2nd PUC Maths Question Bank Chapter 4 Determinant

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Karnataka 2nd PUC Maths Question Bank Chapter 4 Determinant

2nd PUC Maths Determinant One Mark Questions and Answers

Question 1.
Evaluate the following determinants:
a. \(\left| \begin{matrix} 2 & 4 \\ -5 & -1 \end{matrix} \right| \)
Answer:
\(\left| \begin{matrix} 2 & 4 \\ -5 & -1 \end{matrix} \right| \) = 2 × (-1) – 9 (-5) × 4 = – 2 + 20 = 18
2nd PUC Maths Question Bank Chapter 4 Determinant 1

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 2.
If A = \(\left[ \begin{matrix} 1 & 2 \\ 4 & 2 \end{matrix} \right] \), then show that |2A| = 4|A|
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 2

Question 3.
If A = \(\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} \right] \) then show that |3A| = 27 |A|.
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 3
|3A| = 3(3 × 12 – 0 × 6) = 3(36) = 108 ……… (1)
|A| = 1(1 × 4 – 0 × 2) = 4
27|A| = 27 × 4 = 108 ………. (2)
From (1) and (2) |3A| = 27|A|.

Question 4.
Evaluate \(\left| \begin{matrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{matrix} \right| \)
Answer:
\(\left| \begin{matrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{matrix} \right| \) = 3(0 – 5) + 1(0 + 3) – 2(0 – 0) = -15 + 3 – 0 = -12.

Question 5.
If A = \(\left[ \begin{matrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{matrix} \right] \) find |A|.
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 4
= 1(-9 + 12) -1 (-18 + 15) – 2(8 – 5) = 3 + 3 – 6 = 0.

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 6.
Find the value of x if, \(\left| \begin{matrix} 2 & 4 \\ 5 & 1 \end{matrix} \right| =\left| \begin{matrix} 2x & 4 \\ 6 & x \end{matrix} \right| \)
Answer:
2 × 1 – 5 × 4 = 2x × x – 6 × 4 ⇒ 2 – 20 = 2x2 – 24
2x2 = -18 + 24 ⇒ x2 = 3 ⇒ x = ±√3 .

Question 7.
If \(\left| \begin{matrix} x & 2 \\ 18 & x \end{matrix} \right| =\left| \begin{matrix} 6 & 2 \\ 18 & 6 \end{matrix} \right| \), find x.
Answer:
x2 – 18 × 2 = 36 – 18 × 2 ⇒ x2 – 36 = 36 – 36
⇒ x2 – 36 = 0 – 36 ⇒ x2 = 36 ⇒ x = ±6

Question 8.
Find the adjoint of the matrix \(\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right] \)
Answer:
Let A = \(\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right] \)
∴ A11 = 4, A12 = -3
A21 = -2 and A22 = 1
2nd PUC Maths Question Bank Chapter 4 Determinant 5

Question 9.
Find the value of x for which \(\left| \begin{matrix} 3 & x \\ 18 & x \end{matrix} \right| =\left| \begin{matrix} 6 & 2 \\ 18 & 6 \end{matrix} \right| \)
Answer:
3 – x2 = 3 – 8 ⇒ x2 = 8 ⇒ x = ±√2

Question 10.
If A = \(\left[ \begin{matrix} 1 & 2 \\ 4 & 2 \end{matrix} \right] \) find |3A|.
Answer:
|3A| = \(\left| \begin{matrix} 3 & 6 \\ 12 & 6 \end{matrix} \right| \) = 18 – 72 = – 54.

Question 11.
If A is a square matrix with |A| = 6, find the value of |AA’I.
Answer:
|A| = 6 ∴ |A’| = 6
|AA’| = |A| × |A| = 6 × 6 = 36

Question 12.
Evaluate : \(\left| \begin{matrix} 2 & 4 & 5 \\ 5 & 7 & 3 \\ 2 & 4 & 5 \end{matrix} \right| \)
Answer:
\(\left| \begin{matrix} 2 & 4 & 5 \\ 5 & 7 & 3 \\ 2 & 4 & 5 \end{matrix} \right| \) = 0 (∵ R1 = R3)

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 13.
If A is a square matrix of order 3 and |A| = 5 then find |Adj A|.
Answer:
|Adj A| = |A|2 = 52 = 25.

Question 14.
Define a singular matrix.
Answer:
A square matrix A is called a singular matrix if |A| = 0.

Question 15.
If A is an invertible matrix of order 2 × 3 such that |A| = 5 then find |A-1|.
Answer:
|A-1| = \(\frac{1}{|A|}=\frac{1}{5}\)

Question 16.
IfA is a square matrix A.(Adj A) = 10I then find |Adj A|.
Answer:
A. (Adj A) = 10I comparing with the result
A. (Adj A) = |A|l
2nd PUC Maths Question Bank Chapter 4 Determinant 6
∴ lAdj A| = |A|2 = 102 = 100

Question 17.
If A = \(\left[ \begin{matrix} 2 & 3 \\ 6 & x \end{matrix} \right] \) is singular then find x.
Answer:
|A| = 0
\(\left| \begin{matrix} 2 & 3 \\ 6 & x \end{matrix} \right| \) = 0 ⇒ 2x – 18 = 0 ⇒ x = 9

Question 18.
If A = \(\left[ \begin{matrix} 2 & 3 \\ 5 & 8 \end{matrix} \right] \) find |Adj A|.
Answer:
Adj A = \(\left[ \begin{matrix} 8 & -3 \\ -5 & 2 \end{matrix} \right] \)
|Adj A| = 16 – 15 = 1

Question 19.
Find the adjoint of A = \(\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right]\)
Answer:
Adj A = \(\left[ \begin{matrix} 4 & -2 \\ -3 & 1 \end{matrix} \right]\)

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 20.
If A = \(\left[ \begin{matrix} 7 & 3 \\ 5 & 2 \end{matrix} \right]\) find A-1.
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 7

2nd PUC Maths Determinant Two Marks Questions and Answers

Question 1.
Find area of the triangle with vertices at the point given in each of the following:
(i) (1, 0), (6, 0), (4, 3)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 8

(ii) (2, 7), (1, 1), (10, 8).
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 9

Question 2.
Show that the points A(a, b + c), B(b, c + a), C (c, a + b) are collinear.
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 10
= \(\frac { 1 }{ 2 }\)[a{c + a(a + b)} – (b + c){b – c} + 1 {b(a + b) – c(c + a)}]
= \(\frac { 1 }{ 2 }\)[ac – ab – (b2 – c2) + ab + b2 – c2 – ca]
= \(\frac { 1 }{ 2 }\)[0] = o
Since the area generated by the three points is 0 they are collinear.

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 3.
Find the value of k is area of triangle is± 4sq units and vertices are (k, 0), (4, 0), (0,2).
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 11
⇒ – 2k + 8 = 8 ⇒ 2k = 0 k = 0
on taking – ve sign we get – 2k + 8 = – 8 ⇒ 2k – 16 ⇒ k = 8 ∴ k = 0, 8.

Question 4.
Find the equation of line joining (1, 2) and (3, 6) using determinants.
Answer:
Let P(x, y) be any point on the line joining A (1, 2) and B (3, 6). If the points A, B and P are in collinear, then the are of the triangle ABP is equal to zero.
2nd PUC Maths Question Bank Chapter 4 Determinant 12
⇒ 6 – y – 6 + 2x + 3y – 6x = 0 ⇒ 2y – 4x = 0 ⇒ y = 2x.

Question 5.
If the area of the triangle is ± 35 sq units with vertices (2, – 6), (5, 4) and (k, 4), find the value of k.
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 13
⇒ 0 + 30 – 6k + 20 – 45 = 70 ⇒ 10k = – 20 ⇒ k = – 2
on taking – ve sing we get k = 12.

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 6.
If each element of a row is expressed as sum of two elements then verify for a third order determinant that the determinant can be expressed as sum of two determinants.
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 14
∴ It is verified that if each element of a row is expressed as sum of two elements in a third order determinant that the determinant can be expressed as sum of two determinants.

Question 7.
Find the inverse of the matrix A = \(\left[ \begin{matrix} 2 & -2 \\ 4 & 3 \end{matrix} \right] \)
Answer:
2nd PUC Maths Question Bank Chapter - 4 Determinant - 15

Question 8.
Examine the consistency of the system of equations given by x + 2y = 2, 2x + 3y = 3 Note : the system of equations of the form AX = B is consistent, if |A| ≠ 0. If |A| = 0 and (Adjoint of A) (B) = 0 then the system of equations is consistent and if (Adjoint of A) (B) = 0. Then the system of equation is inconsistent.
Answer:
|A| = \(\left| \begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix} \right| \) = 3 – 4 = -1 ≠ 0
∴ the system of equation is consistent.

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 9.
Prove that the value of the determinant remains unchanged if its rows and columns are interchanged.
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 16

Question 10.
Prove that if any two rows (or columns) of a determinant are interchanged, then sign of determinant changes.
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 17
2nd PUC Maths Question Bank Chapter - 4 Determinant 18

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 11.
Prove that if each element of a row (or a column) of a determinant is multiplied by k, then its value gets multiplied by k.
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 19

Question 12.
Prove that if to each element of any row or column of a determinant, the equimultiples of corresponding elements of other row (or column) are added, then value of determinant remains the same
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 20
where Δ1 is obtained by (he operation. R1 → R1 + kR3
Here we have multiplied the elements of the third row (R3) by a constant k and added them to the corresponding elements of the first row (R1).
Symbolically, we write this operation as R1 → R1 + kR3.
2nd PUC Maths Question Bank Chapter 4 Determinant 21
= Δ + 0 (since R, and R, are proportional)
Hence Δ = Δ1

2nd PUC Maths Question Bank Chapter 4 Determinant

2nd PUC Maths Determinant Four Marks Questions and Answers

Question 1.
Prove that \(\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { b }^{ 2 } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| \) = (a – b) (b – c) (c – a)
Answer:
2nd PUC Maths Question Bank Chapter - 4 Determinant - 22
= (a – b) (b – c) [o{c2 – c{b + c)} -1 {0 – 1 (b + c)} + (a + b){0 -1}]
= (a – b) (b – c) [0 + b + c – a – b]
= (a – b) (b – c) (c – a) = RHS

Question 2.
Prove that \(\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ { a }^{ 2 } & { b }^{ 2 } & { c }^{ 2 } \end{matrix} \right| \) = (a – b) (b – c) (c – a) (a + b + c)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 23
Now, expanding along R1, we get
= (a – b) (b – c) [1 × (b2 + bc + c2) -1 × (a2 + ab + c2)
= (a – b) (b – c) [b2 + bc + c2 – a2 – ab – b2]
= (a – b) (b – c) (bc – ab + c2 – a2)
= (a – b) (b – c) [b(c – a) + (c – a)(a + c)]
= (a – b) (b – c) (c – a) (a + b + c)
= RHS . Hence proved.

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 3.
Prove that \(\left| \begin{matrix} x & { x }^{ 2 } & yz \\ y & { y }^{ 2 } & zx \\ z & { z }^{ 2 } & xy \end{matrix} \right| \) = (x – y) ( y – z) (z – x) (xy + yz zx)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 24
2nd PUC Maths Question Bank Chapter 4 Determinant 25
= (y + x) (y – x) (z – x) (z3 + x2 + xz) – (z + x) (z – x) (y – x) (y2 + x2 + xy)
= (y – x) (z – x) [(y – x) (z3 + x2 + xz) – (z + x) (y2 + x2 + xy)]
= (y – x) (z – x) [(yz2 + yx2 + xyz + xz2 + x3 + x2z – zy2 – zx2 – xyz – xy2 – x3 – x2y]
= (y – x) (z – x) [yz2 + zy2 + xz2 – xy2] .
= (y – x) (z – x) [yz (z – y) + x (z – y)(z + y)]
= (y – x) (z – x) [(z – y) (xy + yz + zx)]
= (x – y) (y – z) (z – x) (xy + yz + zx) = RHS
Hence proved.

Question 4.
2nd PUC Maths Question Bank Chapter 4 Determinant 26
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 27
2nd PUC Maths Question Bank Chapter 4 Determinant 28
Expanding along C1, we get
= (5x + 4) {1(4 – x)(4 – x)} = (5x + 4)(4 – x)2 = RHS.
Hence proved.

Question 5.
2nd PUC Maths Question Bank Chapter 4 Determinant 29
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 30
Expanding along C3, we get
= (3y + k) (1 × k.k) = k2(3y + k) = RHS. Hence proved.

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 6.
2nd PUC Maths Question Bank Chapter 4 Determinant 31
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 32
2nd PUC Maths Question Bank Chapter 4 Determinant 33
Expanding along R1, we get
= (a + b + c) {1(- b – c – a)(- c – a – b)}
= (a + b + c) [-(b + c + a)×(-)(c + a + b)]
= (a + b + c) (a + b + c) (a + b + c) = (a + b + c)3 = RHS.
Hence proved.

Question 7.
2nd PUC Maths Question Bank Chapter 4 Determinant 34
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 35
Expanding along R3, we get
= 2(x + y + z)3 [(1)(1 – o)] = 2(x + y + z)3 = RHS
Hence proved.

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 8.
Prove that \(\left| \begin{matrix} 1 & x & { x }^{ 2 } \\ { x }^{ 2 } & 1 & x \\ x & { x }^{ 2 } & 1 \end{matrix} \right| \) = (1 – x3)2
Answer:
LHS = \(\left| \begin{matrix} 1 & x & { x }^{ 2 } \\ { x }^{ 2 } & 1 & x \\ x & { x }^{ 2 } & 1 \end{matrix} \right|\)
2nd PUC Maths Question Bank Chapter 4 Determinant 36
Expanding along C1, we get
= (1 + x + x2)(1 – x)[(1 × (1 + x) – (-x)(x))]
= (1 + x + x2)(1 – x)(1 – x)(1 + x + x2)
= [(1 – x3)(1 – x3)] = (1 – x3)2 = RHS [∵ 1 – x3 = (1 – x)(1 + x + x2)]
Hence proved.

Question 9.
2nd PUC Maths Question Bank Chapter 4 Determinant 37
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 38
2nd PUC Maths Question Bank Chapter 4 Determinant 39
Expanding along R1, we get
(1 + a2 + b2)2 [1(1 + a2 + b2) – 0 + 0] = (1 + a2 + b2)3 =RHS

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 10.
2nd PUC Maths Question Bank Chapter 4 Determinant 40
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 41
= -1 × (-c2) + 1[1(a2 + 1) + 1(b2)] = (a2 + b2 + c2 + 1) = 1 + a2 + b2 + c2 = RHS.
Hence proved.

Question 11.
2nd PUC Maths Question Bank Chapter 4 Determinant 42
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 43

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 12.
2nd PUC Maths Question Bank Chapter 4 Determinant 44
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 45
2nd PUC Maths Question Bank Chapter 4 Determinant 46

Question 13.
2nd PUC Maths Question Bank Chapter 4 Determinant 47
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 48

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 14.
2nd PUC Maths Question Bank Chapter 4 Determinant 49
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 50
2nd PUC Maths Question Bank Chapter 4 Determinant 51

Question 15.
Evaluate Δ = \(\left| \begin{matrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{matrix} \right| \)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 52
= (b – a) (c – a) [(-b + c)] (Expanding along first column)
= (a – b) (b – c) (c – a)

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 16.
Prove that \(\left| \begin{matrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{matrix} \right| \) = 4abc (March 2014)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 53
= 2c(ab + b2 – bc) – 2b(bc – c2 – ac)
= 2abc + 2cb2 – 2bc2 – 2b2c – 2b2c + 2bc2 + 2abc
= 4 abc

Question 17.
2nd PUC Maths Question Bank Chapter 4 Determinant 54
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 55
= (1 + xyz) (y – x) (z – x)(z – y)(on expandingalongCl)
Since D = O and x, y, z are all different, i.e., x – y ≠ 0, we get 1 + xyz = 0

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 18.
2nd PUC Maths Question Bank Chapter 4 Determinant 56
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 57

Question 19.
\(\left| \begin{matrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{matrix} \right| \) = 0
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 58

Question 20.
\(\left| \begin{matrix} x & a & x+a \\ y & b & y+b \\ z & c & z+c \end{matrix} \right| \) = 0
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 59

Question 21.
Show that \(\left| \begin{matrix} x & p & q \\ p & x & x \\ p & q & x \end{matrix} \right| \) =(x – p) (x2 + px – 2q2)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 60
Applying R11 = R1 + R2
Expanding along C1
Δ = (x – p) (px + x2 – 2q2) = (x – p)(x2 + px – 2q2)

2nd PUC Maths Question Bank Chapter 4 Determinant

2nd PUC Maths Determinant Five Marks Questions and Answers

Solve the following equations by matrix method.

Question 1.
3x – 2y + 3z = 8 ; 2x + y – z = 1; 4x – 3y + 2z = 4
Answer:
The system of equation can be written in the form AX = B, where
2nd PUC Maths Question Bank Chapter 4 Determinant 61
We see that
|A| = (2 – 3) + 2(4 + 4) + 3(-6 – 4) = -17 ≠ 0 .
Hence, A is nonsingular and so its inverse exists. Now
2nd PUC Maths Question Bank Chapter 4 Determinant - 62
2nd PUC Maths Question Bank Chapter 4 Determinant - 63

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 2.
If A = \(\), find A-1. using A-1 solve the system.
Answer:
The given system can be written as AX = B, where
2nd PUC Maths Question Bank Chapter 4 Determinant 64
Thus, A is non-singular,. Therefore, its inverse exists.
Therefore, the given system is consistent and has a unique solution given by X = A-1 B. Cofactors of A are
2nd PUC Maths Question Bank Chapter 4 Determinant 65
2nd PUC Maths Question Bank Chapter 4 Determinant 66

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 3.
For the matrix A = \(\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{matrix} \right] \). show that A3 – 6A2 + 5A + 11I = 0. Hence, find A-1.
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 67
2nd PUC Maths Question Bank Chapter 4 Determinant 68
2nd PUC Maths Question Bank Chapter 4 Determinant 69

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 4.
If A = \(\left[ \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right] \) verify that A3 – 6A2 + 9A – 4I = 0 and hence, find A-1.

Question 5.
Let A = \(\left[ \begin{matrix} 3 & 7 \\ 2 & 5 \end{matrix} \right] \) and B = \(\left[ \begin{matrix} 6 & 8 \\ 7 & 9 \end{matrix} \right] \) verify that (AB)-1 = B-1 A-1.
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 70
2nd PUC Maths Question Bank Chapter 4 Determinant 71

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 6.
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.
Answer:
Let first, second and third numbers be denoted by x, y and z, respectively. Then, according to given conditions, we have
x + y + z = 6
y + 3z = 11
x + z = 2y or x – 2y + z = 0
This system can be written as AX = B, where
2nd PUC Maths Question Bank Chapter 4 Determinant 72
2nd PUC Maths Question Bank Chapter 4 Determinant 73

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 7.
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. The cost of 6 kg onion, 2 kg wheat and 3 kg rice is ₹70. Find cost of each item per kg by matrix method.
Answer:
Let the prices (per kg) of onion, wheat and rice be ₹x, ₹y and ₹z, respectivelyihen
4x + 3y + 2z = 60, 2x + 4y + 6z = 90, 6x + 2y + 3z = 70.
This system of equations can be written as AX = B, where
2nd PUC Maths Question Bank Chapter 4 Determinant 74
Thus, A is non-singular, Therefore, its inverse exists.
Therefore, the given system is consistent and has a unique solution given by X = A-1 B.
Cofactors of A are
2nd PUC Maths Question Bank Chapter 4 Determinant 75
2nd PUC Maths Question Bank Chapter 4 Determinant 76
Here, x = 1, y = \(\frac { 1 }{ 2 }\) and z = \(\frac { -3 }{ 2 }\)
∴ X = 5, y = 8 and z = 8. Hence, price of onion per kg is ₹5, price of wheat per kg is ₹8 and that of rice per kg is ₹8.

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 8.
2nd PUC Maths Question Bank Chapter 4 Determinant 77
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinant 78
2nd PUC Maths Question Bank Chapter 4 Determinant 79

2nd PUC Maths Question Bank Chapter 4 Determinant

Question 9.
2nd PUC Maths Question Bank Chapter - 4 Determinant 80
Answer:
2nd PUC Maths Question Bank Chapter - 4 Determinant 81

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