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Properties of DeterminantDeterminant: To each square matrix A we can associate a expression or number (real or complex) known as its determinant denoted by det A or ∣A∣If A=[ a ] then ∣A∣=a .Determinant of a square matrix of order 2:
If A=[a bc d ] then ∣A∣=ad – bc
For example ∣2 3– 3 4∣=2×4– 3×–3=89=17
Determinant of a square matrix matrix of order 3:
Let A=[a b cx y zu v w ] then we can find its determinant in six possible ways
Expansion Along First Row:
∣A∣= a∣y zv w∣– b∣x z
u w∣ c∣x yu v∣
Expansion Along Second Row:
∣A∣=−x∣b cv w∣ y∣a c
u w∣− z∣a bu v∣
Expansion Along Third Row:
∣A∣=u∣b cy z∣−v∣a c
x z∣w∣a bx y∣
Expansion along first column:
∣A∣=a∣y zv w∣– x∣b c
v w∣u∣b cy z∣
Expansion along second column:
∣A∣=−b∣x zu w∣ y∣a c
u w∣– v∣a cx z∣
Expansion along third column:
∣A∣=c∣x yu v∣– z∣a b
u v∣w∣a bx y∣
Note: We expand the determinant along a row or column which has maximum number of zeros.
Example-1: Evaluate the determinant ∣2 −3 13 −2 41 2 5∣ by expanding it along first row.
Solution: ∣2 −3 13 −2 41 2 5∣=2∣−2 4
2 5∣−−3∣3 41 5∣1∣3 −2
1 2 ∣⇒∣2 −3 1
3 −2 41 2 5∣=2−10−83 15−462=−36338=5
2
Example-2: Find the value of the determinant ∣ 2 3 −1−1 2 −42 −1 3 ∣ by expanding it along the second row.
Solution:
∣2 3 −1−1 2 −42 −1 3 ∣=−−1∣ 3 −1
−1 3 ∣2∣2 −12 3 ∣−−4∣2 3
2 −1∣=9−12 624 −2−6
⇒∣2 3 −1−1 2 −42 −1 3 ∣=816−32=−8
Example-3: Find the value of the determinant ∣3 2 −1−2 1 14 −2 3 ∣ by expanding along it third row.
Solution: ∣ 3 2 −1−2 1 14 −2 3 ∣=4∣2 −1
1 1 ∣−−2∣ 3 −1−2 1 ∣3∣ 3 2
−2 1∣⇒∣ 3 2 −1
−2 1 14 −2 3 ∣=42123−23 34=12221=35
Example-4: Find the value of the determinant ∣2 1 13 −2 31 2 1∣ by expanding along first column.
Solution: ∣2 1 13 −2 31 2 1∣=2∣−2 3
2 1∣−3∣1 12 1∣1∣ 1 1
−2 3∣
∣2 1 13 −2 31 2 1∣=2−2−6−31−232=−1635=−8
Example-5: Find the value of the determinant ∣2 −1 34 2 −15 2 1 ∣ by expanding it along second column.
Solution: ∣2 −1 34 2 −15 2 1 ∣=−−1∣4 −1
5 1 ∣2∣2 35 1∣−2∣2 3
4 −1∣⇒∣2 −1 3
4 2 −15 2 1 ∣=4522−15−2 −2−12=9−2628=11
Example-6: Find the value of the determinant ∣3 −1 −12 2 −12 3 −1∣ by expanding it along the third column.
Solution: ∣3 −1 −12 2 −12 3 −1∣=−1∣2 2
2 3∣−−1∣3 −12 3 ∣−1∣3 −1
2 2 ∣
3
∣3 −1 −12 2 −12 3 −1∣=−6−492−62=−211−8=1
Exercise-4.1
1. Find the value of the determinant ∣3 1 22 −1 34 1 2∣ by expanding it along the first row. Also find
its value by expanding it along first column.
2. Find the value of the determinant ∣ 3 −1 1−2 1 14 2 1∣ by expanding it along the second row. Also
find its value by expanding it along second column.
3. Find the value of the determinant ∣ 3 4 0−1 2 32 4 5∣ by expanding it along the third row. Also find
its value by expanding it along third column.
4. Find the value of the determinant ∣ a 0 11 2 1−1 1 2∣
5. Find the value of the determinant ∣ 2 0 1−1 0 21 2 1∣
6. Find the value of the determinant ∣cos −sinsin cos ∣
7. Find the values of x for which ∣3 xx 1∣=∣3 2
4 1∣8. Find the value of thew determinant ∣x2−x1 x−1
x1 x1∣9. Find the value of the determinant ∣ 0 sin −cos
−sin 0 sincos −sin 0 ∣
Properties of Determinant
Property-1: Let A=[ aij ] be a square matrix of order n then ∣A∣=∣AT∣
For example ∣a b cx y zu v w∣=∣
a x ub y vc z w∣
Property-2: Let A=[ aij ] be a square matrix of order n n≥2 and B be matrix obtained from A by interchanging any two rows(columns) of A , then ∣A∣=–∣B∣
4
For example ∣a b cx y zu v w∣= – ∣x y z
a b cu v w∣ or ∣a b c
x y zu v w∣= – ∣b a c
y x zv u w∣
Property-3: If any two rows(columns) of a square matrix A=[ aij ] of order n are identical, then its determinant is zero i.e. ∣A∣=0
For example ∣a b cx y za b c∣= 0 or ∣a b a
x y xu v u∣= 0
Property-4: Let A=[ aij ] be a square matrix of order n , and let B be the matrix obtained from A by multiplying each element of a row(column) of A by a scalar then ∣B∣=k∣A∣
For example ∣ka kb kcx y zu v w ∣= k∣a b c
x y zu v w∣ or ∣ka b c
kx y zku v w∣= k∣a b c
x y zu v w∣
Property-5: Let A be a square matrix such that each element of a row(column) of A is expressed as the sum of two or more terms. Then, the determinant of A can be expressed as the sum of determinants of two or more matrices of of the same order.
For example ∣a a ' b b ' c c 'x y zu v w ∣=∣a b c
x y zu v w∣∣
a ' b ' c 'x y zu v w ∣
or ∣ a b cx x ' y zu u ' v w∣=∣
a b cx y zu v w∣∣
a ' b cx ' y zu ' v w∣
Property-6: Let A be any square matrix and B be a matrix obtained from A by adding to a row(column) of A a scalar multiple of another row(column) of A , then ∣B∣=∣A∣
For example ∣a b cx y zu v w∣=∣
a ku b kv c kwx y zu v w ∣ or
∣a b cx y zu v w∣=∣
a kc b cx kz y zu kw v w∣
Example1. Prove that ∣ 1 1 1a b c
bc ca ab∣=0
5
Solution: Let =∣ 1 1 1a b c
bc ca ab∣Applying R2R2R3 we get
=∣ 1 1 1abc abc abcbc ca ab ∣
Taking abc common from R2 we get
=abc ∣ 1 1 11 1 1
bc ca ab∣
⇒=0 .̈ R1and R2are identical
Example-2: Prove that ∣a−b b−c c−ab−c c−a a−bc−a a−b b−c∣=0
Solution: Let =∣a−b b−c c−ab−c c−a a−bc−a a−b b−c∣
Applying R1R1C 2R3
=∣ 0 0 0b−c c−a a−bc−a a−b b−c∣
⇒ =0 ( .̈ all elements in R1 are zero)
Example-3 Prove that ∣ 1 1 1ab bc ca
c ab a bc b ca ∣=0
Solution: Let =∣ 1 1 1ab bc ca
c ab abc bca∣Applying R2R2R3 we get
=∣ 1 1 1abbcca abbcca abbccac ab a bc bca ∣
⇒ =abbcca∣ 1 1 11 1 1
c ab a bc bca∣⇒=0 (As R1 and R2 are identical)
6
Example-4: Prove that ∣1 1 1a b ca2 b2 c2∣=a−bb−c c−a
Solution: Let =∣1 1 1a b ca2 b2 c2∣
Applying C1C1– C2 and C2C2−C3 we get
=∣ 0 0 1a−b b−c ca2−b2 b2−c2 c2∣
Taking a – b and b – c common from C1 and C2 respectively we get
=a−bb−c ∣ 0 0 11 1 c
ab bc c2∣
Expanding along R1 we get =a – bb– c {bc – ab}=a−b b−cc−a
Example-5 Prove that ∣1 1 1a b ca3 b3 c3∣=a – bb – cc – aabc
Solution: Let =∣1 1 1a b ca3 b3 c3∣
Applying C1C1−C2 andC 2C2−C3 we get
=∣ 0 0 1a−b b−c ca3−b3 b3−c3 c3∣
⇒=a – bb – c∣ 0 0 11 1 c
a2abb2 b2bcc2 c3∣ Applying C1C1– C2 we get
=a−bb−c ∣ 0 0 10 1 c
a2ab−bc−c2 b2bcc2 c2∣=a−bb−c ∣ 0 0 1
0 1 ca−cabc b2bcc2 c2∣
=a−bb−ca−cabc ∣0 0 10 1 c1 b2bcc2 c2∣ (Taking out a−c abc from C1 )
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=a−bb−ca−cabc 1⋅∣0 11 b2bcc2∣ (Expanding along R3 )
=a – bb– c a−cabc×−1=a−bb−cc−aabc
Example 6: Prove that ∣1a 1 11 1b 11 1 1c∣=abc 11
a1b1c =abcbccaab
Solution: Let =∣1a 1 11 1b 11 1 1c∣
Taking a ,b , c common from R1, R2 and R3 we get
=abc∣1a1 1
a1a
1b
1b1 1
b1c
1c
1c1∣
Applying R1 R2R3 we get
=abc∣11a1b1c
1 1a1b1c
11a 1b1c
1b
11b
1b
1c
1c
11c
∣Taking 11
a1b1c common from R1 we get
=abc11a1b1c ∣1 1 1
1b
11b
1b
1c
1c
11c∣
Applying C1 C1– C2 and C2 C2 –C3 we get
=abc11a1b1c ∣ 0 0 1
– 1 1 1b
0 – 1 11c∣
Expanding along R1 we get
=abc11a1b1c 1⋅∣–1 1
0 –1∣=abc11a1b1c =abcbccaab
8
Example-7 Prove that ∣1a2– b2 2ab – 2b2ab 1– a2b2 2a2b – 2 a 1 – a2 – b2∣=1a2b23
Solution: Let =∣1a2 – b2 2 ab – 2b2ab 1 – a2b2 2a2b – 2a 1– a2– b2∣
Applying C1C1−bC3 and C2C2aC3 we get
=∣ 1a2b2 0 – 2b0 1a2b2 2 a
b 1a2b2 – a1a2b2 1– a2– b2∣Taking 1a2b2 common from both C1 and C2 we get
=1a2b22∣1 0 – 2b0 1 2ab – a 1– a2– b2∣
Applying R3 R3– bR1aR2 we get
=1a2b22∣1 0 – 2b0 1 2a0 0 1a2b2∣
Expanding along R3 we get
=1a2b221a2b2⋅∣1 00 1∣
⇒ =1a2b23
Example 8: Prove ∣bc 2 a2 a2
b2 ca2 b2
c2 c2 ab2∣=2abc abc3
Solution: Let =∣bc2 a2 a2
b2 ca2 b2
c2 c2 ab2∣Applying C1C1– C3 and C2C2 –C 3 we get
=∣bc2 –a2 0 a2
0 ca2– b2 b2
c2– ab2 c2– ab2 ab2∣Taking abc common from C 1 and C3 we get
=abc2∣bc – a 0 a2
0 ca – b b2
c – a – b c – a – b ab2∣Applying R3R3 – R1R2 we get
=abc2∣bc – a 0 a2
0 ca – b b2
– 2 b – 2a 2ab∣ 9
⇒ =abc 2
ab ∣abac – a2 0 a2
0 bcab – b2 b2
– 2ab – 2ab 2 ab∣ C1aC1 andC 2 bC2
Applying C1C1C3 and C2C 2C3
=abc 2
ab ∣abac a2 a2
b2 bcab b2
0 0 2ab∣=
abc 2
ab2ab⋅∣abbc a2
b2 bcab∣ (By Expanding along R3 )
Taking a common from R1 and b common from R2 we get
=2ab abc 2∣bc ab ca∣
⇒ =2ab abc2 {bcbac2ac−ab }⇒=2abc abc3
Example 9: Prove that ∣1 1 1a b cbc ca ab∣=a – bb – cc – a
Solution: Let =∣1 1 1a b cbc ca ab∣=a – bb – cc – a
Applying C1C1– C2 and C2C2 –C 3 we get
=∣ 0 0 1a – b b –c c
– c a – b – ab– c ab∣⇒ =a – bb– c ∣ 0 0 1
1 1 c– c – a ab∣ (Taking a – b common from C1 and b – c from C2 )
Expanding along R1 we get
=a – bb– c c – a1⋅∣ 1 1– c – a∣
⇒=a – bb – cc – a
Example-10 Prove that ∣y z x xy zx yz z x y∣=4 xyz
Solution: Let =∣yz x xy zx yz z x y∣
Applying C1C1−C3 and C2C2−C3 we get
=∣yz−x 0 x0 z x− y y
z−x− y z− x− y x y∣ 10
Applying R3R3 – R1R2 we get
=∣yz−x 0 x0 z x− y y
−2y −2x 0∣Expanding along R1 we get= yz – x2 xyx 2 y z x – y ⇒=2 xy yz – xzx – y =4 xyz
Example-11 Prove that ∣bc ca abqr r p pqy z z x x y∣=2∣a b c
p q rx y z∣
Solution: Let =∣bc ca abqr rp pqyz zx x y∣
Applying C1C2C3 we get
=∣2 abc ca ab2 pqr r p pq2 x yz z x x y∣
⇒=2∣abc ca abpqr rp pqx yz zx x y∣
Applying C2C2 –C 1 and C3C3−C1 we get
=2∣abc −b −cpqr −q −rx yz − y −z∣
Applying C1C1C2C3 we get
=2∣a −b −cp −q −rx − y −z∣
⇒=∣a b cp q rx y z∣
Example-12: Prove that ∣a21 ab acab b21 bcca cb c21∣=1a2b2c2
Solution: Let =∣a21 ab acab b21 bcca cb c21∣
⇒= 1abc∣aa
21 a2b a2 cab2 bb21 b2 cc2a c2b c c21∣ ( R1
1aR1 , R2
1bR2 and R3
1cR3 )
11
⇒=abcabc∣a21 a2 a2
b2 b21 b2
c2 c2 c21∣⇒=∣a21 a2 a2
b2 b21 b2
c2 c2 c21∣Applying R1R1R2R3 we get
=∣1a2b2 1a2b2 1a2b2
b2 b21 b2
c2 c2 c21 ∣=1a2b2∣1 1 1
b2 b21 b2
c2 c2 c21∣ (By taking 1a2b2 coomon from R1 )
Applying C1C1– C2 and C2C 2–C 3 we get
=1a2b2∣ 0 0 1–1 1 b2
0 –1 c21∣Expanding along R1 we get
=1a2b2∣–1 10 –1∣=1a2b2
Example-13: Prove that ∣a – b – c 2a 2 a2b b – c – a 2 b2c 2c c – a – b∣=abc3
Solution: Let =∣a – b – c 2a 2a2b b – c – a 2b2c 2c c –a – b∣
Applying R1R1R2R3 we get
=∣abc abc abc2b b – c – a 2b2c 2c c – a – b∣
⇒=abc∣1 1 12 b b – c – a 2b2c 2c c – a – b∣
Applying C1C1– C2 and C2C 2–C 3 we get
=abc∣ 0 0 1abc – abc 2b
0 abc c−a−b∣=abc3∣0 0 1
1 – 1 2b0 1 c – a – b∣ (Taking abc common from C1 and C2 )
12
=abc3∣1 – 10 1 ∣ (Expandibg along R1 )
⇒=abc3
Example 14: Prove that ∣sin cos cos sin cos cos sin cos cos ∣=0
Solution: Let =∣sin cos cos sin cos cos sin cos cos ∣
⇒=∣sin cos cos cos – sin sinsin cos coscos – sin sinsin cos cos cos – sin sin∣
By applying C3C3sinC1– cosC 2
⇒=∣sin cos 0sin cos 0sin cos 0∣
⇒=0
Example-15: Prove that ∣ a2 bc acc2
a2ab b2 acab b2bc c2 ∣=4a2b2 c2
Solution: Let =∣ a2 bc acc2
a2ab b2 acab b2bc c2 ∣
⇒=a bc∣ a c acab b ab bc c ∣
Applying R1R1R2R3 we get
=abc∣2ab 2 bc 2 caab b ab bc c ∣
=2abc∣ab bc caab b ab bc c ∣
Applying R1R1– R2 and we get
=2abc∣ 0 c cab b ab bc c∣
Applying C2C 2–C 3 we get
13
=2abc∣ 0 0 cab b – a ab b c∣
=2abc {c abb2– b2ab }=4 a2b2 c2
Example-16 If a ,b , c are positive and unequal, show that the value of the determinant
=∣a b cb c ac a b∣
is negative.Solution: Applying C1C1C2C3 we get
=∣abc b cabc c aabc a b∣=abc ∣1 b c
1 c a1 a b∣
⇒=abc∣0 b – c c – a0 c – a a –b1 a b ∣ ApplyingR1R1−R2 and R2 R2−R3
⇒=abc {b−ca−b−c−a2}⇒=abcba– b2– cabc – c2 – a22ca⇒=abc– a2– b2– c2abbcca=– abca2b2c2– ab – bc – ca
⇒=−12abc2a22b22c2– 2 ab– 2bc – 2ca
⇒=−12abc {a – b2b – c 2c – a2}
As a ,b , c are positive and unequal therefore abc0 and a – b2b – c2c – a20⇒0
Example 17: Solve ∣a x a – x a – xa – x ax a – xa – x a – x ax∣=0
Solution: Let =∣ax a – x a – xa – x ax a – xa – x a – x a x∣
Applying C1C1C2C3 we get =∣3a – x a – x a – x3a – x a x a – x3a – x a – x ax∣
⇒=3a– x ∣1 a– x a – x1 a x a – x1 a– x ax∣
Applying R2R2– R1 and R3R3 – R1 we get
=3a – x ∣1 a– x a – x0 2 x 00 0 2 x ∣
14
⇒=3a– x ×1×∣2 x 00 2 x∣ (Expanding along C1 )
⇒=3a– x 4 x2
Thus =0⇒3a – x4 x2=0⇒ x=0,3a
Example 18: Prove that =∣x x2 1 px3
y y 2 1 py3
z z 2 1 pz3∣Solution: We have =∣x x2 1 px3
y y 2 1 py3
z z 2 1 pz3∣⇒=∣x x 2 1
y y2 1z z2 1∣ pxyz∣1 x x2
1 y y2
1 z z2∣ {As each element of C3 is sum of two elements}
⇒ = – ∣1 x2 x1 y2 y1 z2 z ∣ pxyz∣1 x x2
1 y y2
1 z z 2∣⇒=∣1 x x 2
1 y y2
1 z z 2∣ pxyz∣1 x x2
1 y y2
1 z z 2∣⇒= 1 pxyz ∣1 x x2
1 y y2
1 z z 2∣⇒= 1 pxyz ∣1 x x2
0 y – x y2 – x2
0 z – x z2 – x2∣ (Applying R2R2– R3 and R3R3 – R1 )
⇒= 1 pxyz y – x z – x ∣1 x x2
0 1 y x0 1 z x∣
⇒=1pxyz y – x z – x∣1 yx1 zx∣ (Expanding along C1 )
⇒=1pxyz y – z z – x z x – y – x=1 pxyz x – y y – z z – x
Example-19: Show that ∣a b ca2 b2 c2
bc ca ab∣=∣1 1 1a2 b2 c2
a3 b3 c3∣=a – bb – c c – aabbcca
Solution: Let =∣a b ca2 b2 c2
bc ca ab∣Applying C1
1aC1 , C2
1bC2 and C3
1cC3 we get
15
= 1abc∣ a2 b2 c2
a3 b3 c3
abc abc abc∣⇒= abc
abc∣a2 b2 c2
a3 b3 c3
1 1 1 ∣=∣a2 b2 c2
a3 b3 c3
1 1 1 ∣Applying R1⇔ R3 we get
= – ∣1 1 1a3 b3 c3
a2 b2 c2∣Applying R2⇔R3 we get
=∣1 1 1a2 b2 c2
a3 b3 c3∣Applying C1C1– C2 and C2C 2–C 3 we get
=∣ 0 0 1a2 – b2 b2 – c2 c2
a3 – b3 b3 – c3 c3∣=∣ 0 0 1a – ba b b – cb c c2
a – ba2 ab b2 b – cb2 bc c2 c3∣⇒= a – bb – c∣ 0 0 1
a b b c c2
a2 ab b2 b2 bc c2 c3∣Applying C2C 2–C 1 we get
= a – bb – c ∣ 0 0 1a b c – a c2
a2ab b2 c – aa b c c3∣⇒= a – bb – cc – a∣ 0 0 1
a b 1 c2
a2 ab b2 a b c c3∣Expanding along R1 we get=a – bb– c c – a {ababc – a2abb2}⇒=a – bb – cc – a a2abacabb2bc – a2– ab−b2⇒=a – bb – cc – a abbcca
Example-20: Prove that ∣ a a b a b c2a 3a 2 b 4a 3b 2c3a 6a 3b 10a 6b 3c∣=a3
Solution: Let =∣a a b a b c2 a 3a 2b 4a 3b 2c3a 6a 3b 10 a 6b 3c∣
Since each element of the second column is sum of two elements we get
16
=∣ a a a b c2a 3a 4a 3b 2c3a 6a 10a 6b 3c∣∣
a b a b c2 a 2b 4a 3b 2c3a 3b 10a 6 b 3c∣
⇒ =∣a a a b c2 a 3a 4 a 3b 2c3a 6a 10a 6b 3c∣ ab⋅∣1 1 a b c
2 2 4a 3b 2 c3 3 10a 6b 3c∣
⇒=∣a a a b c2a 3a 4a 3b 2c3a 6a 10 a 6b 3c∣ab⋅0 [ .̈ C2 and C3 are identical in second determinant)
As each element of C3 is sum of three elements therefore
=∣a a a2 a 3a 4 a3a 6a 10a∣∣
a a b2a 3a 3b3a 6 a 6b∣∣
a a c2a 3a 2c3a 6a 3c∣
⇒=a3∣1 1 12 3 43 6 10∣ a2b∣1 1 1
2 3 33 6 6∣ a2 c∣1 1 1
2 3 23 6 3∣
As C2 and C3 are identical in second determinant and C1 and C3 are identical in third determinant thus we get
=a3∣1 1 12 3 43 6 10∣ a2 b ⋅0 a2 c ⋅0
⇒=a3∣1 0 02 1 23 3 7∣ [Applying C2C 2–C 1 and C3C3– C1 ]
⇒=a37– 6 [Expanding along R1 ]⇒ =a3
Exercise 4.2Using Properties of determinants, prove the following
1. ∣x a x ay b y bz c z c∣= 0
2. ∣1 bc a b c1 ca b c a1 ab c a b∣= 0
3. ∣0 a – b– a 0 – cb c 0 ∣= 0
4. ∣– a2 ab acba – b2 bcca cb – c2∣=0
17
5. ∣b2 c2 bc b cc2a2 ca c aa2b2 ab a b∣= 0
6. ∣x x2 yzy y2 zxz z2 xy∣= x – y y – z z – xxy yz zx
7. ∣x 4 2 x 2 x2 x x 4 2 x2 x 2 x 2 x 4∣=5 x 4 4 – x 2
8. ∣y k y yy y k yy y y k∣= k23y k
9. ∣1 a a2 – bc1 b b2 – ca1 c c2 – ab∣= 0
10. ∣b c a ab c a bc c a b∣= 4abc
11. ∣a – b – c 2a 2 a2b b – c – a 2 b2c 2c c – a – b∣= a b c 3
12. ∣x y 2 z x yz y z 2 x yz x z x 2 y∣= 2 x y z 3
13. ∣1 x x2
x 2 1 xx x2 1 ∣= 1 – x23
14. ∣x y x x5 x 4 y 4 x 2 x10 x 8 y 8 x 3 x ∣= x3
15. ∣b c a – b ac a b – c ba b c – a c ∣= 3abc – a3 – b3 – c3
16. ∣ y z2 xy zxxy x z 2 yzxz yz x y 2∣= 2 xyz x y z 3
17. ∣b2 c2 ab acba c2 a2 bcca cb a2 b2∣= 4 a2b2c2
18
18. ∣3a – a b – a c– b a 3b – b c– c a – c b 3c ∣= 3a b c ab bc ca
19. ∣1 1 p 1 p q2 3 2 p 4 3 p 2q3 6 3 p 10 6 p 3q∣= 1
20. Prove that ∣x sin cos– sin – x 1cos 1 x ∣ is independent of
21. Evaluate ∣cos cos cos sin – sin – sin cos 0sin cos sinsin cos ∣
22. Solve the equation ∣x a x xx x a xx x x a∣= 0 , a≠0
23. If x≠ y≠z and ∣x x2 1 x3
y y2 1 y3
z z2 1 z3 ∣= 0 , then prove that xyz=–1
24. Solve ∣x – 2 2 x – 3 3 x – 4x – 4 2 x – 9 3 x – 16x – 8 2 x – 27 3 x – 64∣=0
25. If a , b , c are in A.P, find value of ∣2 y 4 5 y 7 8 y a3 y 5 6 y 8 9 y b4 y 6 7 y 9 10 y c∣
Area of TriangleResult : Area of triangle whose vertices are x1, y1, z 1 , x2, y2, z2; x3, y3, z 3 is given by
=±12∣x1 y1 1x2 y 2 1x3 y3 1∣
Collinearity of three points: Three points x1, y1, z 1 , x2, y2, z2; x3, y3, z 3 will be collinear if and only
if ∣x1 y1 1x 2 y2 1x3 y3 1∣= 0
Example-1: Find the area of the triangle whose vertices are 1,2 ,2 ;3 , 3; – 1
Solution: =±12∣1 2 1
2 3 13 −1 1∣=±1
2 {131−22−31−2−9}
⇒=±1242 – 11=5
2 square units
Example-2: Show that the points 1,2; 4,−1; 5.−2 are collinear.
19
Solution: ∣1 2 14 – 1 15 – 2 1∣=1 –12– 2 4 –51 – 85=12 – 3=0
Example-3: If points a ,0 ,0,b ,x , y are collinear then show that xa yb=1
Solution: As points a ,0 ,0,b ,x , y are collinear ∣a 0 10 b 1x y 1∣=0
⇒ab – y – bx=0 ⇒ab– ay – bx=0⇒aybx=abDivide both sides by ab we getxa yb=1
Exercise 4.31. Find the area of the triangle whose vertices are
(i) 2,3 ,4,– 1 ,0,2 (ii) 3,– 1 ,2,4 ,9,2 (iii) 4, –1 , 4,1 ,7,32. Show that the points 3,1 ,4,0 ,6, – 2 are collinear.3. Show that the points 3,1 ,2, – 1 ,1,– 3 are collinear.4. If the points a ,0 ,0,b ,1,1 are collinear then show ab=ab5. Using determinants prove that the points a ,b ,a ' , b ' and a – a ' , b – b ' are collinear if
ab '=ba '6. Find the value of such that 2,3 , ,1 , – 1,2 are collinear.7. Using determinants find the equation of the line joining the points 2,3 and –3,1
Minors and Co-FactorsMinors: Let A=[ aij ] be a square matrix of order n . Then we define minor of a ij as a determinant of order n – 1 obtained from A be deleting its ith row and jth column. Minor of a ij is denoted by M ij
Co-Factor: Let A=[ aij ] be a square matrix of order n . Then we define co-factor of a ij as –i j M ij
. We denote the co-factor of a ij as Aij or C ij
Thus Aij=–1i j M ij
or Aij={M ij if i j is even−M ij if i j is odd
Result: If A=[ aij ] be a square matrix of order n
∣A∣=∑j=1
n
aij Aij (Expansion along ith row)
∣A∣=∑j=1
n
aijC ij (Expanding along jth column)
To be more specific let A=[ aij ] be a square matrix of order 3.Then ∣A∣=a11 A11a12 A12a13 A13 (Expansion along first column)∣A∣=a21 A21a22 A22a23 A23 (Expanding along second row)∣A∣=a31 A31a32 A32a33 A33 (Expanding along third row)∣A∣=a11 A11a21A21a31C31 (Expansion along first column)∣A∣=a12 A12a22 A22a32 A32 (Expanding along second column)∣A∣=a13 A13a23 A23a33 A33 (Expanding along third column)Note: If we multiply the elements of a row(column) by the co-factors of elements of different row(column) and then add the value of such expression is zero.
20
For example for a square matrix of order 3 a11 A21a12 A22a13 A23=0
Example-1: Find the minor and co-factor of each element of the matrix [ 2 3 1– 1 4 22 5 1]
Solution: M 11=∣4 25 1∣=4−10=−6 ; A11=– 6
M 12=∣−1 22 1∣=−1−4=−5, A12=5
M 13=∣−1 42 5∣=−5−8=−13, A13=−13
M 21=∣3 15 1∣=3−5=−2, A21=2
M 22=∣2 12 1∣=2−2=0, A22=0
M 23=∣2 33 5∣=10−9=1,A23=−1
M 31=∣3 14 2∣=6−4=2, A31=2
M 32=∣ 2 1−1 2∣=41=5, A32=−5
M 33=∣ 2 3−1 4∣=83=11, A33=11
Example-2: Find the determinant of the matrix [ 2 3 4– 1 2 33 2 5] and verify the relation
∣A∣=a11 A11a12 A12a13 A13
Solution: ∣A∣=2∣2 32 5∣– 3∣–1 3
3 5∣4∣– 1 23 2∣=210 – 6–3 – 5 –94 – 2 – 6=842– 32=18
A11=−111∣2 32 5∣=10−6=4 , A12=−112∣−1 3
3 5∣=−−5−9=14
A13=−113∣−1 23 2∣=−2−6=−8
Thus a11 A11a12 A12a13 A13=243144−8=842−32=18Thus ∣A∣=a11 A11a12 A12a13 A13
Example-3: For the matrix [ 2 3 11 2 4– 1 2 1] , verify that a21A11a22 A12a23 A13=0
Solution: A11=−111∣2 42 1∣=2– 8=– 6
A12=−112∣1 4−1 2∣=−14=−5
21
A13=−113∣ 1 2−1 2∣=22=4
Thus a21A11a22 A12a23 A13=1−62−544=−6−1016=0
Exercise-4.4
1. Write the minors and co-factors of each element the determinant ∣2 31 4∣
2. Write the minors and cofactors of each element of the determinant ∣ 2 −1 3−2 4 −15 1 2 ∣
3. Using co-factors of the elements of the 2nd column find the value fo the determinant
∣2 – 1 03 2 –12 1 4 ∣
4. Using co-factors of elements of 3rd row find the value of the determinant ∣3 – 1 1– 1 0 23 5 1∣
5. For the determinant ∣2 −1 34 1 51 0 2∣ show that a11 A21a12A22a13 A23=0
6. For the determinant ∣– 1 2 51 2 – 12 5 7 ∣ , show that a13A12a23A22a33 A23=0
Adjoint of a MatrixDefinition: : The adjoint of a square matrix A=[ aij ]n×n is defined as the transpose of the matrix [ Aij ]n× n , where Aij is the co-factor of the element a ij . Adjoint of the matrix A is denoted by adjA
Example-1: Find the adjoint of the matrix [2 34 1]
Solution: A11=1, AA12=−4,A21=−3, A22=2
Hence adj A= transposeof [ 1 – 4– 3 2 ]=[ 1 – 3
– 4 2 ]Note: For a 2×2 matrix A=[a b
c d ] adjA=[ d – b– c a ]
Thus the rule to write the adjoint of 2×2 matrix is that (i) interchange the position of diagonal elements and(ii) change the sign of the non-diagonal elements Result: ∣adj A∣=∣A∣n−1
Result: A adjA=∣A∣I
22
Example-2: Find the adjoint of the matrix [ 2 1 3– 1 4 12 5 1]
Solution: A11=−1, A12=3, A13=−13 ; A21=14, A22=−4, A23=−8 ; A31=−11, A32=−5, A33=9
adjA=[ – 1 14 –113 – 4 –5
– 13 – 8 9 ]Example 3: Find the adjoint of the matrix A=[2 3 –1
2 1 –14 1 2 ] and verify that A adjA =∣A∣I . Also
verify that ∣adj A∣=∣A∣n−1
Solution: A11=3, A12=−8, A13=−2 ; A21=−7, A22=8, A23=10 ; A31=−2, A32=0, A33=−4
Thus adjA=[ 3 –7 – 2– 8 8 0– 2 10 – 4]
and ∣A∣=233−8−−2=6−242=−16
Now A adj A=[2 3 −12 1 −14 1 2 ][ 3 −7 −2
−8 8 0−2 10 −4]=[– 16 0 0
0 – 16 00 0 – 16]=–16 I=∣A∣I
Now
∣adj A∣=∣3 −7 −2−8 8 0−2 10 −4∣=3−32−0−−732−0−2 −8016=−96224128=256=−162
Thus ∣adjA∣=∣A∣2=∣A∣3−1=∣A∣n−1 (Here n=3 )
Exercise 4.5
1. Find the adjoint of the matrix [ 2 3– 1 5] and verify the result A adjA=∣A∣I
2. Find the adjoint of the matrix [ 3 –1 2– 3 1 21 3 1] and verify the result A adjA=∣A∣I
3. Find the adjoint of the matrix [2 31 –3] and verify the result ∣adj A∣=∣A∣n−1
4. Find the adjoint of the matrix [3 –1 02 3 14 1 2] and verify the result ∣adj A∣=∣A∣n−1
5. For the matrix A=[2 3 11 2 43 5 5 ] show that A adj A=0
Inverse of a MatrixSingular Matrix: A square matrix A is said to be singular if ∣A∣=0Non-Singular Matrix: A square matrix A is said to be non-singular if ∣A∣≠0Result: If A and B are nonsingular matrices of the same order, then AB and BA are also non
23
singular matrices of the same order.Result: ∣AB∣=∣A∣∣B∣where A and B are square matrices of the same order.Invertible Matrix: A square matrix A is invertible if and only if A is nonsingular.
and A–1= 1∣A∣
adj A
Example-1 Show that the matrix [2 31 3] is invertible and find its inverse.
Solution: Let A=[2 31 3] . Then
∣A∣=6−3=3≠0 ⇒ A is nonsingular. Thus A is invertible.
A11=3, A12=–1, A21=– 3, A22=2 ⇒adj A=[ 3 −3−1 2 ]
Thus A−1= 1∣A∣
adj A=13 [ 3 −3−1 2 ]=[ 1 −1
−13
23 ]
Example-2: Show that A=[2 3 –11 4 22 –1 3 ] is invertible and find A–1
Solution: ∣A∣=∣2 3 −11 4 22 −1 3 ∣=2 122– 33– 4– 1– 1– 8=2839=40
A11=14, A12=1, A13=– 9, A21=– 8, A22=8, A23=8, A31=10, A32=−5, A33=5
Thus adj A=[14 −8 101 8 −5−9 8 5 ]
.̇. A−1= 1∣A∣
adj A
⇒ A−1= 140 [14 −8 10
1 8 −5−9 8 5 ]=[
720
– 15
14
140
15
−18
– 940
15
18]
Example-3:: If A=[2 31 – 4] and B=[ 1 – 2
– 1 3 ] , then verify that AB −1=B−1 A−1
Solution: We have AB=[2 31 – 4 ][ 1 – 2
–1 3 ]=[–1 55 –14]
Since ∣AB∣=14−25=−11≠0 , AB −1 exists and is given by
AB−1= 1∣AB∣
adjAB=− 111 [– 14 – 5
– 5 – 1]= 111 [14 5
5 1]Further ∣A∣=−11≠0 and ∣B∣=1≠0 . Therefore, A−1 and B–1 both exist and are given by
A−1=− 111 [−4 −3
−1 2 ], B−1=[3 21 1]
24
Therefore B−1 A−1=−111 [3 2
1 1][−4 −3−1 2 ]= 1
11 [14 55 1]
Hence AB−1=B−1 A−1
Example-4: If A=[2 34 5] then show that A2 – 7 A– 2 I=0 and hence find A–1 .
Solution: A2=[2 34 5][2 3
4 5]=[16 2128 37] , 7A=7[2 3
4 5]=[14 2128 35] , 2I=[2 0
0 2]Thus A2−7A−2I=[16 21
28 37 ]−[14 2128 35]−[2 0
0 2]=[0 00 0]
Thus A2 – 7 A– 2 I=0 Pre-Multiply by A−1 we getA–1 A2– 7 A–1 A– 2 A–1 A=0 ⇒ A –7 I – 2 A–1=0
⇒2A−1=A−7I=[2 34 5]−[7 0
0 7]=[– 5 34 – 2]
⇒ A−1=12 [– 5 3
4 – 2]⇒ A−1=[– 5
232
2 –1]Example-5: If A=[2 1 1
1 2 11 1 2] and B=[ 3 –1 –1
– 1 3 –1– 1 –1 3 ] then find AB and hence find A−1
Solution: AB=[2 1 11 2 11 1 2][ 3 – 1 – 1
– 1 3 – 1– 1 – 1 3 ]=[4 0 0
0 4 00 0 4]
Thus AB=4 I
Pre-Multiply by A–1 AB=4 A–1 I ⇒ IB=4 A–1 ⇒B=4 A– 1 ⇒ A–1= 14B
Thus A−1=[34
– 14
– 14
– 14
34
– 14
– 14
– 14
34]
Example:6 Find the inverse of the matrix [cos −sin 0sin cos 0
0 0 1]Solution: Let A=[cos −sin 0
sin cos 00 0 1]
∣A∣=cos2sin2=1 (By expanding along third row)A11=cos , A12=−sin , A13=0 ; A21=sin , A22=cos , A23=0 ; A31=0, A32=0, A33=1
25
adj A=[ cos sin 0−sin cos 0
0 0 1]Thus A−1= 1
∣A∣adj A
⇒ A−1=[ cos sin 0−sin cos 0
0 0 1]Example: -7: If A= [2 4
3 – 2 ] then show that A−1= 116
A
Solution: We have ∣A∣=−4−12=−16≠0 Thus A−1 exists.
A−1= 1∣A∣
adj A=− 116 [−2 −4
−3 2 ]= 116 [2 4
3 −2]= 116
A
Example -8: If A=[ 1 tan x−tan x 1 ] then show that AT A−1=[cos 2 x −sin 2 x
sin 2 x cos 2 x ]Solution: ∣A∣=1tan 2 x≠0 , thus A−1 exists and is given by
A−1= 1∣A∣
adj A= 11tan2 x [ 1 −tan x
tan x 1 ]AT=[ 1 −tan x
tan x 1 ]Thus AT A= 1
1tan2 x [ 1 −tan xtan x 1 ][ 1 −tan x
tan x 1 ]= 11tan2 x∣1−tan2 x −2tan x
2tan x 1−tan2 x∣⇒ AT A−1=[1−tan 2 x
1tan 2 x−
2tan x1tan 2 x
2 tan x1tan 2 x
1−tan2 x1tan2 x
]=[cos 2 x −sin 2 xsin 2 x cos 2 x ]
Example-9: For the matrix A=[2 34 –1] , find the values of a ad b such that A2aAbI=0 .
Hence find A– 1
Solution: We have A=[2 34 −1]
Therefore A2=[2 34 −1][2 3
4 −1]=[16 34 13]
As A2aAbI=0
Therefore [16 34 13][2a 3a
4a −a][b 00 b]=[0 0
0 0]⇒[162ab 33a
44a 13−ab]=[0 00 0]
⇒162ab=0, 33a=0, 44a=0,13−ab=0⇒a=−1 and b=−14
Thus A2a Ab I=0⇒ A2−A−14I=0
26
Pre-Multiply by A−1 we getA−1 A2−A−1 A−14 A−1 I=0
⇒ A−I−14A−1=0 ⇒14A−1=A−I ⇒ A−1= 114 {[2 3
4 −1]−[1 00 1]}
⇒ A−1=[ 114
314
414
− 214 ]=[
114
314
27
−17 ]
Example-10: Find a 2×2 matrix B such that B[1 −21 4 ]=[6 0
0 6]Solution: Let A=[1 −2
1 4 ] and C=[6 00 6 ]
Then, ∣A∣=42=6≠0 . Thus A−1 exits and is given by
A−1= 1∣A∣
adj A =16 [ 4 2−1 1]
Now B[1 −21 4 ]=[6 0
0 6] ⇒B A=C
Post-Multiply both side by A−1 we get B A A−1=C A−1 or B I=C A−1 or
B=C A−1=16 [6 0
0 6 ][ 4 2−1 1]=1
6 [24 12−6 6 ]
Thus B=[ 4 2−1 1]
Example:11 Find a matrix A such that [2 31 2]A[1 – 2
3 1 ]=[2 11 3]
Solution: Let B = [2 31 2 ],C = [1 – 2
3 1 ],D = [2 11 3 ]
Thus we have BAC=DPre-Multiply by B–1 and post multiply by C–1 we get B– 1 BAC C– 1=B– 1 DC– 1 ⇒ IAI=B– 1 DC–1
⇒A=B– 1D C –1
∣B∣=1,∣C∣=7
Thus B–1 = 1∣B∣
adjB = [ 2 – 3– 1 2 ]= [ 2 – 3
– 1 2 ]and C–1= 1
∣C∣adjC=1
7 [ 1 2– 3 1]
Thus A = 17 [ 2 – 3
– 1 2 ][2 11 3 ][ 1 2
– 3 1]Thus A=1
7 [1 – 70 5 ] [ 1 2
– 3 1]
27
Thus A=17 [ 22 – 5
– 15 5 ]= [ 227
– 57
– 157
57 ]
Exercise 4.6
1. Show that [2 3 41 2 11 1 4] is non singular.
2. Show that the matrix [3 1 22 1 35 2 5] is singular
3. Show that the matrix [a – b b – c c – ab– c c – a a – bc – a a – b b – c] is singular.
4. Find the value of such that the matrix [2 31 2 30 1 2 ] is singular.
5. Find the inverse of the matrix [2 31 – 2 ]
6. Find the inverse of the matrix [ 3 –1– 2 4 ]
7. Find the inverse of the matrix [cos – sinsin cos ]
8. Find the inverse of the matrix [a bc d ]
9. Find the inverse of the matrix [2 3 – 11 2 52 0 – 1]
10. Find the inverse of the matrix [3 –1 – 22 1 34 1 2 ]
11. If A=[2 34 – 1] and B=[4 1
2 – 1] then verify that AB– 1=B– 1 A– 1
12. If A = [ 1 2 22 1 3
– 1 2 1 ] and B = [0 – 1 22 1 12 – 1 1 ] then verify that AB–1=B– 1 A– 1
13. If A = [2 34 – 1] then prove that A2 – A – 14I=0 and hence find A^{-1}
28
14. If A = [2 13 2 ] then find a and b such that A2aAbI=0 and hence find A– 1
15. Find the inverse of the matrix [1 0 00 cos sin 0 sin – cos]
16. For the matrix A=[ 2 – 1 1– 1 2 – 11 – 1 2 ] , show that A3 – 6A29A – 4 I=0 . Hence find A– 1
17. For the matrix A=[1 1 11 2 – 32 – 1 3 ] show that A3 – 6A25A11 I=0
18. Show that the matrix A=[ 2 3– 1 4] is zero of the polynomial f x=x2 – 6x11 and hence find
A– 1
19. Find the matrix X for which [3 27 5]X=[1 −2
1 3 ]20. Find the matrix X for which X[3 – 1
2 4 ]= [1 23 4 ]
21. Find the matrix A for which [3 12 2 ]A [ – 1 2
3 4 ]=[ 1 3−1 0]
22. Find the inverse of the matrix A = [a b
c 1 bca ]and show that aA–1=A2bc1I –aA
23. Given A = [ 2 – 3– 4 7 ] , compute A– 1 and show that 2A –1=9I – A
24. If A = [4 52 1 ] , then show that A−3I=2I3A–1
25. Given A = [5 0 42 3 21 2 1 ], B– 1 = [1 3 3
1 4 31 3 4 ] . Compute AB–1
26. Show that [ 1 – tan 2
tan 2
1 ][ 1 tan 2
– tan 2
1 ]– 1
= [cos – sinsin cos ]
27. If A = 19 [ – 8 1 4
4 4 71 – 8 4 ] , prove that A– 1=AT
28. If A = [3 – 3 42 – 3 40 – 1 1 ] , prove that A– 1=A3
29. If A = [ – 1 2 0– 1 1 10 1 0 ] , show that A2=A–1
29
System of Linear Equations
Consider a system of linear equations {a1 xb1 y=c1
a2 xb2 y=c2.
Using matrix notations this can be written as [a1 b1
a2 b2][xy]=[c1
c2] or AX=B where A=[a1 b1
a2 b2] ,
X=[xy ] and B=[c1
c2]If ∣A∣≠0 then A– 1 existsPre-Multiplying AX=B by A– 1 we get A– 1 AX=A – 1 B ⇒ I X=A – 1 B⇒X=A–1 B
Consider the system of equations {a1 xb1 yc1 z=d1
a2 xb2 yc2 z=d 2
a3 xb3 yc3 z=d3
This can be written as [a1 b1 c1
a2 b2 c2
a3 b3 c3][xyz]=[d1
d2
d3]
or A X=B where A=[a1 b1 c1
a2 b2 c2
a3 b3 c3] , X=[xyz ] and B=[d1
d 2
d 3]
If ∣A∣≠0 then A−1 exists.Pre-Multiplying A X=B by A−1 we get A−1 A X=A−1B⇒ I X=A−1B⇒ X=A−1 B
Note: The system of equations A X=B will have(i) unique solution if ∣A∣≠0 and the solution is given by X=A−1B(ii) infinitely many solutions if adj AB=0(iii) no solution if adj AB≠0
Example-1: Solve the system of linear equations {2 x4 y=103 x− y=8
Solution: The given system of equations can be written as [2 43 −1][xy]=[10
8 ]or A X=B where A=[2 4
3 −1], X=[xy ], B=[108 ]
Now ∣A∣=∣2 43 −1∣=−2−12=−14≠0 , thus A−1 exists.
A11=−1, A12=−3, A21=−4 , A22=2
Thus adj A=[−1 −4−3 2 ]
.̇. A−1= 1∣A∣
adj A=− 114 [−1 −4
−3 2 ]
30
Now X=A−1B ⇒ X=− 114 [−1 −4
−3 2 ][108 ]
Thus X=−114 [−42
−14]=[31] ⇒[xy]=[31]Hence solution of the system of equations is x=3, y=1
Example:2 Solve the system of equations {3 x−2 y z=3x2 y−z=5x− y z=1
Solution: The above system of equations can be written as A X=B where
A=[3 −2 11 2 −11 −1 1 ], X=[
xyz ], B=[
351]
∣A∣=∣3 −2 11 2 −11 −1 1 ∣=32−1−−2111−1−2=4≠0 . Thus A−1 exists.
Now A11=1, A12=−2, A13=−3, A21=1, A22=2, A23=1, A31=0, A32=4, A33=8
Thus adj A=[ 1 1 0−2 2 4−3 1 8]
Using A−1= 1∣A∣
adj A we get
A−1=14 [ 1 1 0
−2 2 4−3 1 8]
Using X=A−1B we get
X=14 [ 1 1 0
−2 2 4−3 1 8][351 ]=1
4 [884]=[221] ⇒[xyz]=[221]
Thus solution of the given system of equations is x=2, y=2, z=1
Example-3: If A=[1 2 11 0 32 −3 0] then find A−1 and hence solve the system of equations
x2 yz=7, x3 z=11, 2x−3y=1
Solution: ∣A∣=∣1 2 11 0 32 −3 0∣=109−2 0−61−3−0=912−3=18≠0
Thus A−1 exists. A11=9, A12=6, A13=−3 ; A21=−3, A22=−2, A23=7, A31=6, A32=−2, A33=−2
Thus adj A=[ 9 −3 66 −2 −2−3 7 −2]
31
⇒ A−1= 1∣A∣
adj A= 118 [ 9 −3 6
6 −2 −2−3 7 −2]
Now given system of equations can be expressed as [1 2 11 0 32 −3 0][
xyz ]=[
7111 ] or
A X=B where A=[1 2 11 0 32 −3 0] , X=[xyz ] and B=[ 7
111 ] .
.̇. X=A−1 B= 118 [ 9 −3 6
6 −2 −2−3 7 −2][
7112 ]= 1
18 [361854]=[
213]
.̇.[xyz ]=[213]Hence solution of the system of equations is x=2, y=1, z=3
Example-5: If A=[1 −1 12 1 −31 1 1 ] , find A−1 and hence solve the system of equations
x2 y z=4,−x yz=0, x−3yz=2
Solution: A=[1 −1 12 1 −31 1 1 ]
.̇.∣A∣=∣1 −1 12 1 −31 1 1 ∣=1 1312312−1=10≠0
So, A is invertible.A11=5, A12=−5, A13=1, A21=2, A22=0, A23=−2, A31=2, A32=5, A33=3
.̇. adj A=[ 4 2 2−5 0 51 −2 3]
⇒ A−1= 1∣A∣
adj A= 110 [ 4 2 2
−5 0 51 −2 3 ]
Now the given system of equations is expressible as [ 1 2 1−1 1 11 −3 1][
xyz ]=[
402]
or, AT X=B where X=[xyz ], B=[402 ]
Now, ∣AT∣=∣A∣=10≠0 . So, the system of equations is consistent with a unique solution.X= AT −1 B=A−1T B
32
[xyz ]= 110 [ 4 2 2
−5 0 51 −2 3 ]
T
[402]⇒[xyz]= 1
10 [4 −5 12 0 −22 5 3 ][402]= 1
10 [18414]=[
952575]
Hence x= 95,
y= 25,
z=75 is the solution of the given system of linear equations.
Example-6: Find the product of the matrices [2 1 11 2 11 1 2] and [ 3 – 1 – 1
– 1 3 – 1– 1 – 1 3 ] and hence solve the
system of equations 2 x y z=4, x2 yz=4, x y2 z=4
Solution:Let A=[2 1 11 2 11 1 2] and B=[ 3 −1 −1
−1 −3 −1−1 −1 3 ]
A B=[2 1 11 2 11 1 2][ 3 −1 −1
−1 −3 −1−1 −1 3 ]=[4 0 0
0 4 00 0 4]=4 I
Pre-Multiplying by A−1 we get A−1 AB=4 A−1 I⇒ I B=4 A−1⇒4A−1=B⇒ A−1= 14B
Thus A−1=14 [ 3 −1 −1−1 3 −1−1 −1 3 ]
Given system of equations can be written as [2 1 11 2 11 1 2][
xyz ]=[
444 ] or A X=C where
X=[xyz ],C=[444].̇.X=A−1C=1
4 [ 3 −1 −1−1 3 −1−1 −1 3 ][4444]=1
4 [444]⇒[xyz]=[
111]
Hence the solution of the given system of equations is x=1, y=1, z=1Example-7: Solve the system of equations 2 x− y z=4, x2 y−z=3,3 x y=7 if consistent.
Solution: The given system of equations can be expressed as [2 −1 11 2 −13 1 0 ][xyz ]=[
437 ]
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or A X=B where A=[2 −1 11 2 −13 1 0 ], X=[xyz], B=[
437]
∣A∣=2011 031 1−6=23−5 =0 ⇒ A−1 does not exist.Thus system of equations does not have unique solution.A11=1, A12=−3, A13=−5, A21=1, A22=−3, A23=−5, A31=−1, A32=3, A33=5
Thus adj A=[ 1 1 −1−3 −3 3−5 −5 5 ]
⇒adj A B=[ 1 1 −1−3 −3 3−5 −5 5 ][437 ]=[000]
Thus system of equations has infinitely many solutions.To find the solutions let us put z=k in the first two equations. Thus we get
2 x− yk=4, x2 y−k=3or 2 x− y=4−k , x2y=3k This can be written as
[2 −11 2 ][xy]=[4−k
3k ]⇒[xy]=[2 −1
1 2 ]−1
[4−k3k ]
⇒[xy]=15 [ 2 1−1 2][4−k
3k ]⇒[xy]=1
5[11−k23k ]
⇒[xy]=[11−k5
23k5
]Thus solution of the given system of equations is x=11−k
5, y=23 k
5,z=k where k is any real
number.Example-8: Show that the system of equations 2 x y z=4 , x y z=1,3 x2 y2 z=4 is inconsistent.
Solution: The given system of equations can be written as [2 1 11 1 13 2 2][
xyz ]=[
414 ]
or A X=B where A=[2 1 11 1 13 2 2], X=[xyz ], B=[
414]
∣A∣=22−2−12−312−3=0 . Thus A is singular matrix. Thus A−1 does not exist.Now A11=0, A12=1, A13=−1, A21=0, A22=1, A23=−1, A31=0, A32=−1, A33=1
Thus adj A=[ 0 0 01 1 −1−1 −1 1 ]
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⇒adj A B=[ 0 0 01 1 −1−1 −1 1 ][414]=[ 0
1−1]
As ∣A∣=0and adj AB≠0 , therefore the system of equations do not have any solution.Thus the given system of equations is inconsistent.Homogeneous system of equations: A system of equations is said to be homogeneous if it can be written as A X=BTrivial Solution: A solution in which each variable is zero is known as trivial solution. Every system homogeneous linear equations always has trivial solution. A homogeneous system of linear equations will have trivial solution only if ∣A∣=0Non-trivial Solution: A homogeneous system of linear equations AX=B will have non-trivial solution if ∣A∣≠0Example-9: Solve the following homogeneous system of linear equations
2x− y z=0, x y− z=0, x− y=0Solution:The given system of equations can be written as A X=0 where
A=[2 −1 11 1 −11 −1 0 ], X=[xyz]
Now ∣A∣=20−11 011−1−1=−21−2=−3≠0Thus the given system of equations has trivial solution.Thus solution is x=0, y=0, x=0
Exercise 4.7Solve the following system of linear equations by matrix method
1. 5 x7 y24 x6 y3=0
2. 3 x4 y−5=0x− y3=0
3.x yz=3
2 x− y z=−12 x y−3z=−9
4.6 x−12 y25 z=44 x15 y−20 z=32 x18 y15 z=10
5.3 x4 y7 z=162 x− y3 z=19x2 y−3z=25
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6.
2x−3
y3
z=10
1x1
y1
z=10
3x−
1y
2z=13
7.3 x4 y2 z=8
2 y−3 z=3x−2 y6 z=−2
8.2 x6 y=23 x−z=−8
2 x− yz=−3
9.8 x4 y3 z=18
2 x y z=5x2 y z=5
10.2 x yz=4x2 yz=4x y2 z=4
11. If A=[1 −1 02 3 40 1 2] and B=[ 2 2 −4
−4 2 −42 −1 5 ] are two square matrices, find AB and hence
solve the system of equations x− y=3, 2 x3 y4 z=17, y2 z=7
12. If A=[2 −3 53 2 −41 1 −2 ] find A−1 and hence solve the system of equations
2 x−3 y5 z=11,3 x2 y−4 z=−5, x y2 z=−3
13. Find A−1 if A=[1 2 51 −1 −12 3 −1] . Hence solve the system of equations
x2 y5 z=10, x− y−z=−2,2 x3 y− z=−11
14. Find A−1 if A=[ 4 5 2−5 −4 2−2 2 8 ] and hence solve the system of equations
4 x−5 y−2 z=2,5 x−4 y2 z=−2, 2 x2 y8 z=−115. Solve the following system of equations if consistent
2 x y z=4, x yz=1,3 x2 y2 z=516. Show that the system of equations 3 x yz=1, x y−z=2,4 x2 y=217. Solve the following homogeneous\system of linear equations
(i) 2 x− yz=0
3 x2 y− z=0x4 y3 z=0
(ii) 2 x− y− z=0x y z=0
3 x2 y− z=0(iii)
x yz=0x− y−5 z=0x2 y4 z=0
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