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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
Question 2.
If A = \(\left[ \begin{matrix} 1 & 2 \\ 4 & 2 \end{matrix} \right] \), then show that |2A| = 4|A|
Answer:
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:
|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:
= 1(-9 + 12) -1 (-18 + 15) – 2(8 – 5) = 3 + 3 – 6 = 0.
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
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)
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
∴ 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 = \(\left[ \begin{matrix} 4 & -2 \\ -3 & 1 \end{matrix} \right]\)
Question 20.
If A = \(\left[ \begin{matrix} 7 & 3 \\ 5 & 2 \end{matrix} \right]\) find A-1.
Answer:
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:
(ii) (2, 7), (1, 1), (10, 8).
Answer:
Question 2.
Show that the points A(a, b + c), B(b, c + a), C (c, a + b) are collinear.
Answer:
= \(\frac { 1 }{ 2 }\)[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.
Question 3.
Find the value of k is area of triangle is± 4sq units and vertices are (k, 0), (4, 0), (0,2).
Answer:
⇒ – 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.
⇒ 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:
⇒ 0 + 30 – 6k + 20 – 45 = 70 ⇒ 10k = – 20 ⇒ k = – 2
on taking – ve sing we get k = 12.
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:
∴ 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:
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.
Question 9.
Prove that the value of the determinant remains unchanged if its rows and columns are interchanged.
Answer:
Question 10.
Prove that if any two rows (or columns) of a determinant are interchanged, then sign of determinant changes.
Answer:
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:
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:
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.
= Δ + 0 (since R, and R, are proportional)
Hence Δ = Δ1
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:
= (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:
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.
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:
= (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.
Answer:
Expanding along C1, we get
= (5x + 4) {1(4 – x)(4 – x)} = (5x + 4)(4 – x)2 = RHS.
Hence proved.
Question 5.
Answer:
Expanding along C3, we get
= (3y + k) (1 × k.k) = k2(3y + k) = RHS. Hence proved.
Question 6.
Answer:
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.
Answer:
Expanding along R3, we get
= 2(x + y + z)3 [(1)(1 – o)] = 2(x + y + z)3 = RHS
Hence proved.
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 = \(\left| \begin{matrix} 1 & x & { x }^{ 2 } \\ { x }^{ 2 } & 1 & x \\ x & { x }^{ 2 } & 1 \end{matrix} \right|\)
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.
Answer:
Expanding along R1, we get
(1 + a2 + b2)2 [1(1 + a2 + b2) – 0 + 0] = (1 + a2 + b2)3 =RHS
Question 10.
Answer:
= -1 × (-c2) + 1[1(a2 + 1) + 1(b2)] = (a2 + b2 + c2 + 1) = 1 + a2 + b2 + c2 = RHS.
Hence proved.
Question 11.
Answer:
Question 12.
Answer:
Question 13.
Answer:
Question 14.
Answer:
Question 15.
Evaluate Δ = \(\left| \begin{matrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{matrix} \right| \)
Answer:
= (b – a) (c – a) [(-b + c)] (Expanding along first column)
= (a – b) (b – c) (c – a)
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:
= 2c(ab + b2 – bc) – 2b(bc – c2 – ac)
= 2abc + 2cb2 – 2bc2 – 2b2c – 2b2c + 2bc2 + 2abc
= 4 abc
Question 17.
Answer:
= (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
Question 18.
Answer:
Question 19.
\(\left| \begin{matrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{matrix} \right| \) = 0
Answer:
Question 20.
\(\left| \begin{matrix} x & a & x+a \\ y & b & y+b \\ z & c & z+c \end{matrix} \right| \) = 0
Answer:
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:
Applying R11 = R1 + R2
Expanding along C1
Δ = (x – p) (px + x2 – 2q2) = (x – p)(x2 + px – 2q2)
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
We see that
|A| = (2 – 3) + 2(4 + 4) + 3(-6 – 4) = -17 ≠ 0 .
Hence, A is nonsingular and so its inverse exists. Now
Question 2.
If A = \(\), find A-1. using A-1 solve the system.
Answer:
The given system can be written as AX = B, where
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
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:
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:
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
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
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
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.
Question 8.
Answer:
Question 9.
Answer:
