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# Solution Determinants Test- 4 Class 12 Notes | EduRev

Created by: Arshdeep Kaur

## Class 12 : Solution Determinants Test- 4 Class 12 Notes | EduRev

``` Page 1

CBSE TEST PAPER-09
CLASS - XII MATHEMATICS (Algebra)
Topic:- Determinants
1. x
2
â€“ 36 = 36 - 36
x
2
= 36
6 x = ±
2.
1  1
4  8
A =
= 8 â€“ 8
= 0
Hence A is singular
3.
2 3 2 3
2 3 2 3
2 3 2 3
a a abc 1      a a
1
b b abc 1      b b
c c abc 1      c c
abc
=
2 3 2 3
2 3 2 3
2 3 2 3
a a 1 1      a a
b b 1 1      b b
c c 1 1      c c
abc
abc
= =
2 3 2 3
1 3 2 3 2 3
2 3
2 3 2 3
1      a a 1      a a
1      b b 1      b b
1      c c 1      c c
C C
C C
? ? ?
=
? ?
?
? ?

Hence Prove
4. 2(0 20) 3( 42 4) 5(30 0) A = - + - - + -
28 = -
2  6  1
' -3    0     5
5     4    -7
' 28
A
A
=
= - Hence prove.
Page 2

CBSE TEST PAPER-09
CLASS - XII MATHEMATICS (Algebra)
Topic:- Determinants
1. x
2
â€“ 36 = 36 - 36
x
2
= 36
6 x = ±
2.
1  1
4  8
A =
= 8 â€“ 8
= 0
Hence A is singular
3.
2 3 2 3
2 3 2 3
2 3 2 3
a a abc 1      a a
1
b b abc 1      b b
c c abc 1      c c
abc
=
2 3 2 3
2 3 2 3
2 3 2 3
a a 1 1      a a
b b 1 1      b b
c c 1 1      c c
abc
abc
= =
2 3 2 3
1 3 2 3 2 3
2 3
2 3 2 3
1      a a 1      a a
1      b b 1      b b
1      c c 1      c c
C C
C C
? ? ?
=
? ?
?
? ?

Hence Prove
4. 2(0 20) 3( 42 4) 5(30 0) A = - + - - + -
28 = -
2  6  1
' -3    0     5
5     4    -7
' 28
A
A
=
= - Hence prove.
5. Taking a, b, c common from R
1
, R
2
and R
3
1 1 1
1
1 1 1
1
1 1 1
1
a a a
abc
b b b
c c c
+
= +
+
1 1 2 3
R R R R ? + +
1 1 1 1 1 1 1 1 1
1 + +  1 + +  1 + +
1 1 1
1
1 1 1
1
a b c a b c a b c
abc
b b b
c c c
+ + +
= +
+
1  1  1
1 1 1 1 1 1
1 + + 1
1 1 1
1
abc
a b c b b b
c c c
? ?
= + +
? ?
? ?
+
1 1 3 2 2 3
,  C C C C C C ? - ? - 0  0  1
1 1 1 1
1 + + 0  1
1
1    1 1
abc
a b c b
c
? ?
= +
? ?
? ?
- - +
Expending along R
1

[ ]
1 1 1
1 + + 1 abc
a b c
abc bc ac bc
? ?
= +
? ?
? ?
= + + +
6.
2 3
2 3
2 3
x      x 1+x
y      y 1+y
z      z 1+z
? =
Page 3

CBSE TEST PAPER-09
CLASS - XII MATHEMATICS (Algebra)
Topic:- Determinants
1. x
2
â€“ 36 = 36 - 36
x
2
= 36
6 x = ±
2.
1  1
4  8
A =
= 8 â€“ 8
= 0
Hence A is singular
3.
2 3 2 3
2 3 2 3
2 3 2 3
a a abc 1      a a
1
b b abc 1      b b
c c abc 1      c c
abc
=
2 3 2 3
2 3 2 3
2 3 2 3
a a 1 1      a a
b b 1 1      b b
c c 1 1      c c
abc
abc
= =
2 3 2 3
1 3 2 3 2 3
2 3
2 3 2 3
1      a a 1      a a
1      b b 1      b b
1      c c 1      c c
C C
C C
? ? ?
=
? ?
?
? ?

Hence Prove
4. 2(0 20) 3( 42 4) 5(30 0) A = - + - - + -
28 = -
2  6  1
' -3    0     5
5     4    -7
' 28
A
A
=
= - Hence prove.
5. Taking a, b, c common from R
1
, R
2
and R
3
1 1 1
1
1 1 1
1
1 1 1
1
a a a
abc
b b b
c c c
+
= +
+
1 1 2 3
R R R R ? + +
1 1 1 1 1 1 1 1 1
1 + +  1 + +  1 + +
1 1 1
1
1 1 1
1
a b c a b c a b c
abc
b b b
c c c
+ + +
= +
+
1  1  1
1 1 1 1 1 1
1 + + 1
1 1 1
1
abc
a b c b b b
c c c
? ?
= + +
? ?
? ?
+
1 1 3 2 2 3
,  C C C C C C ? - ? - 0  0  1
1 1 1 1
1 + + 0  1
1
1    1 1
abc
a b c b
c
? ?
= +
? ?
? ?
- - +
Expending along R
1

[ ]
1 1 1
1 + + 1 abc
a b c
abc bc ac bc
? ?
= +
? ?
? ?
= + + +
6.
2 3
2 3
2 3
x      x 1+x
y      y 1+y
z      z 1+z
? =
2 2 3
2 2 3
2 2 3
x      x 1 x      x x
y      y 1 y      y y
z      z 1 z      z z
? = + ? =
2 2
2 2
2 2
1      x      x 1      x      x
1     y      y 1     y      y
1     z      z 1     z      z
xyz = +
( )
2
2
2
1      x      x
1 1     y      y
1     z      z
xyz = +
( )
2
2 2 1 2 2
3 3 1
2 2
1      x  x
1 0     y-x      y
0     z-x      z
R R R
xyz x
R R R
x
? - ? ?
= + - ? ?
? - ? ?
-
( )( )( )
2
1  x  x
1 0  1  y
0  1  z
xyz y x z x x
x
= + - - +
+
( )( )( )( ) 1
0( )
xyz y x z x z y
given
= + - - - ? =
x, y, z all are different
0,  0,  0
1 0
x y y z z x
xyz
- ? - ? - ?
? + =
7. Let P (x, y) be any point on AB. Then area of ? ABP is zero
0  0  1
1
1  3  1 0
2
x     y     1
=
3 y x =
Area ? ABD =3 square unit
1  3  1
1
0  0  1 3
2
K  0  1
2 k
= ±
= ±
8.
2
2  3 2  3
.
1  2 1  2
A
? ? ? ?
=
? ? ? ?
? ? ? ?

Page 4

CBSE TEST PAPER-09
CLASS - XII MATHEMATICS (Algebra)
Topic:- Determinants
1. x
2
â€“ 36 = 36 - 36
x
2
= 36
6 x = ±
2.
1  1
4  8
A =
= 8 â€“ 8
= 0
Hence A is singular
3.
2 3 2 3
2 3 2 3
2 3 2 3
a a abc 1      a a
1
b b abc 1      b b
c c abc 1      c c
abc
=
2 3 2 3
2 3 2 3
2 3 2 3
a a 1 1      a a
b b 1 1      b b
c c 1 1      c c
abc
abc
= =
2 3 2 3
1 3 2 3 2 3
2 3
2 3 2 3
1      a a 1      a a
1      b b 1      b b
1      c c 1      c c
C C
C C
? ? ?
=
? ?
?
? ?

Hence Prove
4. 2(0 20) 3( 42 4) 5(30 0) A = - + - - + -
28 = -
2  6  1
' -3    0     5
5     4    -7
' 28
A
A
=
= - Hence prove.
5. Taking a, b, c common from R
1
, R
2
and R
3
1 1 1
1
1 1 1
1
1 1 1
1
a a a
abc
b b b
c c c
+
= +
+
1 1 2 3
R R R R ? + +
1 1 1 1 1 1 1 1 1
1 + +  1 + +  1 + +
1 1 1
1
1 1 1
1
a b c a b c a b c
abc
b b b
c c c
+ + +
= +
+
1  1  1
1 1 1 1 1 1
1 + + 1
1 1 1
1
abc
a b c b b b
c c c
? ?
= + +
? ?
? ?
+
1 1 3 2 2 3
,  C C C C C C ? - ? - 0  0  1
1 1 1 1
1 + + 0  1
1
1    1 1
abc
a b c b
c
? ?
= +
? ?
? ?
- - +
Expending along R
1

[ ]
1 1 1
1 + + 1 abc
a b c
abc bc ac bc
? ?
= +
? ?
? ?
= + + +
6.
2 3
2 3
2 3
x      x 1+x
y      y 1+y
z      z 1+z
? =
2 2 3
2 2 3
2 2 3
x      x 1 x      x x
y      y 1 y      y y
z      z 1 z      z z
? = + ? =
2 2
2 2
2 2
1      x      x 1      x      x
1     y      y 1     y      y
1     z      z 1     z      z
xyz = +
( )
2
2
2
1      x      x
1 1     y      y
1     z      z
xyz = +
( )
2
2 2 1 2 2
3 3 1
2 2
1      x  x
1 0     y-x      y
0     z-x      z
R R R
xyz x
R R R
x
? - ? ?
= + - ? ?
? - ? ?
-
( )( )( )
2
1  x  x
1 0  1  y
0  1  z
xyz y x z x x
x
= + - - +
+
( )( )( )( ) 1
0( )
xyz y x z x z y
given
= + - - - ? =
x, y, z all are different
0,  0,  0
1 0
x y y z z x
xyz
- ? - ? - ?
? + =
7. Let P (x, y) be any point on AB. Then area of ? ABP is zero
0  0  1
1
1  3  1 0
2
x     y     1
=
3 y x =
Area ? ABD =3 square unit
1  3  1
1
0  0  1 3
2
K  0  1
2 k
= ±
= ±
8.
2
2  3 2  3
.
1  2 1  2
A
? ? ? ?
=
? ? ? ?
? ? ? ?

7  12
4
1  7
? ?
=
? ?
? ?
2
7  12 8  12 1  0
4
1  7 4  8 0  1
A A I
? ? ? ? ? ?
- + = - +
? ? ? ? ? ?
? ? ? ? ? ?
0  0
0  0
0
? ?
=
? ?
? ?
=
2
2
4 0
4
A A I
A A I
- + =
- = - 1 1 1
1 1
4
4
AAA AA IA
AI I IA AA I
- - - - - - = - ? ? - = - =
? ?
?
1
4
2     -3
-1     2
A I A
- = - ? ?
=
? ?
? ?
9. The system of equation be written in the form AX = B, whose
3     -2     3 8
2      1    -1 1
4     -3     2 4
x
A X y B
z
? ? ? ? ? ?
? ? ? ? ? ?
= = =
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
17 0 A = - ?
1
1     -5     -1
1
-8      -6      9
17
-10     1      7
A
- - ? ?
? ?
=
? ?
? ?
? ?
1
X A B
- =
1     -5     -1 8
1
-8      -6      9 1
17
-10     1      7 4
- ? ?? ?
? ?? ?
=
? ?? ?
- ? ?? ?
? ?? ?
1
2
3
x
y
z
? ? ? ?
? ? ? ?
=
? ? ? ?
? ? ? ?
? ? ? ?
1,  2,  3 x y z = = =
10. I â€“ x II â€“ y  II â€“ y
x + y + z = 6
y + 3z + 11
Page 5

CBSE TEST PAPER-09
CLASS - XII MATHEMATICS (Algebra)
Topic:- Determinants
1. x
2
â€“ 36 = 36 - 36
x
2
= 36
6 x = ±
2.
1  1
4  8
A =
= 8 â€“ 8
= 0
Hence A is singular
3.
2 3 2 3
2 3 2 3
2 3 2 3
a a abc 1      a a
1
b b abc 1      b b
c c abc 1      c c
abc
=
2 3 2 3
2 3 2 3
2 3 2 3
a a 1 1      a a
b b 1 1      b b
c c 1 1      c c
abc
abc
= =
2 3 2 3
1 3 2 3 2 3
2 3
2 3 2 3
1      a a 1      a a
1      b b 1      b b
1      c c 1      c c
C C
C C
? ? ?
=
? ?
?
? ?

Hence Prove
4. 2(0 20) 3( 42 4) 5(30 0) A = - + - - + -
28 = -
2  6  1
' -3    0     5
5     4    -7
' 28
A
A
=
= - Hence prove.
5. Taking a, b, c common from R
1
, R
2
and R
3
1 1 1
1
1 1 1
1
1 1 1
1
a a a
abc
b b b
c c c
+
= +
+
1 1 2 3
R R R R ? + +
1 1 1 1 1 1 1 1 1
1 + +  1 + +  1 + +
1 1 1
1
1 1 1
1
a b c a b c a b c
abc
b b b
c c c
+ + +
= +
+
1  1  1
1 1 1 1 1 1
1 + + 1
1 1 1
1
abc
a b c b b b
c c c
? ?
= + +
? ?
? ?
+
1 1 3 2 2 3
,  C C C C C C ? - ? - 0  0  1
1 1 1 1
1 + + 0  1
1
1    1 1
abc
a b c b
c
? ?
= +
? ?
? ?
- - +
Expending along R
1

[ ]
1 1 1
1 + + 1 abc
a b c
abc bc ac bc
? ?
= +
? ?
? ?
= + + +
6.
2 3
2 3
2 3
x      x 1+x
y      y 1+y
z      z 1+z
? =
2 2 3
2 2 3
2 2 3
x      x 1 x      x x
y      y 1 y      y y
z      z 1 z      z z
? = + ? =
2 2
2 2
2 2
1      x      x 1      x      x
1     y      y 1     y      y
1     z      z 1     z      z
xyz = +
( )
2
2
2
1      x      x
1 1     y      y
1     z      z
xyz = +
( )
2
2 2 1 2 2
3 3 1
2 2
1      x  x
1 0     y-x      y
0     z-x      z
R R R
xyz x
R R R
x
? - ? ?
= + - ? ?
? - ? ?
-
( )( )( )
2
1  x  x
1 0  1  y
0  1  z
xyz y x z x x
x
= + - - +
+
( )( )( )( ) 1
0( )
xyz y x z x z y
given
= + - - - ? =
x, y, z all are different
0,  0,  0
1 0
x y y z z x
xyz
- ? - ? - ?
? + =
7. Let P (x, y) be any point on AB. Then area of ? ABP is zero
0  0  1
1
1  3  1 0
2
x     y     1
=
3 y x =
Area ? ABD =3 square unit
1  3  1
1
0  0  1 3
2
K  0  1
2 k
= ±
= ±
8.
2
2  3 2  3
.
1  2 1  2
A
? ? ? ?
=
? ? ? ?
? ? ? ?

7  12
4
1  7
? ?
=
? ?
? ?
2
7  12 8  12 1  0
4
1  7 4  8 0  1
A A I
? ? ? ? ? ?
- + = - +
? ? ? ? ? ?
? ? ? ? ? ?
0  0
0  0
0
? ?
=
? ?
? ?
=
2
2
4 0
4
A A I
A A I
- + =
- = - 1 1 1
1 1
4
4
AAA AA IA
AI I IA AA I
- - - - - - = - ? ? - = - =
? ?
?
1
4
2     -3
-1     2
A I A
- = - ? ?
=
? ?
? ?
9. The system of equation be written in the form AX = B, whose
3     -2     3 8
2      1    -1 1
4     -3     2 4
x
A X y B
z
? ? ? ? ? ?
? ? ? ? ? ?
= = =
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
17 0 A = - ?
1
1     -5     -1
1
-8      -6      9
17
-10     1      7
A
- - ? ?
? ?
=
? ?
? ?
? ?
1
X A B
- =
1     -5     -1 8
1
-8      -6      9 1
17
-10     1      7 4
- ? ?? ?
? ?? ?
=
? ?? ?
- ? ?? ?
? ?? ?
1
2
3
x
y
z
? ? ? ?
? ? ? ?
=
? ? ? ?
? ? ? ?
? ? ? ?
1,  2,  3 x y z = = =
10. I â€“ x II â€“ y  II â€“ y
x + y + z = 6
y + 3z + 11
x + z + 2y
This system can be written as AX = B whose
1  1  1 6
0  1  3 11
1    -2    1 0
x
A X y B
z
? ? ? ? ? ?
? ? ? ? ? ?
= = =
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
9 0 A = ?
11 12 13
21 22 23
31 32 33
7,  3,  1
3,  0,  3
2,  3,  1
A A A
A A A
A A A
= = = - = - = =
= = - =
7     -3     2
3      0    -3
-1    3      1
aiJ A
? ?
? ?
=
? ?
? ?
? ?
1
7     -3     2
1 1
3      0    -3
9
-1    3      1
A aiJ A
A
- ? ?
? ?
= =
? ?
? ?
? ?
1
X A B
- =
7     -3     2 6
1
3      0    -3 11
9
-1    3      1 0
x
y
z
? ? ? ?? ?
? ? ? ?? ?
=
? ? ? ?? ?
? ? ? ?? ?
? ? ? ?? ?
1
2
3
x
y
z
? ? ? ?
? ? ? ?
=
? ? ? ?
? ? ? ?
? ? ? ?
1,  2,  3 x y z = = =
```
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