Writing given equations in matrix form: The matrix form of given equations are 1-223x1y = 41 or,  AX = B .......................i) We have, A = 1-223= 3 + 4= 7  We know that, A-1 = 1A Adj(A) = 1732-21 = 1732-21 We know that, .....

Here, A = 1-13-3 and B = 2-52-5 Now,        AB = (1-13-32-52-5              = 2-2-    5+56-6   -   15+15 ∴   AB = 0000. which show that AB is a null .....

Here, M = 536K             So, |(M)| = 536K  or,  27 = 5K – 18 or,  45=5K ∴    K = 9. Thus, the value of k is .....

Here, A = 20-3x, B = 40-91 and A2 = B, So, 20-3x20-3x = 40-91 or, 4+00+0-6-3x0+x2 = 40-91 Equating the corresponding elements     x2 = 1 ∴ x  =  ± 1 Also, – 3x = -9 or, -3x  =  -9 + 6 or, -3x .....

Here, m273×3-2-7m= 1001                      [ AB = I.] or, 3m-14-2m+2m 21-21-14+3m = 1001 or, 3m-140 0-14+3m  = 1001 Equating the corresponding components; 3m -14 = 1 or, 3m = .....

Here, K = 2-35m So, K= 2-35m|2m + 15 or, 23 = 2m + 15 or, 8 = 2m ∴ m = 4 Thus, the value of m is .....

Here, P = 2345 and K = 100-1 We know that, PK = 2345100-1       = 2+00-34+00-5       = 2-34-5 ∴ PK = .....

Here, 5x + 3y = 7 ……….. (i) 7y - 10x = 12 ……. (ii) Writing equations (i) and (ii) in the matrix form, 53-107 1xy = 712 or, AX = B...............................i) We have, A = 53-107         .....

Here, M = -5-8p12 and |(M)| = 20. We have, |(M)| = -5-8p12 or, 20 = -60 + 8p or, 80 = 8p ∴   p  = 10. Thus, the value of p is .....

Here, Given equations are:         5x – 9y = 47;          2y + 3x = 6. Writing these equations in matrix form then, 5-932xy = 476 or, AX = B .....

Here, X = 2-3ab         and X2 = I. So, 2-3ab 2-3ab = I or, 2 ×2-3 ×a2 ×-3+ -3b2 ×a+b ×aa ×-3+ b ×b = I  or, 4-3a-6a-3b2a+ab-3a+b2 = 1001 Equating the corresponding .....

Here, D = 1231 We have D2 = 12311231       = 1×1+2 ×3 1 ×2+2 ×13 ×1+1 ×33×2+1 ×1        = 7467 ∴ D2 = 7467 Thus, D2 is 7467. .....

Here,   x = 23y      or, 3x – 2y = 0, and 4x - 3y = 1. Writing above equation in the matrix form Then, 3-24-3xy = 01 We have, A = 3-24-3= -9 + 8= -1 Adj (A) = -32-43 We know that, A-1 = Adj(A) = 1-1-32-43 = .....

Here, given equations are: 10x + 12y = 6………. (i) 25x - 2y = 2 ………... (ii) The matrix form of given equations are 101225-21x1y = 62 or, AX = B................................i) We have,A = 101225-2= -20 - .....

Here, A = 2113, B = 231m and |(BA)| = -5 So, BA = 231m2113       = 4+32+92+m1+3m       = 7112+m1+3m Now, |(BA)| = 7112+m1+3m or, - 5 = 7 + 21m – 22 – 11m or, - 5 = 10 m – 15 or, 10 = 10 .....

Here, A = 31-12, AT = 3-112 Now, AAT = 31-123-112          = 9+1-3+2-3+21+4          = 10-1-15 Thus, the AAT is .....

Here, A-1 = 3254 We have, |(A-1)| = 3254                            = 12 – 10                          .....

Inverse matrix:      If A and B are two square matrices of same order, I is an identity matrix of same order and AB = BA = I, then A and B are said to be inverse to each other. The inverse of A is denoted by A-1. So, A-1 = B. Here, .....

Here, AB = 3423 3-4-23                  =  3 ×3+4 ×-23×-4+4×3 2 ×3+3 ×-22 ×-4+ 3 ×3,                 = .....

Here, A = 12-3-205and B = 23-405-1 Now, AB = 12-3-20523-405-1        = 2- 8-153+0+3 -4+0+25-6+0-5        = -21621-11 ∴ AB = .....

Here, Given matrix = x342 So, x342 = 0                      [ If inverse of A can not defined, determinant of A = 0.] or,  2x – 12 = 0 or,  2x = 12 ∴    x = .....

Here, 3x-642 = 0 or, 6x + 24 = 0 or, 6x = -24 ∴ x = -4. Thus, the value of x is .....

Here,  Functional value at x = 3 is f(3) =  x +2 = 3 + 2 = 5. For left land limit of f(x) at x = 3,     lim⁡_x→3⁡f(x) =    lim⁡_x→3⁡( x2 - 9 )/(x - 3) = x + 3 = 3 + 3 = .....

Here,  Functional value at x = 1 is f(1) = 2. For left land limit of f(x) at x = 1,     lim⁡_x→1⁡f(x) =    lim⁡_x→1⁡( x2 - 1 )/(x - 1) = x + 1 = 1 + 1 = 2 . For right hand .....

Here,  Functional value at x = 2 is f(2) =5.2 + 2 = 12. For left land limit of f(x) at x = 2,     lim⁡_x→2⁡f(x) =    lim⁡_x→2⁡( 5x + 2 ) = 5.2 + 2 = 12 . For right .....

Here,  Functional value at x = 3 is f(3) =32 + 1 = 10. For left land limit of f(x) at x = 3,     lim⁡_x→3⁡f(x) =    lim⁡_x→3⁡( x2 + 1 ) = 32 + 1 = 10 . For right hand .....

Here,  Functional value at x = 2 is f(2) =2 -  (2)2 = - 2. For left land limit of f(x) at x = 2,     lim⁡_x→2⁡f(x) =    lim⁡_x→2⁡(2 - x2 ) = 2 -  22 = - 2. For .....

Here,  Functional value at x = 5 is f(5) = (5)2 + 2 = 27. For left land limit of f(x) at x = 5,     lim⁡_x→5⁡f(x) =    lim⁡_x→5⁡(x2 + 2) = 52 + 2 = 27 For right hand limit of .....

Here,  Functional value at x = 2 is f(2) = 3. For Left Hand Limit of f(x) at x = 2,     lim⁡_x→2⁡f(x) =    lim⁡_x→2⁡(2x - 1) = 2 ⨯ 2 - 1 = 3. For Right Hand Limit of f(x) at x = .....

Real numbers are all the numbers on the continuous number line with no .....

Function to be continuous at x = a is         Left hand limit = Right hand limit = Functional value. i.e; LHL = RHL = .....

The required condition is   limx→a- ⁡f(x)   = limx→a +⁡  = .....

Set of rational numbers are discontinuous in a number .....

Required notation is [-4, .....

In the interval (a, b], the interval does not contain point a but it contain the another end point b. This interval called left-open .....

The interval notation for -4 < x < 5 is (-4, .....

If the graph of any function has jump or break in a point then the function is called .....

An interval containing the end point a but not containing another end point b is called right open interval. This is denoted as [a, .....

An interval containing both the end points a & b is called closed interval. It is denoted by [a, .....

The required notation is [-4, .....

An interval not containing both the end points a & b is known as an open interval. It is written as (a, .....

If the graph of a function can be drawn without lifting the pencil from the paper this is called continuity. Example: Moving .....

Here,  y = x2, y = 2 - x Value table of y = x2 

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Here,  Given equation is x2 + 7x + 12 = 0  or,  x2 = -7x – 12 Let y = x2 = -7x – 12 Then,      y = x2 …….. (i)      y = -7x – 12 ……… (ii). From .....

Here,  x2 – 3x + 2 = 0  Let, x2 = 3x - 2  Let y = x2 = 3x – 2 then, y = x2 and y = 3x – 2  Taking the equation y = x2 then,

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Here,  x2 + x - 2 = 0  or,  x2 = 2 – x Let y = x2 = 2 - x Taking y = x2 , then

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Here,  x2 – 3x + 2 = 0  Let, x2 = 3x - 2 = y So that,  y = x2 ……… (i)  and  y = 3x – 2 ……….. (ii) From (i),  y = x2 

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Here,  The corresponding equation of x + y ≤ 6 is x + y = 6 

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Here,  x2 – 5x + 6 = 0  Let, x2 = 5x - 6 = y So that,  y = x2 ……… (i)  and  y = 5x – 6 ……….. (ii) From (i),  y = x2 

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Here, x2 – 2x – 3 = 0 or, x2 = 2x + 3 = y Let, y = x2 = 2x + 3 So, y = x2 ……… (i)

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