its order and degree (if defined).

### indicate its order and degree (if defined). (i) (ii)

Therefore, its degree is three.

### Differentiate with respect to 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}...

### Choose the correct answer Let A =(Fig 1) where 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...

### Choose the correct answer If x,y,z are non-zero real numbers, then the inverse of matrix A =(fig1) is

(A) (B) (C) (D) SOLUTION: Given: Matrix A = exists and unique solution is ……….(i) Now and and adj. A = = And = = = Therefore, option (A) is...

### Choose the correct answer in If a,b,c are in A.P., then the determinant (fig 1) 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...

### Solve the system of the following equations: (Using matrices):

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....

### Using properties of determinants in , prove that:

=0 SOLUTION: L.H.S. = = = = = = [ C2 and C3 have become identical] = 0 = R.H.S.

### Using properties of determinants in , prove that:

SOLUTION: L.H.S. = = = = = = 1 = R.H.S.

### Using properties of determinants in , prove that:

SOLUTION: L.H.S. = = = = = = = = = = R.H.S.

### Using properties of determinants in , prove that:

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. = = =...

### Using properties of determinants in , prove that:

SOLUTION: L.H.S. = = = = Expanding along third column, = = = = = = = R.H.S.

### Evaluate:

SOLUTION: Let = = = =

### Evaluate:

SOLUTION: Let = = = = = = = = =

### Let A=(fig 1), prove that (i) (adj A)^-1 = adj(A)^-1 (ii) (A^-1)^-1=A

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...

### If A^-1 and B (in fig 1) is given , find (AB)^-1.

fig 1: , B= solution: Given: and B = Since, [Reversal law] ……….(i) Now = = Therefore, exists. and and adj. B = = From eq. (i), ...

### Prove that : fig 1

fig 1: solution: L.H.S. = = = = = = = = = R.H.S. Proved.

### Solve the equation: fig1

fig1: Either ……….(i) Or But this is contrary as given that . Therefore, from eq. (i), is only the solution.

### If a, b and c are real numbers, and fig 1 Show that either a+b+c=0 or a=b=c

fig 1= Given: Here, Either ……….(i) Or [Expanding along first row] and and and and ……….(ii) Therefore, from eq. (i) and (ii),...

### Evaluate: fig1

fig 1: Expanding along first row, = = = = = 1

### Without expanding the determinant, prove that

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 $\mathrm{R} 3$...

### Prove that the determinant 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 &...

### The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 60. The cost of 2 kg onion, 4 kg wheat and 2 kg rice is Rs. 90. The cost of 6 kg onion, 2 k wheat and 3 kg rice is Rs. 70. Find cost of each item per kg by matrix method.

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...

### If 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 &...

### Solve system of linear equations, using matrix method, in

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...

### Solve system of linear equations, using matrix method, in

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...

### Solve system of linear equations, using matrix method, in 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...

### Solve system of linear equations, using matrix method, in 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 \\...

### Solve system of linear equations, using matrix method, in 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...

### Solve system of linear equations, using matrix method, in 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 &...

### Solve system of linear equations, using matrix method, in 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...

### Solve system of linear equations, using matrix method, in 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...

### Examine the consistency of the system of equations in & &

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...

### Examine the consistency of the system of equations in 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...

### Examine the consistency of the system of equations in 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...

### Examine the consistency of the system of equations in 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 \\...

### Examine the consistency of the system of equations in 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...

### Examine the consistency of the system of equations in 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...

### If is an invertible matrix of order 2 , then is equal to: (A) det A (B) det A (C) 1 (D) 0

The correct answer is Option (B). Explanation: $A A^{-1}=\mid$ $\operatorname{det}\left(A A^{-1}=I\right)$ $\operatorname{det}(A) \operatorname{det}\left(A^{-1}\right)=1$...

### Let 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 $n=3)$

### If a=(fig 1) find and hence find A inverse.

fig(1) A = = Now = = L.H.S. = = = = = = = R.H.S. Now, to find , multiplying by =

### For the matrix A = (FIG 1) show that =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...

### For the matrix , 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 &...

### If , 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 &...

### Let and verify that

$A=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]$ $|\mathrm{A}|=\left|\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right|=1 \neq 0$ $\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|}$...

### Find the inverse of each of the matrices (if it exists) given in

Let A = = exists. A11 = , A12 = , A13 = , A21 = , A22 = , A23 = , A31 = , A32 = , A33 = adj. A =

### Find the inverse of each of the matrices (if it exists) given in

Let A = = exists. A11 = , A12 = , A13 = , A21 = , A22 = , A23 = , A31 = , A32 = , A33 = adj. A =

### Find the inverse of each of the matrices (if it exists) given in

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 = ,...

### Find the inverse of each of the matrices (if it exists) given in

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. }...

### Find the inverse of each of the matrices (if it exists) given in

$\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]$ Solution: Let $\mathrm{A}=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0...

### Find the inverse of each of the matrices (if it exists) given in

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...

### Find the inverse of each of the matrices (if it exists) given in

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,...

### Verify in

. 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...

### Verify in

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...

### Find adjoint of each of the matrices in

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...

### Find adjoint of each of the matrices in

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....

### Using Cofactors of elements of third column, evaluate .

Elements of third column of are A13 = Cofactor of = A23 = Cofactor of = A33 = Cofactor of = = = = = = = = = =

### Using Cofactors of elements of second row, evaluate

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...

### Write Minors and Cofactors of the elements of following determinants: (i)(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...

### Write Minors and Cofactors of the elements of following determinants: (i) (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 $\mathrm{a}_{12}=0$...

### If area of triangle is 35 sq units with vertices 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,

### (i) Find equation of line joining 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...

### Find values of 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} &...

### Show that points: 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...

### Find area of the triangle with vertices at the point given in each of the following:(iii)

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

### Find area of the triangle with vertices at the point given in each of the following: (i) (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)...

### Choose the corrrect answer 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 correct option is Option (C) Since, Determinant is a number associated to a square matrix.

### Choose the corrrect answer Let 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)]

### By using properties of determinants, show that:

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...

### By using properties of determinants, show that:

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...

### By using properties of determinants, show that:

L.H.S. = = = = = = = = = = = = = R.H.S. Proved.

### By using properties of determinants, show that: (i) (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...

### By using properties of determinants, show that:

(i) LHS = [operating and ] = = = R.H.S. (ii) L.H.S. = = = = [operating and ] = = = = R.H.S. Proved.

### By using properties of determinants, show that:

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...

### By using properties of determinants, show that:(i) (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...

### Using the properties of determinants and without expanding, prove that

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...

### Using the properties of determinants and without expanding, prove that

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,...

### Using the properties of determinants and without expanding, prove that

LHS: Applying: $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ $\left|\begin{array}{ccc}b+c+c+a+a+b & q+r+r+p+p+q & y+z+z+x+x+y \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right|$...

### Using the properties of determinants and without expanding, prove that:

$\left|\begin{array}{ccc}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|$ Applying: $\mathrm{C}_{3}->\mathrm{C}_{3}+\mathrm{C}_{2}$...

### Using the properties of determinants and without expanding, prove that:

$\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|$ Applying: $\mathrm{C}_{3}->\mathrm{C}_{3}-\mathrm{C}_{1}$ $\left|\begin{array}{lll}2...

### Using the properties of determinants and without expanding, prove that:

$\left|\begin{array}{ccc}a-b & b-c & a-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$ Applying: $\mathrm{C}_{1} \rightarrow...

### Using the properties of determinants and without expanding, prove that:

$\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=0$ L.H.S. $\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z &...

### If then is equal to (A) 6 (B) (C) (D) 0

The correct option is Option (B). Reason : $x^2{\prime} =36$ $x=\pm 6$

### Find values of , 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...

### If A = 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 &...

### Evaluate the determinants(iii) (iv)

(iii) = 0 + 6 – 6 = 0 (iv) =5

### Evaluate the determinants (i) (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 &...

### If A = 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...

### If 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 \\...

### Evaluate the following determinants &

(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|...

### Evaluate the following determinants

$\left|\begin{array}{cc}2 & 4 \\ -5 & -1\end{array}\right|=2(-1)-4(-5)=18$