Determinants Class 12
Master Determinants Class 12 with comprehensive NCERT Solutions, Practice Questions, MCQs, Sample Papers, Case Based Questions, and Video lessons.
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Determinants Class 12 – NCERT Solutions
Each question below opens its complete step-by-step Teachoo solution.
Ex 4.1
13 questionsEx 4.1,1
Ex 4.1, 1 teachoo.com
Evaluate the determinants [4 4
[4s §l
—-5 -1
= 2(-1) - (-5)(4)
=-2+20
=18
Ex 4.1, 2 (i)
Ex 4.1, 2 teachoo.com
Evaluate the determinant
(i) |cos 6 —sin °|
sin@ cos@
[cos 8 —sin |
sin® cosé
= cos @(cos 8) - (sin 8) (-sin 8)
= cos? § + sin’ 6
=1 (As sin? 8 + cos? 8 = 1)
Ex 4.1, 2 (ii)
Evaluate the determinant
(ii) |■8(x2−x+1&x−1@x+1&x+1)|
Ex 4.1, 3
Ex 4,1, 3 teachoo.com
1 2
IfA= lL | , then show that |2A| = 4|A|
We need to prove |2A| = 4|A|
Solving L.H.S
|2A|
First calculating 2 A
fl 2
2a=2|, 5]
=|4 x1 2x ek 4]
2x4 2x2 8 4
Ex 4.1, 4
Ex 4.1, 4 teachoo.com
101
IfA= 0 1 | , then show that |3A| = 27 [Al
0 0 4
We have to prove
|3A] =27 |A]
Solving L.H.S
3A]
First Calculating 3A
101 3x1 3x0 3x1 3.0 «3
2a-a)o 1 a|-[3 0 3x1 3x 2|-[o 3 |
00 4 3x0 3x0 3x4!) lO 0 12
Ex 4.1, 5 (i)
Ex 4.1, 5 teachoo.com
Evaluate the determinant
3-1 -2
(i) b 0 =|
3 -5 0
3-1 -2
f 0 =|
3 -5 O
_,)/0 -1 0 -1 0 0
=3[65 ol- ells sl
= 3 (0(0) — (—5) (-1) ) + 1 (0(0)— 3(— 1)) — 2 (O(-5) — 3(0))
=3(0-5)+1(0+3)-2 (0-0)
= 3(-5)+1(3)+0
= -15+3
=-12
Ex 4.1, 5 (ii)
Evaluate the determinant
(ii) |■8(3&−4&5@1&1&−2@2&3&1)|
Ex 4.1, 5 (iii)
Evaluate the determinants
(iii) |■8(0&1&2@−1&0&−3@−2&3&0)|
Ex 4.1, 5 (iv)
Evaluate the determinant
(iv) |■8(2&−1&−2@0&2&−1@3&−5&0)|
∆ = |■8(2&−1&−2@0&2&−1@3&−5&0)|
= 2 |■8(2&−1@−5&0)| – (– 1) |■8(0&−1@3&0)| + (– 2) |■8(0&2@3&−5)|
= 2(2(0) – (−5)(−1)) +1 (0(0) – 3 (−1)) – 2 (0(-5) – 3 (2))
= 2(0 – 5) + 1 (0 + 3) – 2 (0 – 6)
= 2 (– 5) + 1(3) – 2(– 6)
= – 10 + 3 + 12
= 2 + 3
= 5
Ex 4.1, 6
Ex 4.1, 6 teachoo.com
11 -2
IfA= F 1 =| find |A|
5 4 -9
11 -2
[Al = p 1 =
5 4 -9
jy. 3). ,|2 -3 21
=], Tol-tl5 ltl
= 1 (1(-9) - 4(-3)) — 1 (2( -9) - 5(-3)) -2(2(4) - 5(1))
=1(-9+12)+1(-18 + 15)—2 (8-5)
= 1(3)- 1 (-3)-2 (3)
=3+3-6
=6-6
=0
Hence, |Al=0
Ex 4.1, 7 (i)
Ex 4.1, 7 teachoo.com
Find values of x, if
(2 4 2x 4
( [5 il “| x
Calculating Calculating
2 4l_ _ 2x 4)_ _
F i|=2 5(4) I? * | = 2x00 6x4
=2-20 =-18 = 2x? -24
Now
|? ‘| - \?* 4
5 1 6 x
Putting values
-18 = 2x?- 24
2x? 24 =-18
2x? = -18 + 24
2x? = 6
Ex 4.1, 7 (ii)
Find values of x, if
(ii) |■8(2&3@4&5)| = |■8(𝑥&3@2𝑥&5)|
Ex 4.1, 8 (MCQ)
Ex 4.1, 8 teachoo.com
x 2/_]6 2 :
If 9 ‘elie «| then x is equal to
(A) 6 (B) +6 (C) -6 (D) 0
Solving Solving
xX 2]_ _ 6 2)_ _
Lig | =x00-18(2) 3, |= 6t6) - 1812)
=x*— 36 =36-36 = 0
. x 2]_]16 2
Given |, IF lis sl
Putting values
x? -36=0
x -6=0
(x — 6) (x + 6) =0
Ex 4.2
9 questionsEx 4.2,1 (i)
Ex 4.2, 1 teachoo.com
Find area of the triangle with vertices at the point given in each
of the following:
(i) (1, 0), (6, 0), (4, 3)
t) uy dd
X1 V1 X2 V2 Xs Vs
3 1 1
The area of triangle is given by A= 3 [x2 Ye 1
Xz Y3 1
Here,
x,=1 7¥1=0
X)=6 1¥2 = 0
x3 =4 ¥3 = 3
1 101
A= z ; 0 ,
43 1
Ex 4.2,1 (ii)
Find area of the triangle with vertices at the point given in each of the following:
(ii) (2, 7), (1, 1), (10, 8)
The area of triangle is given by ∆ = 1/2 |■8(x1&y1&1@x2&y2&1@x3&y3&1)|
Ex 4.2,1 (iii)
Find area of the triangle with vertices at the point given in each of the following:
(iii) (−2, −3), (3, 2), (−1, −8)
The area of triangle is given by ∆ = 1/2 |■8(x1&y1&1@x2&y2&1@x3&y3&1)|
Ex 4.2, 2
teachoo.com
Ex 4,2, 2 (Introduction)
Show that points A (a, b + c), B (b,c + a), C (c,a + b) are collinear
B
-—__+__ oo,
A B c A Cc
3 points collinear Area of triangle # 0
Area of triangle = 0
Ex 4.2, 3 (i)
Ex 4.2, 3 teachoo.com
Find values of k if area of triangle is 4 square units and vertices are
(i) (k, 0), (4, 0), (0, 2)
1 x ¥, 1
The area of triangle is given by A= 3 kz Y2 1
X3 y3 1
Here
Area of triangle 4 square units
Since area is always positive,
A can have both positive & negative signs
~A=t4,
Putting x, =k, y, =0, x, =4, y, =0, x,=Oy; =2
Ex 4.2, 3 (ii)
Find values of k if area of triangle is 4 square units and vertices are
(ii) (-2, 0), (0, 4), (0, k)
The area of triangle is given by ∆ = 1/2 |■8(x1&y1&1@x2&y2&1@x3&y3&1)|
Ex 4.2, 4 (i)
Ex 4.2, 4 teachoo.com
(i) Find equation of line joining (1, 2) and (3, 6) using determinants
Let L be the line joining the points (1, 2) & (3, 6)
+e—_e——e— |
(1,2) (% y) (3, 6)
Let (x, y) be the third point on line
Since all the there point lie on the same line,
they do not from a triangle
B
oe a™.
Hence, A B Cc A c
Area of triangle = 0 3 points collinear Area of triangle # 0
Thus, A=0 Area of triangle = 0
Ex 4.2, 4 (ii)
Find equation of line joining (3, 1) and (9, 3) using determinants
View solutionEx 4.2, 5 (MCQ)
Ex 4,2,5 teachoo.com
If area of triangle is 35 square units with vertices
(2, -6), (5, 4), and (k, 4). Then k is
A. 12 B. -2 Cc, -12, -2 D. 12, -2
1 : ¥1 i
The area of triangle is given by A = 5 | Ye 1
x3 y3 1
Here,
Area of triangle is 35 square unit
Since area is always positive ,
So A can have positive & negative value
=> A=+35 square unit
Ex 4.3
7 questionsEx 4.3, 1 (i)
Ex 4.3, 1 teachoo.com
Write Minors and Cofactors of the elements of following determinants|
[2 —4
ti) lo 3 |
Minor of a,, = M,, = 3 | =3
Minor of a,, = M,, = 0 | =0
Minor of a, = Mp, = ra =-4
Minor of a,.=M,, = 2 |= 2
Cofactor of ai; = Ay = (-1)'*7. My
Cofactor of a,, = A,, = (-1)""" M,,=(-1)? x3 =3
Cofactor of a) = Ay) = (-1)"* M,, = (-1)? x0 = 0
Cofactor of a,, = A, = (-1)?* M,, = (-1)° (-4) = -1(-4) =4
Cofactor of a.) = Aj. = (-1)?** M,, = (-1)* (2) =2
Ex 4.3, 1 (ii)
Write Minors and Cofactors of the elements of following determinants:
(ii) |■8(a&c@b&d)|
Ex 4.3, 2 (i)
Ex 4.3, 2 teachoo.com
Write Minors and Cofactors of the elements of determinants:
100
()]O 1 0
0 01
Minor of a,,=M,, = 1 0 =| Hl =1(1)-O=1
0 1
rst ==] | =f | =0-0 =0
0 1
orton =Mo=f 1 |. p|=0-0=0
0 0
0 0
Minor of a,, = M>, = = | =0-0=0
0 1
Ex 4.3, 2 (ii)
Write Minors and Cofactors of the elements of determinants:
(ii) |■8(1&0&4@3&5&−1@0&1&2)|
Ex 4.3, 3
Ex 4.3, 3 teachoo.com
5 3 8
Using Cofactors of elements of second row, evaluate A=|2 0 1
1 2 3
A= a2, Az, + a2 Az2 + A23Ap3
a1 = 2, a, = 0, a, = 1,
Calculating cofactor of second row
ie. Ay, ,A., And A),
3.8
May = -B 8] =3x3-2«8 =9-16 = -7
2 3
5 8 P
My = -/[? §| =5x3-8x1-15-8 =7
1 3
Ex 4.3, 4
Ex 4.3, 4 teachoo.com
1 x yz
Using Cofactors of elements of third column, evaluate A=|1 y zx
1 z xy
1 x yz
A=|1 y 2x
1 a xy
A= aj3 Aj3 + az3 Azg + agg Agg
413 = YZ, a93= 2X, ag, = XY,
Calculating cofactors of third column
ie. Ay, A3,And A;
Ex 4.3, 5 (MCQ)
Ex 4,3,5 teachoo.com
Ay, Ay2 Ay
IfA= fs ax 2 and A,is Cofactor of a; then value of A is
43, 432 33
given by
(A) ay1 Agi + Aq2 Aga + 843 Ags
(B) gy Aga + Aq. Ani + 43 Ast
(C) aa, Ang + Ag Aya + 893 Ay3
(D) ay, Ara + Ap, Ant + 31 Ast
Ayr Ay2 AY
IfA= fs az2 2,
43, 432 433
A= Sum of products of elements of row (or column) with their
corresponding cofactors
Ex 4.4
18 questionsEx 4.4, 1
teachoo.com
Ex 4.4, 1 (Method 1)
. wo . 1 2
Find adjoint of each of the matrices. [ |
3.4
1 2
aE al
Interchange Change sign
| 4 7]
—3 #1
Ex 4.4, 2
Ex 4.4, 2 teachoo.com
1 -1 2
Find adjoint of each of the matrices. | 2 3 |
—2 0 1
1 -1 2
LetA= 2 3 5|
—2 90 1
Au An Agi
adjA=|A,, Ay. Az.
Ay3 Ao3 A33
Step 1: Cacluating minors
355
Mir={5 |= 3(1)-0(5) = 3-0-3
-|2 5)o5 7-49) = -
Ma=|“, | =2 (-10) =2+10 =12
-|2 3! e9_-;_e= =
M3=| 4, ol =° (-6)=0+6=6
Ex 4.4, 3
Ex 4.4, 3 teachoo.com
, . ; 2 3
Verify A (adj A) = (adj A) A= |A| I, where A= [4 “.
Let A= [4 a
hor Change sign
n- | we
adj A= ae
- ("2 |
4 2
_|2 3
Mi=|", “6
=2x (-6)-3x(-4) =-12+12=0
Ex 4.4, 4
teachoo.com
Ex 4.4, 4
1 -1 2
Verify A (adj A) = (adj A) A= |A|T for A= F 0 |
1 O 3
Calculating |A|
1 -1 2
JAl=|3 0 -2
1 0 3
_,|0 —2)_,4,|3 —2],5]3 0
=15 3 | cfs 3 12h; al
=1(0-0)+1(9+2)+2 (0-0)
=11
Ex 4.4, 5
Ex 4.4, 5 teackoo.com
. . . gs . 2 -2
Find the inverse of each of the matrices (if it exists) ( 3 |
_f2 -2
Let A = l, 3
We know that
a1 ag
A ial adj A
exists when |A|#0
2 2
Ial=|, |
= 2x3-4x(-2) =6+8= 14
Since |A|#0, A exists
Ex 4.4, 6
Ex 4.4, 6 teachoo.com
. . . fie act -—1 5
Find the inverse of each of the matrices (if it exists) [3 |
_f-1 5
LetA = = |
We know that
ae1 ag
A ial adj A
exists when |A|#0
_|-1 5
lal=|-3 3
= -1x2-(-3)x5 = -2+15 = 13
Since |A|#0, A exists
Ex 4.4, 7
Ex 4.4, 7 teachoo.com
12 3
Find the inverse of each of the matrices (if it easy 2 |
0 0 5
1 2 3
LetA= ° 2 |
00 5
We know that
ae tag
A Tal (adj A)
exists if [AJ #0
Ex 4.4, 8
Ex 4.4, 8 teachoo.com
10 0
Find the inverse of each of the matrices (if it exists).]3. 3 0
5 2 -1
10 0
LetA=|3 3 0
5 2 -1
We know that
ae fag
A ia AG AD
exists if [AJ #0
Step 1: Calculate |A|
10 0
3.420 3.0 3 3
i= 3 oes _,|-0 +0
33 oles Sl-olf Slof 3
=1(-3-0)+0+0 =-3
Since |A| #0, A exists
Ex 4.4, 9
Ex 4.4, 9 teachoo.com
2 1 3
Find the inverse of each of the matrices (if it exists).| 4 —1 0
—7 2 1
2 1 3
LetA=|4 -1 0
—7 2 «1
We know that
a1 taqj
At= Tal (adj A)
exists if |A]#0
Ex 4.4, 10
Ex 4.4, 10 teachoo.com
1 -1 2
Find the inverse of each of the matrices (if it exists). f 2 3
3 -2 4
1 -1 2
wea 2 “i
3 -2 4
We know that
ae tag
A ia 20 A)
exists if |A| #0
Ex 4.4, 11
Ex 4.4, 11 teachoo.com
1 0 0
Find the inverse of each of the matrices|Q cosa sina
0 sina —cosa@
1 0 0
LetA=|0 cosa sina
0 sina —cosa@.
We know that
1
ae fag
A Tal (adj A)
exists if [AJ #0
Ex 4.4, 12
teachoo.com
Ex 4.4, 12
_[3 7 _[6 8 . 4eplad
Let A= [5 sl and B = [ al verify that (AB)? = B1 A
Taking L.H.S
(ABy*
First calculating AB
AB = 2 7] [° 8) - Ee +77) 3(8) + |
2 SII7 9 2(6)+5(7}) 2(8)+5(9)
_ [18 +49 244 | - [°? |
“12+35 16+451 147 61
Ex 4.4, 13
Ex 4.4, 13 teachoo.com
IFA = [3 1] show that A*-— 5A + 7] = O. Hence find A*.
Calculating A?
AV=AA
-[3 [3 BO tO |
—1 211-1 2! [-1@)4+2(-1) -1(1)+2()
-[°>1 342 )-[8 °]
—3-2 -14+4! |--5 3
Solving L.H.S
A?-5A+ 7]
78 5 3 41 1 0
-[E 3 L al*7\o |
Ex 4.4, 14
Ex 4.4, 14 teachoo.com
. 3.2 .
For the matrix A= [; ‘| , find the numbers a and b such that
A?+aA+bl=0.
Finding A2
AP=AA
_f3 2] 78 2
~ [; 1 [; 1
_ hex +21) 3(2)+ Her _ [? +2 6+ 2)
~[1@)+10) 1(2)4+1@)) 18+1 241
- 7; |
4 3
Ex 4.4, 15
Ex 4.4, 15 teachoo.com
1 #1 1
For the matrix A= f 2 = show that A? - 6A2 + 5A + 111 =0.
2 -1 3
Hence, find A
Calculating A?
AV=AA
1 #1 1 1 #1 1
= I: 2 3 f 2 3
2 -1 3 2 -1 3
101) +14) +102) 1(1) + 1(2) + 1(-1) 11) + 1(-3) +103)
=]11) + 20) + (-3)(2) 10) + 2(@2)+ ©3)(-1)_ 10) + 2(-3) + (-3))
2) + (-1)G) +32) 20) +©1)@)+3C-1) 20) + (-1)C-3) + 38)
14+1+2 142-1 1-343 4 2 1
-[peans 1+4+3 i-6-s]=[-3 8
2-1+6 2-2-3 24349 7 -3 14
Ex 4.4, 16
Ex 4.4, 16 feachoo.com
2 -1 1
IFA= F 2 =| verify that A? - 6A? + 9A - 41 = O and hence
1 -1 2
find At
Calculating A2
AV=AA
2 -1 1772 -1 #1
= F 2 ] F 2 =|
1 -1 2 1 -1 2
22)+(-DE-D)+10) 26D4+CD@)+16-1) 20) + F-)(-) + 102)
=|-1(2)+ 20-1) + (-1)0) —1(-1) + 22) + (-1)-1),_ 10). + 2(-1) + (2)
1(2)+(-1)-1) +20) 10-1) +(-1)2) + 2-1) 10) + (1) (1) + 202)
44141 —-2-2-1 24142 6 -5 5S
[2-2-3 1+4+1 1-3-9) -|-s 6 “5
241+2 -1-2-2 14+1+4 5 -5 6
Ex 4.4, 17 (MCQ)
Ex 4.4, 17 (Method 1) teachoo.com
Let A be a nonsingular square matrix of order 3 x 3. Then
|adj Al is equal to
A.|A| B. |A|? c. |A/? D. 3 |Al
We know that Nonsingular: Where |A|# 0
ladj A|=|A|"~*
where n is the order of Matrix A
Here, n=3
ladj Al = |A|?~*
= |Al?
Hence, B is the correct answer
Ex 4.4, 18 (MCQ)
Ex 4.4, 18 teachoo.com
If Ais an invertible matrix of order 2, then det(A™) is equal to
1
A. det (A) B. det (a) c.1 D.0
We know that
AAt=I
Taking determinant both sides
JAA |= ||
JALIA*T=1Il = (ABs = JA} [B/)
JA] |A*]=1 ( {i =1)
At} so
Ana a
Since [A] # 0
Ex 4.5
16 questionsEx 4.5, 1
Ex 4.5, 1 teackoo.com
Examine the consistency of the system of equations.
X+2y=2
2x + 3y=3
X+2y=2
2x + 3y =3
Step 1
Write equation AX = B
1 2)/*)_[2
[; 31 [ ~ [3]
t f ft
A xX B
Ex 4.5, 2
teackhoo.com
Ex 4.5, 2
Examine the consistency of the system of equations.
2x-y=5
x+y=4
2x-y=5
x+y=4
Step 1
Write equation as AX = B
2 —1)[*]_75
bl Gl
tf ft fT
A xX B
_f2 -1 _[* 75
Hence A =[4 x=) 8= Ef]
Ex 4.5, 3
Ex 4.5, 3 teachoo.com
Examine the consistency of the system of equations.
X+3y=5
2x + 6y =8
X+3y=5
2x + 6y =8
Writing equation as AX = B
1 3) /*)_]5
L lbl-ll
t ft tf
A xX B
_fi 3 _[* _[5
Hence A = [; al X= [5 &B= [al
Ex 4.5, 4
Ex 4.5, 4 teachoo.com
Examine the consistency of the system of equations.
xtytz=1
2x + 3y+2z=2
ax +ay + 2az=4
Simplifying 3" equation
ax + ay + 2az =4
a(x+y+2z)=4
4
xty+2z= a
Ex 4.5, 5
Ex 4.5,5 teachoo.com
Examine the consistency of the system of equations.
3x-y-2z2=2
2y-z=-1
3x -5y =3
The system of equations can be written as
3x-y-2z=2
Ox+2y-z=-1
3x - 5y +0z=3
Writing equation as AX = B
3-1 -—2);x) [2
b= =p. FI
3 -5 OJjlz 3
Ex 4.5, 6
Ex 4.5, 6 teachoo.com
Examine the consistency of the system of equations.
5x-y+4z=5
2x + 3y+5z=2
5x - 2y+6z=-1
The system of equations is
5x-y+4z=5
2x + 3y+5z=2
5x - 2y +6z=-1
Writing equation as AX = B
5 -1 4);* 5
E 3 sl l-|2 |
5 —2 6\'z —1
Ex 4.5, 7
Ex 4.5, 7 teachoo.com
Solve system of linear equations, using matrix method.
5x+ 2y =4
7x+3y=5
The system of equations is
5x+2y=4
7X+3y=5
Writing equation as AX = B
5 2] [*]_[4
[ 3] b] ~ [s|
ft ft f
A X B
_f5 2 _[* _[4
Hence A = [> sh X= 5 &B= [s|
Ex 4.5, 8
teachoo.com
Ex 4.5, 8
Solve system of linear equations, using matrix method.
2x-y=-2
3x +4y=3
The system of equation is
2x-y=-2
3x + 4y =3
Step 1
Write above equation as AX =B
2 —1] [*]_[-2
Ea lbl-[s]
ft tf ft
A x B
_f2 -1) )_[* _[-2
Hencea=[; 4 beB] &B=| 3 |
Ex 4.5, 9
Ex 4.5, 9 teackoo.com
Solve system of linear equations, using matrix method.
4x -3y=3
3x-—5y =7
The system of equation is
4x-3y=3
3x-—5y =7
Writing equation as AX = B
4 —3)][*]_]3
E -slbI-bI
ft ft ft
A xX B
_f4 —3] y_]* _ [3
Hence A = [5 =) X= [5 &B= [7]
Ex 4.5, 10
Ex 4.5, 10 teackoo.com
Solve system of linear equations, using matrix method.
5x+2y=3
3x+2y=5
The system of equation is
5x + 2y=3
3x+2y=5
Writing above equation as AX = B
5S 2)/*)]_]J3
E albl-[s!
ft ft fT
A xX B
_[5 2) )_]* _ [3
Hence A = [3 ‘Ik: X= [5 &B= [=|
Ex 4.5, 11
Ex 4.5, 11 teachoo.com
Solve system of linear equations, using matrix method.
Qxtytz=1
3
x-2y-z= 3
3y-5z=9
The system of equation is
Qxtytz=1
3
x-2y-z= 3
3y-52=9
Writing above equation as AX = B
2 1 1) I
1-2 -1 b| =15
0 3 —SIlz 9
t f fT
A xX B
Ex 4.5, 12
teachoo.com
Ex 4.5, 12
Solve system of linear equations, using matrix method.
x-y+z=4
2x+y-3z=0
X+y+z2=2
The system of equations is
Xx-y+z=4
2x+y-3z=0
Xty+z=2
Step 1
Write equation as AX = B
Ex 4.5, 13
Ex 4.5, 13 teachoo.com
Solve system of linear equations, using matrix method.
2x+3y+3z=5
x-2y+z=-4
3x- y-2z=3
The system of equations is
2x+3y+3z=5
x-2y+z=-4
3x-y-2z=3
Ex 4.5, 14
Ex 4.5, 14 teackoo.com
Solve system of linear equations, using matrix method.
x-y+2z=7
3x + 4y -5z2=-5
2x-y+3z=12
The system of equations are
X-y+2z=7
3x + 4y -5z=-5
2x-y+3z=12
Writing equation as AX = B
Ex 4.5, 15
Ex4.5, 15 teachoo.com
2 -3 #5
IfA=13. 2 —A4l, find A+. Using A* solve the system of equations
1 1 -2
2x-3y+5z=11
3x + 2y-4z2=-5
X+y —2z=-3
Writing equation as AX = B
2 -3 S yx 11
1 1 -—2)'z —3
Ex 4.5, 16
Ex 4.5, 16 teachoo.com
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 6 kg rice is Rs 90. The cost of 6 kg onion
2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by
matrix method
Let the cost of onion, wheat & rice per kg be x, y, z respectively.
Given,
Cost of 4 kg onion, 3 kg wheat & 2 kg rice is Rs 60.
4x + 3y + 2z = 60
Cost of 2 kg onion, 4 kg wheat & 6 kg rice is Rs 90.
2x + 4y + 6z=90
Cost of 6 kg onion, 2 kg wheat & 3 kg Rice is Rs 70
6x + 2y + 3z=70
Examples
19 questionsExample 1
teachoo.com
Example 1
2 4
Evaluate 4 4
ne
-1 2
=2x2—-(-1)x4
=4+4
=8
Example 2
Example 2 teachoo.com
Evaluate | * ad
x—-1 x
| x aad
x-1 x
=xxx-(x-1)x(x+1)
=x? —(x* — 12)
=x? —y7 +1
=1
Example 3
Example 3 teachoo.com
124
eluate the determinant =| 3 i
4 10
12 4
a 3 y
4 10
3.0 -1 0 -1 3
=) ol -2 14 ol t4 Ty il
= 1(3(0)-1(0)) —2 (-1(0) - 4(0)) + 4 (-1(1) - 4(3)}
= 1(0) — 2 (0) + 4 (-1- 12)
=0+0+ 4(-13)
=-52
Example 4
Example 4 teachoo.com
0 sina —cosa
Evaluate A= |— sing 0 sinB
cosa —sinB 0
0 sing —cosa@
A=|-sina 0 sin B
cosa —sinB 0
_ 0 sinB] | |-sina sing [coon 0 |
=Ol_ sing 0 | sina| oa 0 | (08 a)) cosa —sin B
= 0 (0(0) — (— sin B) sin B) —sin @ (— sin @ {0} — cos @ sin B)— cos @ (-sin @ {-sin B ) — cos @ (0)}
= 0(0 + sin? B) —sin a (O—cosa sin B)—- cosa (sin a sin B—-0)
=0-sin a@ (-cos @ sin B )—cos a (sin @ sin B)
Example 5
Example 5 teachoo.com
; (3 x{_|3 2
Find values of x for which |" " = I; ‘|
Calculating Calculating
3 xI_ _ 3 2) _ _
[P t]= 3-69 al 3-4)
=3-% =3-8=-5
Given
|? =P ‘|
x 1 4 1
Putting values
3-x7=-5
-x2=-5-3
-x*=-8
x°=8
Example 6
Example 6 teachoo.com
Find the area of the triangle whose vertices are
(3, 8), (— 4, 2) and (5, 1).
1 X, y, 1
The area of triangle is given by A = 5pm Ye 1
X3 3 1
Here
x, =3, y, = 8,
= 4, Y. = 2,
%3=5, y3=1
1 3 81
A=>|-4 2 1
5 11
Example 7
Example 7 teachoo.com
Find the equation of the line joining A(1, 3) and B(0, 0) using
determinants and find k if D(k, 0) is a point such that area of
triangle ABD is 3 sq units.
Equation of line
—+—__+——_+— L
(x,y) A(1,3) BO, 0)
Let L be the line joining the A(1, 3) & B(0, 0)
Let (x, y) be the third point on line
Since all the three points lie on the same line,
they do not from a triangle
Example 8
Example 8 teackoo.com
12 3
Find the minor of element 6 in the determinant A = : 5 s
7 8 9
Since 6 lies in the second row and third column,
i.e. a, =6
1 2
Minor of ay, = M53 = = F *|=8-14 = -6
7 8
7 8
Thus, minor of element 6 is -6.
Example 9
Example 9 teachoo.com
Find minors and cofactors of all the elements of the determinant
al
4 3
Minor of a,,=M,, = | 3 =3
Minor of ay. = My = |4 | =4
. —2
Minor of a, =M,, = | = -2
. 1 -
Minor of a,,=M,, = |= 1
Cofactor of a; = Ay = (-1)'7. My
Cofactor of a,,=A,, =(-1)*" M,, =(-1).3 =3
Cofactor of a,, =A,, =(-1)** M,, =(-1)? .4=(-1) (4)= -4
Cofactor of a,, = A), =(—1)?*1. M,, =(— 1)? (—2)= (-1)(-2)=2
Cofactor of a,, = Aj. =(-1)?*?7. Mj, =(-1)*.(1)=(1) (1)=1
Example 10
teachoo.co
Example 10 ma
Find minors and cofactors of the elements a,,, a), in the
1 12 13
determinant A= ap, a22 a3
Ag, 32 33
. Az2 Ag3
Minor of ay, =My,= [41 A223] = | = 92 433 — 423 Aga
A327 33
1 932 33
Cofactor of aij = Ay = (-1)'*/. My
Air Ay2 Ay3 a a
Minor of a,,=M>,=}@21 a2. ag3/=|°'2 “Mla -
aa = Moi = f821 22 23] = = 12 433— 432 A743
a a a A327 33
31 432 433
Cofactor of a,, = Ay, = (- 1). My, = (-1)? . (a9) a33 — a3 a3)
Example 11
Example 11 teachoo.com
Find minors and cofactors of the elements of the determinant
2-3 5
6 0 4 | and verify that a,, A3, + a.) As, + a43 A33 = 0
1 5 -7
First, finding minors and cofactors
Minor of a,, = May | 0 4 1 Al 0- 5(4) = -20
5 -7
. 6 4
Minor of ay, = M,> =|6 4]= I; “| = 6(-7) — 1(4) = - 46
1 —7
Minor of a;3 = Mis -| 0 | -|° ¢ | = 515) -0= 30
1 5 7
Example 12
Example 12 (Method 1) teachoo.com
Find adj A for A= [; 3
Step1: Calculate minors
mol es
oo“
mel 1-3
Example 13
Example 13 teackoo.com
13 3
IfA=|1 4 3], then verify that A adj A= |A| I. Also find A+.
13 4
Solving L.H.S
A (adj A)
First Calculating adj A
Ay. Ag, Agi
adjA=|Ai2 Az. Ago
Aig Ag3 A33
Example 14
Example 14 teachoo.com
[2 3 _f1 -2 F t= ptat
IfA= [; “1 and B= i 3 ] , then verify that (AB) =BtA
Solving L.H.S
(AB) *
First calculating AB
_[2 3 1 -2
ap=[F LS]
-| 2(1) + 3(-1) 2(—2) + 3(3)
“U0) + (-4)-1) 1-2) + (-4)3
=? 73 re le[e |
1+4 -—2-12 5 -14
Example 15
Example 15 teachoo.com
on 2 3B] ccc: ;
Show that the matrix A = [; | satisfies the equation
A?-4A+1=0, where lis 2 x 2 identity matrix and O is 2 x 2 zero
matrix. Using this equation, find A™.
First calculating A?
AZ=A.A
_f2 3772 3
~ [; a1 [; a
_ Pes +31) 2@6)+ | _ (3 +3 64+ Al
1(2)+24) 1(3)+2(2)) l2+2 344
- [7 2
4 $7
Example 16
Example 16 teachoo.com
Solve the system of equations
2x+5y=1
3x+2y=7
Step 1
Write equation as AX = B
2 5)/7*]_71
I albl-bI
t ft ft
A xX B
_f2 5 _[* _fi
asl; 2) x= []-8=[
Step 2
_ {2 5
Calculate [A], A= [5 |
Example 17
Example 17 teachoo.com
Solve the following system of equations by matrix method.
3x—2y+3z=8
2x+y-z=1
4x—3y+2z=4
The system of equation is
3x—2y+3z=8
2x+y-z=1
4x —3y+2z=4
Writing equation as AX=B
3 -2 3)/*) [8
2 1 —-1);/y/=]1
4 -3 2/12 4
t f tf
A xX R
Example 18
Example 18 teachoo.com
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 the first, second & third number be x, y, z respectively
Given,
x+y+z=6
y+3z=11
X+z=2y orx—2y+z=0
Step 1
Write equation as AX = B
Example 19
Example 19 teachoo.com
1 -1 2)]/-2 0 1
Use product 0 2 “ | 9 2 3 to solve the system of
3-2 4 6 1 -2
equations
x-y+2z=1
2y-3z=1
3x-2y+4z2=2
Consider the product
1-1 2)]/7-2 0 1
0 2 | | 9 2 -
3-2 4 6 1 —-2
1(—2) + (-1)(9) + 2(6) 10) + (-1)(2) + 20): 14) + (-1)(-3) + 2(-2)
=/0(—2) + 2(9) + (—3)(6) 0(0) + 2(2) + (-3)(1) 04) + 0(—3) + (—3)(-2)
3(-2) + (-2)) + (6) 3(0) + (—2)(2) + 41) 31) + (—2)(-3) + 4(-2)
Miscellaneous
9 questionsMisc 1
Misc 1 teachoo.com
x sin@ cosé
Prove that determinant |—sin@ —x 1. | is independent of 6.
cos6@ 1 x
x sin@ cosé
LetA=|—sin@ —-x 1
cos 6 1 x
—x 1 . —sind 1 —sinO —-x
A= - +
xy 1 sin | oso A 058 | oo |
=x (-x?-1)-sin 0 (-xsin 8 — cos 8) + cos 0 (-sin@ +x cos 6)
=-x?-x+xsin?0+ sin @ cos 8 —sin 6 cos 6 + x cos?6
= —-?-x+x sin? 6+xcos?6
= -x?-x +x (sin? 6 + cos? 6)
= -x? -x+x(1) (As sin? 8 + cos? 8 = 1)
= -?
Misc 2
Misc 2 (Method 1) teachoo.com
cosacosB cosasinB —sina
Evaluate | —sinB cosB 0
sinacosB sinasingB C8
cosacosB cosasinB —sina
—sinB cosB 0
sinacosB sinasinB CS&
Expanding Determinant along C,
_ cos B 0 . —sing 0 . sing cos 8
=cos a cos B lsinasing cos al-o a sin B loinc cosB cos alos" o |sinacos 6 sina ‘inal
=cosa cosB (cos B cosa —0) —cosa sin B(-sin B cosa —0)—sina (-sin?B sina —cos*B sina)
= cosa cosB(cosBcosa) —cosa sinB(-sinB cosa) +sin?a sin? B + cos? B sin?a
= cos’ a cos*B + cos*a sin? B + sin? a ( sin? B + cos” B)
= cos? a (1) + sin? a (1) (Using sin? x + cos? x = 1}
=cos? a+ sin’a (Using sin? x + cos? x = 1)
Misc 3
Misc 3 teachoo.com
3 -1 1 1 2 -2
AT=|—15 6 -s and B = F 3 0 | , Find (AB)*
5 -2 2 0 -2 1
We know that
(AB)? =B7A7
We are given A“, so calculating B?
Calculating B*
We know that
Bt =— adj
Bi adj (B)
exists if |B] #0
Misc 4
teachoo.com
Misc 4 (i)
1 -2 1
Let A= 2 3 ] verify that
1 1 #5
(i) [adj A]* = adj (A*)
First we will calculate adj (A) & A?
t
Aw Aq Aig Aw Aa Agi
adjA=|Az, Azz Ag3)= |Ar2 Azz Azz
Az, Ag. Ag3 Aig Ag3 A33
1 -2 1
A= F 3 ]
1 1 #5
Misc 5
Misc 5 (Method 1) teachoo.com
x y x+y
Evaluate | y x+y x |
x+y x y
x y x+y
Let A= | y x+y x |
x+y x y
We will solve it without using
Properties in this Method
=a[@+yy—x"]-yly? — xe +y)]1+@+y)y-O+y)7I
=x[xy + y? — x7] — yly? —x? — xy] + + y) ey — x? —y? — 2xy]
=z[xy + y? - 2°] —y[y? — 2? - xy] + @ + I? - 9? - 29]
Misc 6
Misc 6 teachoo.com
1 x y
Evaluate|1 x+y y
1 x xty
1 x y
1 x+y y
1 x xty
Applying R,> R,—-R>
1-1 x-x-y y
=| 1 x+y y
1 x x+y
0 -y y
=|1 xty y
41 ye you
Misc 7
Misc 7 teackhoo
Solve the system of the following equations
2 + 3 + 10 = 4
X y Z
4 - 6 + 5 = 1
x y Z
6,9 _20_,
x y Zz
The system of equations are
2,34i0-4
X Yy Zz
4 6,5 1
X y Zz
6,9 _ 20_ 2
Xx y Zz
Now let==u,>=v,&—=w
x y z
Misc 8 (MCQ)
Misc 8 teachoo.com
Choose the correct answer.
If x, y, Z are nonzero real numbers, then the inverse of matrix
x 0 0
a-[o y | is
0 0 2
xt 0 0 xt 0 0
A}0O y? 0 B.xyz} 0 yt O
0 0 zt 0 0 zt
1 0 0 1 100
Cc. =a y | o.<./o 1 |
XYZ XYZ
0 0 z 00 1
x 0 0
cuena=[o y |
0 0 2
Misc 9 (MCQ)
Misc 9 teachoo.com
Choose the correct answer.
1 sin 1
Let A=|—sin@ 1 sin 6| , where 0 < @< 2n, then
-1 —sin8 1
A. Det (A) =0
B. Det (A) € (2, 9)
C. Det (A) € (2, 4)
D. Det (A)é [2, 4]
1 sin® 1
A=]-sin0 1 sin®
-1 —sin® 1
1 sin® 1
|A| =]—sin® 1 sin @
-1 —sin 9 1
Why Learn This With Teachoo?
Determinants associates a scalar value with a square matrix and uses it to test invertibility, solve equations and calculate areas. Students evaluate determinants, use properties, find minors and cofactors, construct adjoints and inverses and solve linear systems. Teachoo offers NCERT exercise solutions, examples, miscellaneous problems and concept-wise explanations for determinant calculations and applications.
Evaluation and properties
For a 2×2 matrix [[a,b],[c,d]], the determinant is ad−bc. A 3×3 determinant can be expanded along any row or column using cofactors. Choosing a row or column containing zeros often reduces work.
Determinant properties allow strategic simplification. Interchanging two rows changes the sign. Multiplying a row by k multiplies the determinant by k. Adding a multiple of one row to another leaves the determinant unchanged. If two rows are equal or proportional, the determinant is zero. Common factors should be extracted explicitly so their effect on the determinant is visible.
Minors, cofactors, adjoint and inverse
The minor Mᵢⱼ is obtained by deleting row i and column j. The cofactor is Aᵢⱼ=(−1)ⁱ⁺ʲMᵢⱼ. The adjoint is the transpose of the cofactor matrix. For a square matrix A with |A|≠0,
A⁻¹=adj A/|A|.
If |A|=0, A is singular and has no inverse. The identity A(adj A)=|A|I provides a useful verification.
Applications
The area of a triangle with coordinate vertices can be expressed using a determinant. The absolute value is required, and zero area indicates collinearity. Systems of linear equations may be solved by inverse matrices or Cramer’s Rule when the relevant determinant is non-zero. If it is zero, consistency requires further analysis rather than division.
Topics and resources on Teachoo
-
NCERT exercises, examples and miscellaneous solutions;
-
determinants of order two and three;
-
simplification using properties;
-
minors, cofactors and cofactor expansion;
-
adjoint and inverse of a matrix;
-
area of a triangle and collinearity;
-
solving linear equations;
-
consistency and parameter-based questions;
-
board and higher-order practice where available.
Learning outcomes
Students should be able to evaluate determinants efficiently, apply properties without altering value incorrectly and calculate minors and cofactors. They should find inverses, solve systems and use determinant methods in coordinate geometry.
Board and entrance-exam preparation
Look for structure before expanding: zeros, repeated patterns and simple row or column transformations can shorten a calculation. Record every operation and its effect. In equation systems, first identify the coefficient matrix and determinant, then choose inverse or Cramer’s method.
Common mistakes to avoid
Do not use determinant notation for a non-square matrix. Cofactor signs alternate in a checkerboard pattern beginning with plus. Interchanging rows changes the sign; adding a multiple of another row does not. Area is non-negative, so take the absolute value. Never apply the inverse formula when |A|=0.
Deeper reasoning and concept connections
A student has understood Determinants only when the idea can be moved between words, diagrams, examples and mathematical notation. Start with a concrete example, identify what changes and what remains fixed, represent the relationship clearly and then state the rule. This movement between representations is important because school and competency questions often present a familiar idea in an unfamiliar form.
The chapter should also be connected to earlier and later mathematics. Definitions supply the language, worked examples reveal the method, and mixed questions test whether the method can be selected without a hint. Instead of memorising the appearance of a solved question, ask what information triggered the method, which condition made it valid and how the answer could be checked. That makes learning transferable to later chapters rather than limited to one exercise.
How to solve unfamiliar and competency-based questions
Read the complete problem before calculating. Underline the quantities, conditions and command word—find, compare, construct, justify, estimate or prove. Rephrase the task in one sentence and choose a representation such as a table, labelled figure, number line, expression or graph. Solve in small steps, keeping units and labels visible.
For an application question, the final line must answer the situation, not only display a number. For an assertion–reason question, test the assertion and reason separately before deciding whether one explains the other. For an MCQ, eliminate options using definitions, signs, size estimates or boundary cases before performing long calculations. If the answer is visual, check it against the stated scale or construction conditions rather than the appearance of the drawing.
What complete mastery looks like
For Determinants, a student should be able to define the central ideas in simple language, recognise them in different representations, solve routine questions accurately and explain the method used. They should also be able to correct a flawed solution, create an example satisfying given conditions and combine two ideas from the chapter in one problem. A reliable mastery test is to solve one direct question, one application question and one reasoning question without looking at notes, then explain all three aloud or in writing.
Keep a compact error log with four labels: concept, interpretation, calculation and presentation. Reattempt each error after a gap instead of rereading the answer immediately. Improvement comes from correcting the decision that caused the mistake, not from repeating questions whose method is already known.
Additional frequently asked questions
What should a student know before starting Determinants?
Revise the definitions, number operations, diagrams or notation used at the beginning of the chapter. The prerequisite list should be short: if an earlier skill blocks progress, repair that skill with two or three focused questions and return to the chapter.
How can a student check an answer in Determinants?
Use an independent check whenever possible: substitute the result, reverse the operation, estimate its size, compare it with the figure, test a simpler case or solve using another representation. A check should examine the mathematical condition, not merely repeat the same arithmetic.
How many questions are enough for strong preparation?
There is no fixed number. Stop counting questions and track coverage: every concept, every standard method, at least one mixed problem, one competency-based problem and every previously incorrect type should be solved independently. Ten varied, analysed questions are more valuable than fifty copied solutions.
How should Teachoo solutions be used without becoming dependent on them?
Attempt the question first and mark the exact step where progress stops. Read only enough of the solution to repair that step, close it and restart the question. Finally, solve a similar problem without help. This turns a solution into feedback rather than a substitute for thinking.
Frequently asked questions
What does a zero determinant mean?
It means the square matrix is singular and has no multiplicative inverse.
Can a determinant be expanded along any row or column?
Yes. A correct cofactor expansion gives the same value; choose the row or column that minimises work.
How are determinants used in coordinate geometry?
They calculate the signed area of a triangle; an absolute value gives area and zero indicates collinear points.