Students can Download 2nd PUC Maths Chapter 4 Determinant Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## 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 = 2x^{2} – 24

2x^{2} = -18 + 24 ⇒ x^{2} = 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:

x^{2} – 18 × 2 = 36 – 18 × 2 ⇒ x^{2} – 36 = 36 – 36

⇒ x^{2} – 36 = 0 – 36 ⇒ x^{2} = 36 ⇒ x = ±6

Question 8.

Find the adjoint of the matrix \(\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right] \)

Answer:

Let A = \(\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right] \)

∴ A_{11} = 4, A_{12} = -3

A_{21} = -2 and A_{22} = 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 – x^{2 }= 3 – 8 ⇒ x^{2} = 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 (∵ R_{1} = R_{3})

Question 13.

If A is a square matrix of order 3 and |A| = 5 then find |Adj A|.

Answer:

|Adj A| = |A|^{2} = 5^{2} = 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} = 10^{2} = 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 – (b^{2} – c^{2}) + ab + b^{2} – c^{2} – 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 Δ_{1} is obtained by (he operation. R_{1} → R_{1} + kR_{3}

Here we have multiplied the elements of the third row (R_{3}) by a constant k and added them to the corresponding elements of the first row (R_{1}).

Symbolically, we write this operation as R_{1} → R_{1} + kR_{3}.

= Δ + 0 (since R, and R, are proportional)

Hence Δ = Δ_{1}

### 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{c^{2} – 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 R_{1}, we get

= (a – b) (b – c) [1 × (b^{2} + bc + c^{2}) -1 × (a^{2} + ab + c^{2})

= (a – b) (b – c) [b^{2} + bc + c^{2} – a^{2} – ab – b^{2}]

= (a – b) (b – c) (bc – ab + c^{2} – a^{2})

= (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) (z^{3} + x^{2} + xz) – (z + x) (z – x) (y – x) (y^{2} + x^{2} + xy)

= (y – x) (z – x) [(y – x) (z^{3} + x^{2} + xz) – (z + x) (y^{2} + x^{2} + xy)]

= (y – x) (z – x) [(yz^{2} + yx^{2} + xyz + xz^{2} + x^{3} + x^{2}z – zy^{2} – zx^{2} – xyz – xy^{2} – x^{3} – x^{2}y]

= (y – x) (z – x) [yz^{2} + zy^{2} + xz^{2} – xy^{2}] .

= (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 C_{1}, we get

= (5x + 4) {1(4 – x)(4 – x)} = (5x + 4)(4 – x)^{2} = RHS.

Hence proved.

Question 5.

Answer:

Expanding along C_{3}, we get

= (3y + k) (1 × k.k) = k^{2}(3y + k) = RHS. Hence proved.

Question 6.

Answer:

Expanding along R_{1}, 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 R^{3}, 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 – x^{3})^{2}

Answer:

LHS = \(\left| \begin{matrix} 1 & x & { x }^{ 2 } \\ { x }^{ 2 } & 1 & x \\ x & { x }^{ 2 } & 1 \end{matrix} \right|\)

Expanding along C_{1}, we get

= (1 + x + x^{2})(1 – x)[(1 × (1 + x) – (-x)(x))]

= (1 + x + x^{2})(1 – x)(1 – x)(1 + x + x^{2})

= [(1 – x^{3})(1 – x^{3})] = (1 – x^{3})^{2} = RHS [∵ 1 – x^{3} = (1 – x)(1 + x + x^{2})]

Hence proved.

Question 9.

Answer:

Expanding along R_{1}, we get

(1 + a^{2} + b^{2})^{2} [1(1 + a^{2} + b^{2}) – 0 + 0] = (1 + a^{2} + b^{2})^{3} =RHS

Question 10.

Answer:

= -1 × (-c^{2}) + 1[1(a^{2} + 1) + 1(b^{2})] = (a^{2} + b^{2} + c^{2} + 1) = 1 + a^{2} + b^{2} + c^{2} = 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 + b^{2} – bc) – 2b(bc – c^{2} – ac)

= 2abc + 2cb^{2} – 2bc^{2} – 2b^{2}c – 2b^{2}c + 2bc^{2} + 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) (x^{2} + px – 2q^{2})

Answer:

Applying R_{1}^{1} = R_{1} + R_{2}

Expanding along C_{1}

Δ = (x – p) (px + x^{2} – 2q^{2}) = (x – p)(x^{2} + px – 2q^{2})

### 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 A^{3} – 6A^{2} + 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 A^{3} – 6A^{2} + 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: