Sponsor Area
Sponsor Area
We are given that | A | = 4 ...(1)
Since A is of order 3 × 3
∴ | 3 A | = 33 |A| = 27 |A|
= 27 × 4 [∵ of (1)]
= 108
(D) None of these.
C.
(C) Determinant is a number associated to a square matrix.We know that determinant is a number which is defined in the case of square matrix.
∴ (C) is the correct answer.
D.
(D) Det(A) ∈ [2, 4]
Sponsor Area
Sponsor Area
Show that:
to 
and dividing by x y z , we get
respectively, we get
we get
, we have
we get
Show that:
and hence factorise.
Let 
by multiplying 
Proceed as in (a) (i) by changing a to x, b to y, c to z.
Sponsor Area
Let the three digit numbers A28,3B9 and 62C where A, B, and C are any integers between 0 and 9, be divisible by a fixed integer k. Show that the determinant
is divisible by k.
Prove that:
...(1)
Putting a = 0 in (1), we get,
Again putting a + b + c = 0 in (1), we get,
Find the area of the triangle, whose vertices are (3, 1), (4, 3) and (-5, 4).
Find the area of the triangle with vertices at the points given in each of the following:
(1, 0) (6,0), (4, 3)
Find the area of the triangle with vertices at the points given in each of the following:
(2, 7), (1, 1), (10, 8)
Find the area of the triangle with vertices at the points given in each of the following:
(–2,–3), (3, 2), (–1,–8)
Let ∆ be the area of the triangle whose vertices are (–2, –3), (3, 2), (–1, –8).
Show that points A(a, b + c), B(b, c + a), C(c, a + b) are collinear.
Let ∆ be the area of the triangle formed by the points
A (a, b + c), B (b, c + a), C(c, a + b)
Using determinants, show that the points (11, 7), (5, 5) and (– 1, 3) are collinear.
Determine x so that the points (3, 2), (x, 2) and (8, 8) lie on a line.
Since area of triangle formed by the points (x , 4), (2, – 6). (5, 4) is 35 sq. cm.
Tips: -
Whenever area of a triangle is given, ± sign should be taken.Find values of k if area of triangle is 4 sq. units and vertices are (k, 0), (4, 0), (0, 2).
Find values of k if area of triangle is 4 sq. units and vertices are (– 2, 0), (0, 4), (0, k)
which is equation of line AB.
Now D is (k, 0).
From given condition,
area of ∆ABD = 3
Find equation of line joining (1, 2) and (3, 6) using determinants.
Find equation of line joining (3, 1) and (9, 3) using determinants.
We know that
Again, 
Similarly, 
12
-2
-12, -2
12, -2
D.
12, -2
Since area of triangle formed by the points (2, -6), (5, 4), (k, 4) is 35 sq. units.
Find the minor of element 6 in the determinant 
Find the minors and co-factors of the elements a11, a21 in the determinant.
If A, B, C are square matrices of the same order such that AB = BA = I and AC = CA = I then B = C.
We have
AB = BA = I ...(1)
and AC = CA = I ...(2)
Now B = BI
= B(AC) [∵ of (2)]
= (BA) C [∵ of associative properties of multiplication]
= IC [∵ of (1)]
= C
∴ B = C
Hence the result.
Note : AB = BA = I ⇒ B is inverse of A
Again AC = CA = 1 ⇒ C is inverse of A
Also B = C
∴ inverse of a matrix A, if it exists, is unique.
i.e. – (– 1 – 0), 1 + 4, – (0 – 2) i.e. 1, 5, 2 respectively.
Cofactors of the elements of third row of | A | are
i.e. – 5 – 6, – (5 – 4), 3 + 2
∴ –11,–1, 5 respectively.
Compute the adjoint of the matrix 
and verify that A (Adj A) = | A | 1.
Verify that A (adj A) = (adj A) = | A | I
where 
...(2)Verify that A (adj A) = (adj A) = | A | I
where
From (1) and (2), we get,
A (adj A) = (adj A) A = | A | I.
Verify that A(adj. A) = (adj. A) A = | A | I where
If 
then verify that A (adj. A) = O.
and its multiplication inverse.
Let 
Co-factors of the elements of first row of | A | are – 7, – 5 respectively.
Co-factors of the elements of second row of | A | are 3, 2 respectively.
If 
, show that 
.
Let A be the matrix 
Find A –1 and verify that 
where I is 2 × 2 unit matrix.
Co-factors of the elements of first row of | A | are 1, – 2 respectively
Co-factors of the elements of second row of | A | are – 8, 3 respectively
If A = 
find 
and verify that 
Here 
Co-factors of the elements of first row of | A | are 6, – 1 respectively
Co-factors of the elements of second row of | A | are – 5, 2 respectively
Given 
compute A–1 and show that 2A–1 = 9 I – A.
Here, 
Co-factors of the elements of first row of | A | are 7, 4 respectively
Co-factors of the elements of second row of | A | are 3, 2 respectively
Find the inverse of 
and verify that AA–1 = I.
If 
find x and y find that A2 + xI = y A. Hence find A–1
From the definition of equality of matrices,
If 
Here, 
Co-factors of the elements of first row are 1, tan x respectively.
Co-factors of the elements of second row are – tan x, 1 respectively.
Find the inverse of the matrix 
and show that a A -1 = (a2 + b c + 1) I – a A.
Here 
Co-factors of the elements of the first row of | A | are 
-c respectively.
Co-factors of the elements of the second row of | A | are b , a respectively.
Now, 
If 
verify (AB)–1 = B –1 A–1
Co-factors of elements of first row of |A| are 7 and – 2 respectively
Co-factors of elements of second row of |A| are – 3 and 1 respectively
Co-factors of elements of first row of | B | are 2 and – 6 respectively.
Co-factors of elements of second row of | B | are – 4 and 3 respectively.
Co-factors of elements of first row of | AB | are 22 and – 48 respectively.
Co-factors of elements of second row of | AB | are – 10 and 21 respectively.
From (1) and (2), we get, (AB) –1 = B –1 A–1
If 
verify that (AB)–1 = B–1 A–1
Co-factors of elements of first row of | AB | and –14, –5 respectively.
Co-factors of elements ofrsecond row of | AB | and –5, –1 respectively.
From (1) and (2), we get, (AB)–1 = B–1 A–1
Let 
verify that (AB)–1 = B–1 A–1.
Co-factors of the elements of first row of | AB j and 61, –47 respectively.
Co-factors of the elements of second row of | AB | and –87, 67 respectively.
From (1) and (2), we get,
(AB)–1 = B–1 A–1.
Co-factors of the elements of first row of | B | are 2, – 3 respectively
Co-factors of the elements of second row of | B | are – 6, 4 respectively
Co-factors of the elements of first row of | AB | are 52, – 43 respectively.
Co-factors of the elements of second row of | AB | are – 22, 18 respectively.
From (1) and (2), (AB)–1 B–1 A–1.
.
Co-factors of the elements of first row of | B | are 9, – 8 respectively
Co-factors of the-elements of second row of | B | are – 7, 6 respectively
Co-factors of the elements of first row of | AB | are 94, – 82 respectively.
Co-factors of the elements of second row of | AB | are – 39, 34 respectively.
.
Co-factors of the first row of | AB | are 37, – 29 respectively.
Co-factors of the second row of | AB | are – 14, 11 respectively.
From (1) and (2), we have,
(AB)–1 = B–1 A–1.
Find (AB)–1 if 
Find (AB)–1 if 
Here 
Co-factors of elements of first row of | A | are 3, – 2 respectively.
Co-factors of elements of second row of | A | are 0, 5 respectively.
Let 
Co-factors of the elements of first row of | A | are
i.e. 2, –9, — 6 respectively.
Co-factors of the elements of second row of | A | are
i.e. 0, – 2, – 1 respectively.
Co-factors of the elements of third row of | A | are
i.e. – 1, 3, 2 respectively.
If 
show that A–1 = A.
Find 
if 
Also show that 
Find the inverse of the matrix 
.
Find the inverse of the matrix 
Find the inverse of the matrix:
Find the inverse of the matrix:
i.e. 3, 1, – 1 respectively
Co-factors of the elements of the second row of | A | are
i.e. I, 3, I respectively
Co-factors of the elements of the third row of | A | are
i.e. 1, 1, 3 respectively
Find the inverse of the matrix:
Find the inverse of the matrix:
Find the inverse of the matrix:
i.e. α cos2 α sin2 α 0 , 0 i.e. – 1, 0, 0 respectively.
Co-factors of the elements of second row of | A | are
i.e. 0, cos α, – sin α, respectively.
Co-factors of the elements of third row of | A | are
i.e. 0, – sin α, cos α respectively.
If 
then compute the inverse of A and verify that A–1 A = I = A A–1
Let A = 
Show that A is inevitable. Find adj. A and A Also verify that AA–1 = A–1 A = I.
i.e., – 10, – 5, – 15 respectively
Co-factors of the elements of the second row of | A | are
i.e., 5, 0, – 20 respectively.
Co-factors of the elements of the third row of | A | are
i.e., – 5, – 5, – 10 respectively.
Also,
From (1) and (2), we get,
AA–1 = I = A–1A.
Verify AA–1 = A–1 A = I where
Here 
Co-factors of the elements of the first row of | A | are
i.e., – 8, 11, –4 respectively
Co-factors of the elements of the second row of | A | are
i.e., 4, – 2, 0 respectively.
Co-factors of the elements of the third row of | A | are
i.e., 4, – 3, 0 respectively.
If 
then A3 = A–1 What is adj. A ?
From (1) and (2),
A3 = A–1
If 
find A–1 and show that A–1 = A2.
From (1) and (2),
A–1 = A2
If A = 
prove that A–1 = A–2 – 6 A + 111.
Here 
Co-factors of the elements of first row of | A | are
i.e. 3, – 15, 5 respectively
Co-factors of the elements of second row of | A | are
i.e. – 1, 6, – 2 respectively
Co-factors of the elements of third row of | A | are
i.e. 1, – 5, 2 respectively
If 
Find A–1 and hence prove that A2 – 4A – 5 I = O.
For the matrix 
show that A3 + 6A2 + 5A + 11 I = 0. Hence, find A–1.
Compute (AB)-1 where 
Here,
Co-factors of the elements of first row of | A | are
i.e. 3 – 4. – (2 – 2). 4 – 3 i.e. – 1,0, I respectively.
Co-factors of the elements of second row of | A | are
i.e. – (0 – 8), 5 – 4 , – (10 – 0) i.e. 8, I, – 10 respectively.
Co-factors of the elements of second row of | A | are
i.e. 0 – 12, – (10 – 8). 15–0 i.e. – 12, 2, 15 respectively.
Compute (AB)1 where:
i.e.. — 8, – 2, 5 respectively
Co-factors of the elements of third row of | A | are
i.e., -7, 3, 2 respectively.
If 
verify that (AB)–1 = B"–1 A–1.
B.
| A |2We know that
I adj A | = | A |n_l where A is of order n
Now A is non-singular matrix of order 3 × 3
∴ | adj A | = | A |2
∴ (B) is correct answer.
If A is an invertible matrix of order 2, then det (A–1) is equal to
det (A)
1
0
B.
We know that
A.
Examine the consistency of the system of equations:
x + 2y = 2
2x + 3y = 3
The given equations are
x + 2y = 2
2x + 3y = 3
These equations can be written as
∴ A–1 exists
∴ system of equations has a unique solution
∴ system of equations is consistent.
Examine the consistency of the system of equations:
2x – y = 5
x + y = 4
The given equations are
2x – y = 5
x + y = 4
These equations can be written as
or 
∴ A–1 exists
∴ system of equations has a unique solution, system of equations is consistent.
Examine the consistency of the system of equations:
x + 3y = 5
2x + 6y = 8
The given equations are
x + 3y = 5
2x + 6y = 8
These equations can be written as
or 
∴ solution does not exist and so system of equations is inconsistent.
Examine the consistency of the system of equations:
x+y+z = 1
2x + 3y + 2z = 2
ax+ay+2az = 4
Examine the consistency of the system of equations:
3x – y – 2z = 2
2y – z = – 1
3x – 5y = 3
The given equations are
3x – y – 2z = 2
0x + 2y – z = – 1
3x – 5y+Oz = 3
These equations can be written as
Cofactors of the elements of first row of | A | are
i.e. 0 – 5, – (0 + 3), 0 – 6 i.e. – 5, – 3, – 6 respectively.
Cofactors of the elements of second row of | A | are
i.e. – (0 – 10), 0 + 6, –(– 15 + 3) i.e. 10, 6, 12 respectively.
Co-factors of the elements of third row of | A | are
i.e. 1 + 4, – (– 3 – 0), 6 – 0 i.e. 5, 3, 6 respectively.
∴ solution does not exist and so system of equations is inconsistent.
Examine the consistency of the system of equations:
5x – y + 4z = 5
2x + 3y + 5z = 2
5x – 2y + 6z = – 1
The given equations are
5x – y + 4z = 5
2x + 3y + 5z = 2
5 x – 2 y + 6 z = – 1
These equations can be written as
∴ given system of equations "has a unique solution and so system of equations is consistent.
Examine the consistencies of the system of equations:
3x – y + 2z = 3
2x + y + 3z =5
x - 2y - z = 1
The given equations are
3x – y + 2z = 3
2x + y + 3z = 5
x – 2 – z = 1
These equations can be written as
or 
Co-factors of the elements of the first row of | A | are
or – 1 + 6, – (– 2, – 3), – 4 – 1 or 5, 5 – 5 respectively.
Co-factors of the elements of second row of | A | are
Co-factors of the elements of the third row of | A | are
or – 5, – 5, 5 respectively.
given equations have no solution.
Examine the consistencies of the system of equations:
x - y+ z = 3
2x - y – z = 2
– x – 2y + 2z = 1
The given equations are
x – y + z = 3 ...(1)
2 x – y – z = 2 ...(2)
– x –2y + 2z = 1 ...(3)
These equations can be written as
or 
Co-factors of the elements of first row of | A | are
or 0, – 3, – 3 respectively.
Co-factors of the elements of 2nd row of | A | are
or 0, 3, 3 respectively.
Co-factors of the elements of third row of | A | are
The given equations are
2x + 5y = 1
3x + 2y = 7
These equations can be written as
or 
The given equations are
5x + 2y = 4
7x + 3y = 5
These equations can be written as
The given equations are
2x – y = –2
3x + 4y = 3
These equations can be written as
or AX = B where 
The given equations are
4x – 3y = 3
3x – 5y = 7
These equations can be written as
The given equations are
5x + 2y = 3
3x + 2y = 5
These equations can be written as
or AX = B where,
Use matrix method to solve the system of equations:
3x – 2y = 7
5x + 3y = 1
The given equations are
3x – 2y = 7
5x + 3y = 1
The equations can be written as
or 
Use matrix method to solve the system of equations:
2x + 3y = – 1
x + 2y = 2
The given equations are
2x + 3y = – 1
x + 2y = 2
The equations can be written as
or 
Use matrix method to solve the system of equations:
5x - 7y = 2
7x - 5y = 3
The given equations are
5x – 7y = 2
1x – 5y = 3
The equations can be written as
or AX = B where 
Use matrix method to solve the system of equations:
x + 2y = 4
2x + 5y = 9
The given equations are
x + 2y = 4
2x + 5y = 9
These equations can be written as
or 
where 
Use matrix method to solve the system of equations:
2x + 3y = 5
3x - y = 2
The given equations are
2x + 3y = 5
3x – y = 2
These equations can be written as
or 
Use matrix method to solve the system of equations:
3x + 5y = 8
2x - y = 1
The given equations are
3x + 5y = 8
2x – y = 1
These equations can be written as
or 
Use matrix method to solve the system of equations:
5x - y = 4
3x + 7y = 10
The given equations are
5x – y = 4
3x + 7y = 10
These equations can be written as
or 
The given equations are
x + 2y – 3z = – 4
2x + 3y + 2z = 2
3x – 3y – 4z = 11
These equations can be written as
Co-factors of the elements of first row of | A | are
Co-factors of the elements of thitd row of | A | are
i.e. 13, – 8, – 1 respectively.
The given equations are
2x – y + z = 0
x + y – z = 6
3x – y – 4z = 7
These equations can be written as
or 
Now 
Co-factors of the elements of first row of | A | are
i.e. – 5, 1, – 4 respectively.
Co-factors of the elements of second row of | A | are
i.e. – 5, – 11, – 1 respectively.
Co-factors of the elements of third row of | A | are
i.e. 0, 3, 3 respectively.
Using matrices, solve the following system of equations:
x + 2y + z =1
2x – y + z = 5
3x + y – z = 0
The given equations are
x + 2y + z = 1
2x – y + z = 5
3x + y – z = 0
These equations can be written as
Co-factors of the elements of first row of | A | are
i.e. 0, 5, 5 respectively.
Co-factors of the elements of second row of | A | are
i.e. 3, – 4, 5 respectively.
Co-factors of the elements of third row of | A | are
i.e. 3, 1, – 5 respectively.
Using matrices, solve the following system of equations:
2x + y – 3r = 13
x + y – z = 6
2x – y + 4z = – 12
The given equations are
2x + y – 3 z = 13
x + y – z – 6
2x – y + 4 z = – 12
These equations can be written as
Co-factors of the elements of first row of | A | are
i.e. 3, – 6, – 3 respectively.
Co-factors of the elements of second row of | A | are
i.e. – 1, 14, 4 respectively.
Co-factors of the elements of third row of | A | are
i.e. 2, – 1, 1 respectively.
Using matrices, solve the following system of linear equations.
3x + 4y + 2z = 8
2y –3z = 3
x – 2y + 6z = –2
The given equations are
3x + 4y + 2z = 8
2y – 3z = 3
x – 2y + 6z = –2
There equations can be written as
Co-factors of the elements of first row of | A | are
i.e. 6, –3, –2 respectively.
Co-factors of the elements of second row of | A | are
i.e. – 28, 16, 10 respectively
Co-factors of the elements of third row of | A | are
i.e. – 16, 9, 6 respectively.
The given equations are
x – y + z = 1
2x+ y – z = 2
x 2 y – z = 4
These equations can be written as
Using matrices, solve the following system of linear equations:
x – y = 3
2x + 3y + 4z = 17
y + 2 z = 7
The given equations are
x – y = 3
2x + 3y + 4z = 17
y + 2z = 7
These equations can be written as
or 
Co-factors of elements of first row of | A | are
i.e. 2, – 4, 2 respectively.
Co-factors of the elements of second row of | A | are
i.e. 2, 2, – 1 respectively.
Co-factors of the elements of third row of | A | are
i.e. – 4, – 4, 5 respectively.
Solve by matrix method:
y + 2z = – 8
x + 2y + 3z = – 14
3x + y + z = – 8
The given equations are
y + 2z = – 8
x + 2y + 3z = – 14
3x + y + z = – 8
These equations can be written as
or 
Co-factors of the elements of first row of | A | are
i.e. –1, 8, –5 respectively.
Co-factors of the elements of second row of | A | are
i.e. 1, –6, 3 respectively.
Co-factors of the elements of third row of | A | are
i.e. –1, 2, –1 respectively.
Solve the following system of equations by matrix method:
3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4
Solve the following equation by matrix method:
The given equations are 
These equations can be written as
The given equations are
x– y + z = 4
2x + y – 3z = 0
x + y + z = 2
These equations can be written as
or 
Co-factors of the elements of first row of | A | are
i.e. 4, –5, 1 respectively.
Co-factors of the elements of second row of | A | are
i.e. 2, 0, –2 respectively.
Co-factors of the elements of third row of | A | are
i.e. 2, 5, 3 respectively
Solve the following system of equations by matrix method:
x – y + z = 2
2x – y = 0
2y – z = 1
The given equations are
x – y + z = 2
2x – y + 0z = 0
0x + 2y– z = 1
These equations can be written as
or 
Now,
Solve the following system of equations by matrix method:
2x – y + z = – 3
3 x – z = – 8
2x + 6y = 2
The given equations are
2x – y + z = –3 i.e. 2x – y + z = – 3
3x – z = –8 i.e. 3x + 0y – z = –8
2x + 6y = 2 i.e. 2x + 6 y + 0z = 2
These equations can be written as
or 
Solve the following system of equations by matrix method:
2x – 3y + 5z =11
3 x + 2y – 4 z = – 5
x + y – 2 z = –3
The given equations are
2x – 3 y + 5 z =11
3x + 2 y – 4 z = – 5
x + y – 2 z = –3
These equations can be written as
Use matrix method to solve the equations:
x + y + z = 3
2x – y + z = 2
x – 2y + 3z = 2
The given equations are
x + y + z = 3
2x – y + r = 2
x – 2y + 3z = 2
These equations can be written as
Solve (Use matrix method):
x + y = 0
y + z = 1
z + x = 3
The given equations are
x + y + Oz =0
Ox + y + z = 1
x + 0y + z = 3
These equations can be written as
or 
Co-factors of the elements of first row of | A | are
i.e. 1, 1 1 respectively.
Co-factors of the elements of second row of | A | are
i.e. -1, 1, 1 respectively.
Co-factors of the elements of third row of | A | are
i.e. 1, -1, 1 respectively.
or 
Co-factors of the elements of first row of | A | are
The given equations are
2x + y + z = 7
x – y – z = – 4
3x + 2y + z = 10
These equations can by written as
Co-factors of the elements of first row of | A | are
i.e. – 1 + 2, – (1 + 3), 2 + 3 i.e. 1, – 4, 5 respectively.
Co-factors of the elements of second row of | A | are
i.e. – (1 – 2), 2 – 3, – (4 – 3) i.e. 1, – 1, –1 respectively.
Co-factors of the elements of third row of | A | are
i.e. – 1 + 1, – (2 – 1), – 2 – 1 i.e. 0, 3, – 3 respectively.
Using matrices, following system of linear equations:
x – y + 2 z = 1
2 y – 3z = 1
3x – 2y + 4z = 2
The given equations are
x - y + 2z = 1
2y – 3z = 1
3x – 2y + 4z = 2
These equations can be written as
or 
Co-factors of the elements of first row of | A | are
i.e. 8 – 6, – (0 + 9), 0 – 6 i.e. 2, – 9, – 6 respectively.
Co-factors of the elements of second row of | A | are
i.e. – (– 4 + 4), 4 – 6, – (– 2 + 7) i.e. 0, – 2, – 1 respectively.
Co-factors of the elements of third row of | A | are
i.e. 3 – 4, – (– 3 – 0), 2+ 0 i.e. – 1, 3, 2 respectively.
Use matrix method to solve the following system of equations:
x – y + 2z = 7
3x + 4 y – 5 z = – 5
2x – y + 3z = 12
The given equations are
x – y + 2z = 7
3x + 4y – 5z= – 5
2 x – y + 3z = 12
These equations can be written as
Co-factors of the elements of first row of | A | are
or 7, 19, – 11 respectively.
Co-factors of the elements of second row of | A | are
or 1, – 1, – 1 respectively.
Co-factors of the elements of third row of | A | are
or – 3, 11, 7 respectively.
Use matrix method to solve the following system of equations:
x – y + 2z = 7
3x + 4 y – 5 z = – 5
2x – y + 3z = 12
The given equations are
x – y + 2z = 7
3x + 4y – 5z= – 5
2 x – y + 3z = 12
These equations can be written as
Co-factors of the elements of first row of | A | are
or 7, 19, – 11 respectively.
Co-factors of the elements of second row of | A | are
or 1, – 1, – 1 respectively.
Co-factors of the elements of third row of | A | are
or – 3, 11, 7 respectively.
Use matrix method to solve the following system of equations:
5x – y + z = 4
3x + 2y – 5z = 2
x + 3 y – 2 z = 5
The given equations are
5x – y + z = 4
3x + 2y – 5z = 2
x + 3y – 2z = 5
These equations can be written as
or AX = B where A = 
Co-factors of the elements of first row of | A | are
or 11, 1, 7 respectively.
Co-factors of the elements of second row of | A | are
Co-factors of the elements of third row of | A | are
or 3, 28, 13 respectively.
Now,
Use matrix method to solve the following system of equations:
4x + 2y + 3z = 2
x + y + z = 1
3x + y – 2z = 5
The given equations are
4x + 2y + 3z = 2
x + y + 2 = 1
3x + y – 2z = 5
These equations can be written as
Co-factors of the elements of first row of | A | are
or – 3, 5, – 2 respectively.
Co-factors of the elements of second row of | A | are
or 7, – 17, 2 respectively.
Co-factors of the elements of third row of | A | are
or – 1, – 1, 2 respectively.
Use matrix method to solve the following system of equations:
2x + 3y + 3z = 5
x – 2 y + z = – 4
3x – y – 2z = 3
The given equations are
2x + 3y + 3z = 5
x – 2 y + z = – 4
3x – y – 2z = 3
These equations can be written as
or 
Co-factors of the elements of first row of | A | are
or 5, 5, 5 respectively.
Co-factors of the elements of second row of | A | are
or 3, – 13, 11 respectively.
Co-factors of the elements of third row of | A | are
or 9, 1, – 7 respectively.
Use matrix method to solve the following system of equations:
2x – 3y + 3z = 1
2x+ 2y+ 3z = 2
3x – 2y + 2z = 3
The given equations are
2x – 3y + 3z = 1
2x+ 2y+ 3z = 2
3x – 2y + 2z = 3
These equations can be written as
or AX = B where 
Co-factors of the elements of first row of | A | are
i.e., 10, 5, – 10 respectively
Co-factors of the elements of second row of | A | are
i.e., 0, – 5, – 5 respectively
Co-factors of the elements of third row of | A | are
i.e., 15, 0, 10 respectively
Use matrix method to solve the following system of equations:
2x – 3y + 3z = 1
2x+ 2y+ 3z = 2
3x – 2y + 2z = 3
The given equations are
2x – 3y + 3z = 1
2x+ 2y+ 3z = 2
3x – 2y + 2z = 3
These equations can be written as
or AX = B where 
Co-factors of the elements of first row of | A | are
i.e., 10, 5, – 10 respectively
Co-factors of the elements of second row of | A | are
i.e., 0, – 5, – 5 respectively
Co-factors of the elements of third row of | A | are
i.e., 15, 0, 10 respectively
Use matrix method to solve the following system of equations:
2x – 3y + 3z = 1
2x+ 2y+ 3z = 2
3x – 2y + 2z = 3
The given equations are
2x – 3y + 3z = 1
2x+ 2y+ 3z = 2
3x – 2y + 2z = 3
These equations can be written as
or AX = B where 
Co-factors of the elements of first row of | A | are
i.e., 10, 5, – 10 respectively
Co-factors of the elements of second row of | A | are
i.e., 0, – 5, – 5 respectively
Co-factors of the elements of third row of | A | are
i.e., 15, 0, 10 respectively
The given equations are
y + 2z = 4 or 0x + y + 2 = z = 4
2z + x = 5 or x + 0y + 2z = 5
x + 2y = 7 or x + 2y + 0z = 7
These equations can be written as
or 
Co-factors of elements of first row of | A | are
i.e., – 4, 2, 2 respectively
Co-factors of the elements of second row of | A | are
i.e., 4, – 2, 1 respectively
Co-factors of the elements of third row of | A | are
i.e., 2, 2, – 1 respectively
The given equations are
3x – 4y + 2z = – 1
2x + 3y + 5z = 7
x + 0y + z = 2
These equations can be written as
or 
Co-factors of the elements of first row of | A | are
i.e., 3, 3, – 3 respectively
Co-factors of the elements of second row of | A | are
i.e., 4, 1, – 4 respectively
Co-factors of the elements of third row of | A | are
i.e., – 26, – 11, 17 respectively
The given equations are
x + y + 0z = 5
0x + y + z = 3
x + 0y + z = 4
These equations can be written as
or 
Co-factors of the elements of first row of |A| are
i.e., 1, 1, – 1 respectively
Co-factors of the elements of second row of |A| are
i.e. – 1, 1, 1 respectively
Co-factors of the elements of third row of | A | are
i.e., 1, – 1, 1 respectively
The given equations are
3x + y + z + 3
2x – y – z = 2
– x – y + z = 1
The given equation can be written as
or 
Co-factors of the elements of first low of | A | are
i.e. – 2, – 1, – 3 respectively
Co-factors of the elements of second row of | A | are
i.e. –2, 4, 2 respectively
Co-factors of the elements of third row of | A | are
i.e. 0, 5, 5 respectively
The given equations are
x + 2y – 3z = 6
3x + 2y – 2z = 3
2x – y + z = 2
These equations can be written as
or AX = B where A = 
Co-factors of the elements of first row of | A | are
i.e. 0, – 7, – 7 respectively
Co-factors of the elements of second row of | A | are
i.e. 1, 7, 5 respectively.
Co-factors of the elements of third row of | A | are
i.e. 2, – 7, – 4 respectively
i.e. 0, –4, –4 respectively
Co-factors of the elements of second row of | A | are
i.e. – 2, – 1, 3 respectively
Co-factors of the elements of third row of | A | are
i.e. 2, 3 – 1 respectively.
The given equations are
x + y – z = 1
3x + y – 2z = 3
x – y – z = – 1
These equations can be written as
Co-factors of the elements of first row of | A | are
i.e. – 3, 1, –4 respectively
Co-factors of the elements of second row of | A | are
i.e. 2, 0, 2 respectively
Co-factors of the elements of third row of | A | are
i.e. – 1, – 1, – 2 respectively
The given equations are
2x – y – z = 7 ...(1)
3x + y – z = 7 ...(2)
x + y – z = 3 ...(3)
These equations can be written as
or 
Co-factors of the elements of first row of | A | are
i.e. 0, 2, 2 respectively
Co-factors of the elements of second row of | A | are
i.e. – 2, – 1, – 3 respectively
Co-factors of the elements of third row of | A | are
i.e. 2, 1, 5 respectively
The given equations are
x + y + z =1
x – 2y + 3z = 2
x – 3y + 5z = 3
These equations can be written as
or 
Co-factors of the elements of first row of | A | are
i.e. – 1, – 2, – 1 respectively
Co-factors of the elements of second row of | A | are
i.e. – 8, 4, 4 respectively
Co-factors of the elements of third row of | A | are
i.e. 5, – 2, – 3 respectively
The given equations are
2x + y + 2z = 3
x + y + 2z = 2
2x + 3y – z = – 2
These equations can be written as
Co-factors of the elements of first row of | A | are
i.e. -7, 5, 1 respectively.
Co-factors of the elements of second row of | A | are
i.e. 7, – 6, – 4 respectively
Co-factors of the elements of third row of | A | are
i.e. 0. – 2, 1 respectively
Now 
The given equations are
x + 2y + z = 7
x + 3z = 11
2x – 3y = 1
These equations can be written as
or 
Co-factors of the elements of first row of | A | are
i.e. 9, 6, – 3 respectively
Co-factors of the elements of second row of | A | are
i.e. – 3, – 2, 7 respectively
Co-factors of the elements of third row of | A | are
i.e. 6, -2, -2 respectively.
The given equations are
x + y + z = 6
2x – y + z = 3
x – 2y + 3z = 6
These equations can be written as
where 
Co-factors of the elements of first row of | A | are
i.e., – 1, – 5, – 3 respectively
Co-factors of the elements of second row of | A | are
Co-factors of the elements of third row of | A | are
i.e., 2, 1, -3 respectively
The given equations are
3x + 14y + 7z = 14
2x – y + 3z = 4
x + 2y – 3z = 0
These equations can be written as
or 
Co-factors of the elements of first row of | A | are
i.e., – 3, 9, 5 respectively.
Co-factors of the elements of second row of | A | are
i.e., 26, – 16, – 2 respectively.
Co-factors of the elements of third row of | A | are
i.e., 19, 5, – 11 respectively.
⇒ x = 1, y = 1, z = 1 is the required solution.
The given equations are
5x + 3y + z = 16
2x + y + 3z = 19
x + 2y + 4z = 25
These equations can be written as
where 
Co-factors of the elements of first row of | A | are
i.e., – 2, – 5, 3 respectively.
Co-factors of the elements of 2nd row of | A | are
i.e., – 10, 9, – 7 respectively.
Co-factors of the elements of 3rd row of | A | are
i.e., 8, – 13, – 1 respectively.
The given equations are
x + y + z = 9
2 x + 5 y + 7 z = 52
2 x + y – z = 0
These equations can be written as
or 
Co-factors of the elements of first row of | A | are
Cofactors of-the elements of 2nd row of | A | are
i.e., 2, – 3, 1 respectively.
Co-factors of the elements of 3rd row of | A | are
i.e., 2, – 5, 3 respectively.
The given equations are
x + y + z = 4
2x – y + z = - 1
2x + y – 3z = – 9
These equations can be written as
Co-factors of the elements of first row of | A | are
i.e. 2, 8, 4 respectively.
Co-factors of the of the elements of second row of | A | are
i.e. 2, 8, 4 respectively.
Co-factors of the of the elements of second row of | A | are
i.e. 4, – 5, 1 respectively.
Co-factors of the elements of third row of | A | are
i.e. 2, 1, – 3 respectively.
The given equations are
x + y – z = 1
3x + y – 2z = 3
x – y – z = – 1
These equations can be written as
or 
Co-factors of the elements of first row of | A | are
i.e., – 3 1, – 4 respectively
Co-factors of the elements of second row of | A | are
i.e. 2, 0, 2 respectively
Co-factors of the elements of third row of | A | are
i.e. – 1, – 1, – 2 respectively
2x – 3y + 5z = 16
3x + 2y – 4z = – 4
x + y – 2z = – 3
These equations can be written as
or 
Now AX = B 
The given equations are
2 x – y – z = 1
x + y + 2z = 1
3x – 2y – 2z = 1
These equations can be written as
or 
Co-factors of the elements of first row of | A | are
i.e. 2, 8, – 5 respectively
Co-factors of the elements of second row of | A | are
i.e. 0, – 1, 1 respectively
Co-factors of the elements of third row of | A | are
The given equations are
2x + 6y = 2
3x – z = –8
2x – y + z = –3
These equations can be written as
or 
The given equations are
x + y + z = 6
x – y + z = 2
2x + y – z = 1
These equations can be written as
or 
Co-factors of the elements of first row of | A | are
Co-factors of the elements of second row of | A | are
i.e. 2, – 3, 1 respectively.
Co-factors of the elements of third row of | A | are
i.e. 2, 0, – 2 respectively.
The given equations are
9x – 5y – 11z = 12
x – 3y + z = 1
2x + 3y – 7z = 2
These equations can be written as
or 
Co-factors of the elements of first row of | A | are
Co-factors of the elements of second row of | A | are
i.e. – 68 – 41, – 37 respectively.
Co-factors of the elements of third row of | A | are
i.e. – 38, – 20, – 22 respectively.
Now,
The given equations are
2x + y – z = 1
x – y + z = 2
3x + y – 2 z = – 1
These equations can be written as
Co-factors of the elements of first row of | A | are
i.e. 1, 5, 4 respectively.
Co-factors of the elements of second row of 
are
i.e. 1. – 1, 1 respectively.
Co-factors of the elements of third row of | A | are
i.e. 0, – 3, – 3 respectively.
The given equations are
6x + y – 3z = 5
x + 3 y – 2z = 5
2x + y + 4 z = 8
These equations can be written as
Co-factors of the elements of first row of | A | are
i.e. 14, – 8, – 5 respectively.
Co-factors of the elements of second row of 
are 
i.e. – 7, 30, – 4 respectively.
Co-factors of the elements of third row of | A | are
i.e. 7, 9, 17 respectively.
The given equations are
2x – y + z = 3
– x + 2y – z = – 4
x – y + 2z = 1
These equations can be written as
Co-factors of the elements of first row of | A | are
i.e. 3, 1, – 1 respectively.
Co-factors of the elements of second row of | A | are
i.e. 1, 3, 1 respectively.
Co-factors of the elements of third row of | A | are
Solve by matrix method:
Solve the system of the following equations:
The given equations are
Put 
∴ given equations become
2a + 3b + 10c = 4
4a – 6b + 5c = 1
6a + 9b – 20c = 2
These equations can be written as
or 
Co-factors of the elements of first row of | A | are
i.e. 120 – 45, – (– 80 – 30), 36 + 36 i.e. 75, 110, 72 respectively.
Co-factors of the elements of second row of | A | are
i.e. – (– 60 – 90), – 40 – 60, – (18 – 18) i.e. 150, 100, 0 respectively.
Co-factors of the elements of third row of | A | are
i.e. 15 + 60, – (10 – 40), – 12 – 12 i.e. 75, 30, –24 respectively.
If 
find A-1, Using A-1, solve the following system of linear equations.
2x – 3y + 5z = 16
3x + 2y – 4z = – 4
x + y – 2z = – 3
Here,
Co-factors of the elements of first row of | A | are
i.e. 0, 2, 1 respectively
Co-factors of the elements of second row of | A | are
i.e., –1, –9 , –5 respectively.
Co-factors of the elements of third row of | A | are
i.e., 2, 23, 13 respectively.
The given equations are
2x – 3y + 5z = 16
3x + 2y – 4z = – 4
x + y – 2z = – 3
These equations can be written as
If 
find A –1 .
Using A–1, solve the following system of linear equations.
2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3
Here, 
Co-factors of the elements of first row of | A | are
i.e., 0, 2, 1 respectively
Co-factors of the elements of second row of | A | are
i.e., –1 –9 , –5 respectively.
Co-factors of the elements of third row of | A | are
i.e., 2, 23, 13 respectively.
The given equations are
2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3
These equations can be written as
or 
Now 
Compute A–1 for the following matrix 
Hence solve the system of equations.
– x + 2y+ 5 z = 2
2x – 3y + Z = 15
– x + y + z = – 3
Here 
Co-factors of the elements of the first row of | A | are
i.e., – 4, – 3, – 1 respectively
Co-factors of the elements of the second row of | A | are
i.e., 3, 4, – 1 respectively
Co-factors of the elements of the third row of | A | are
The given equations are
– r + 2y + 5z = 2
2x – 3y + z = 15
– x + y + z = – 3
These equations can be written as
where 
Compute A–1 for the following matrix
Hence, solve the system of equations:
x + 2y + 5z = 10
x – y – z = – 2
2x + 3y – 2z = – 1
i.e., 19, – 12, 1 respectively
Co-factors of the elements of third row of | A | are
i.e., 3, 6, – 3 respectively
The given equations are
x + 2y + 5z = 10
x – y – z = – 2
2x + 3y – 2z = – 1
These equations can be written as
or 
If 
Find A–1.
Using A solve the following systems of linear equations:
3x – 2y + z = 2
2y + y – 3z = – 5
– x + 2y + z = 6.
Here,
Co-factors of the elements of first row of | A | are
i.e., 7, 1, 5 respectively
Co-factors of the elements of second row of | A | are
i.e. 4, 4, – 4 respectively
Co-factors of the elements of third row of | A | are
i.e. 5, 11, 7 respectively
The given equations are
3x – 2y + z = 2
2y + y – 3 z = – 5
– x + 2y + z = 6
These equations can be written as
or 
Now 
Investigate for what values of a and b the simultaneous equations:
x + y + z = 6
x + 2y + 3z = 10
x + 2y + az = b have a unique solution.
The given equations are
x + y + z = 6
x + 2y + 3z = 10
x + 2y + az = b
These equations can be written as
or 
Given system of equation has a unique solution
when | A | ≠ 0 i.e., a – 3 ≠ i.e., a ≠ 3, and b and may have any value.
The sum of three numbers is 6. If we multiply third number by 3 and add second number to it, we get 11. By adding first and third numbers, we get double of the second number. Represent it algebraically and find the numbers using matrix method.
Let first, second and third numbers be denoted by x, y and z respectively.
From given conditions,
x + y + z = 6
y + 3z = 11 or 0x + y + 3z = 11
x + z = 2y or x – 2y + z = 0
These equations can be written as
or 
Co-factors of the elements of first row of | A | are
i.e., 1 + 6, – (0 - 3), 0 – 1 i.e. 7, 3, – 1 respectively.
Co-factors of the elements of second row of | A | are
i.e. – (1 + 2), 1 – 1, – (– 2 – 1) i.e. - 3, 0, 3 respectively
Co-factors of the elements of third row of | A | are
i.e. 3 – 1(3 – 0), 1 – 0 i.e. 2, – 3, 1 respectively.
Now, 
Let the cost of onion, wheat and rice per kg. be Rs. ,v, Rs. y and Rs. r respectively.
From given conditions,
4x + 3y + 2z = 60
2x + 4y + 6z = 90
6x + 2y + 3z = 70
These equations can be written as
Co-factors of the elements of first row of | A | are
i.e. (12 – 12), – (6 – 36), (4 – 24) i.e. 0, 30, – 20 respectively.
Co-factors of the elements of second row of | A | are
i.e. – (9 – 4), (12 – 12), – (8 - 18) i.e. – 5, 0, 10 respectively.
Co-factors of the elements of third row of | A | are
i.e. (18 – 8), – (24 – 4), (16 – 6), i.e. 10, – 20, 10 respectively.
∴ the cost of onion, wheat and rice per kg. are Rs. 5, Rs. 8, Rs. 8 respectively.
Use Properties of determinants, prove that:
Consider the determinant
Taking abc common outside, we have
Applying the transformation, 
Apply the transformations, 
Expanding along 
, we have
Using properties of determinants prove the following:
we have
Use product 
to solve the system of equations x + 3z = 9,
–x + 2y – 2z = 4, 2x – 3y + 4z = –3
Since A X B = I,
therefore, B = A-1..... (i)
Now, the given system of equations is
x+3z =9
-x+2y-2z =4
2x-3y+4z = -3
This can also be represented as,
Hence x=0 y=5 and =3
Find the co-factor of a12 in the following:
Cofactor of a12 = ( -1 )1+2 [ 6(-7) - 4(1) ]
= (-1) [ -42 - 4 ]
= 46
Find the minor of the element of second row and third column ( a23 ) in the following determinant:
Given determinant =
Minor of the element a23 is M23
Obtained by deleting III column and II row
Using properties of determinants show the following:
Consider,
Now, taking a, b, c common from C1, C2, and C3
Expanding along R3
Using properties of determinants, prove that
[ Taking out a, b, and c common from R1, R2, and R3 respectively]
[ Taking out a, b, and c common from C1, C2, and C3 respectively]
Hence proved.
Using matrix method, solve the following system of equations:
The given system of equation can be written as
Hence, the unique solution of the system of equation is given by X = A- 1 B
Now, the cofactors of A are computed as:
Thus, solution of given system of equation is given by x = 2, y = 3, z = 5.
If A = 
and A adj A = AAT, then 5a +b is equal to
-1
5
4
5
B.
5
Given, A =
and A adj A = AAT, Clearly, A (adj A) = |A|In|
Now, substituting the value of b in Eq. (iii) we get
5a = 2
Hence, 5a + b = 2 +3 = 5
The system of linear equations x+λy−z=0; λx−y−z=0; x+y−λz=0 has a non-trivial solution for
infinitely many values of λ.
exactly one value of λ.
exactly two values of λ.
exactly three values of λ.
D.
exactly three values of λ.
Given system of linear equations is
x+λy−z=0;
λx−y−z=0;
x+y−λz=0
Note that, given system will have a non-trivial solution only if the determinant of the coefficient matrix is zero, ie.
Hence, given system of linear equation has a non-trivial solution for exactly three values of λ.
A = 
is a matrix satisfying the equation AAT = 9I, Where I is 3 x 3 identity matrix, then the ordered pair (a,b) is equal to
(2,-1)
(-2,1)
(2,1)
(-2,-1)
D.
(-2,-1)
Given, 
.
It is given that,
On comparing we get,
a+ 4 +2b = 0
a+ 2b = -4 ... (i)
2a + 2-2b = 0
a-b= -1 ... (ii)
a2 + 4 +b2 = 9 ... (iii)
On solving eqs. (i) and (ii) we get
a = - 2, b = - 1
Hence, (a,b) ≡ (-2,-1)
If α, β ≠ 0 and f(n) = αn+ βn and 
= K(1-α)2(1-β)2(α- β)2, then K is equal to
αβ
1/αβ
1
-1
C.
1
f(n) = αn + βn
f(1) = α + β
f(2) = α2 + β2
f(3) =α3 + β3
f(4) = α4 + β4
Let Δ = 
If A is a 3x3 non- singular matrix such that AAT = ATA, then BBT is equal to
B-1
(B-1)T
B.
lIf A is non - singular matrix then |A| ≠0
AAT = ATA and B = A-1AT
BBT = (A-1AT)(A-1AT)T
= A-1ATA(A-1)T [∵ (AB)T= BTAT]
=A-1AAT(A-1)T [∵ AAT = ATA]
=AT(A-1)T [ ∵A-1A = l]
=A-1A)T [∵ (AB)T = BTAT]
lT = l
Let P and Q be 3 × 3 matrices with P ≠ Q. If P3= Q3 and P2Q = Q2P, then determinant of(P2+ Q2) is equal to
-2
1
0
-1
C.
0
P3= Q3
P3– P2Q = Q3– Q2P
P2(P – Q) = Q2(Q – P)
P2(P – Q) + Q2(P – Q) = O
(P2+ Q2)(P – Q) = O
⇒ |P2+ Q2| = 0
The number of values of k for which the linear equations
4x + ky + 2z = 0
kx + 4y + z = 0
2x + 2y + z = 0
posses a non-zero solution is:
3
2
1
0
B.
2
⇒ 8 - k(k -2) - 2(2k - 8) = 0
⇒ 8 - k2 + 2k - 4k + 16 = 0
⇒ -k2 - 2k + 24 = 0
⇒ k2 + 2k - 24 = 0
⇒ (k + 6)(k - 4) = 0
⇒ k = - 6, 4
Number of values of k is 2
Consider the system of linear equation
x1 + 2x2 + x3 = 3
2x1 + 3x2 + x3 = 3
3x1 + 5x2 + 2x3 = 1
The system has
infinite number of solutions
exactly 3 solutions
a unique solution
no solution
D.
no solution
Subtracting the Eq. (ii) – Eq. (i)
We get x1 + x2 = 0
Subtract equations
Eq. (iii) – 2 × eq. (ii)
x1 + x2 = 5
Therefore, no solutions
Let A be a 2 × 2 matrix with non-zero entries and let A2 = I, where I is 2 × 2 identity matrix. Define Tr(A) = sum of diagonal elements of A and |A| = determinant of matrix A.
Statement-1: Tr(A) = 0.
Statement-2: |A| = 1.
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Statement-1 is true, Statement-2 is true; statement-2 is not a correct explanation for Statement-1.
Statement-1 is true, Statement-2 is false.
Statement-1 is false, Statement-2 is true.
C.
Statement-1 is true, Statement-2 is false.
A satisfies A2 -Tr(A). A + (det A) l = 0
on comparing with A2-I = 0,
we get
Tr (A) = 0, |A| = - 1
Let A be a square matrix all of whose entries are integers. Then which one of the following is true?
If det A = ± 1, then A–1 exists but all its entries are not necessarily integers
If detA ≠ ± 1, then A–1 exists and all its entries are non-integers
If detA = ± 1, then A–1 exists and all its entries are integers
If detA = ± 1, then A–1 need not exist
C.
If detA = ± 1, then A–1 exists and all its entries are integers
Each entry of A is integer, so the cofactor of every entry is an integer and hence each entry in the adjoint of matrix A is integer. Now detA = ± 1 and A–1 =(1/ det(A)) (adj A)
⇒ all entries in A–1 are integers
Let a, b, c be any real numbers. Suppose that there are real numbers x, y, z not all zero such that x = cy + bz, y = az + cx and z = bx + ay. Then a2 + b2 + c2 + 2abc is equal to
2
-1
0
1
D.
1
The system of equations x – cy – bz = 0, cx – y + az = 0 and bx + ay – z = 0 have non-trivial solution if
If D = 
for x ≠ 0, y ≠ 0 then D is
divisible by neither x nor y
divisible by both x and y
divisible by x but not y
divisible by y but not x
B.
divisible by both x and y
Let 
a , b ∈ N. Then
there cannot exist any B such that AB = BA
there exist more than one but finite number of B’s such that AB = BA
there exists exactly one B such that AB = BA
there exist infinitely many B’s such that AB = BA
D.
there exist infinitely many B’s such that AB = BA
If a1, a2, a3,…, an,… are in G.P., then the determinant
1
0
4
2
B.
0
C1 – C2, C2 – C3 two rows becomes identical Answer: 0
If the system of linear equations
x + ky + 3z = 0
3x + ky – 2z = 0
2x + 4y – 3z = 0
has a non-zero solution (x,y,z), then xz/y2 is equal to
30
-10
10
-30
C.
10
System fo equation has non-zero solution
Sponsor Area
Sponsor Area
, find the values of x.
- 24
changes.
If a, b, c are in A.P. find value of
Without expanding, prove that
,where a,b,c and in A.P
respectively
show that either 
respectively
,
...(2)
...(1)
.
and 
show that 
show that 
show that 
Hence find 
show that 
satisfies the equation 
show that 
find x and y so that 
then verify that 
show that 
find 
find the value of x.
then write the value of x. 
then adj (3A2 + 12A) is equal to
If |A2| = 25, then |α| equals