2nd PUC Basic Maths Question Bank Chapter 1 Matrices and Determinants

Students can Download Basic Maths Question Bank Chapter 1 Matrices and Determinants Questions and Answers, Notes Pdf, 2nd PUC Basic 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 Basic Maths Question Bank Chapter 1 Matrices and Determinants

2nd PUC Basic Maths Matrices and Determinants One Mark Questions and Answers

Question 1.
If \(\mathbf{A}=\left[\begin{array}{c}3 \\-2 \\5\end{array}\right]\). Find AA’
Answer:

Question 2.
If \({ A }=\left[ \begin{array}{l} 1 \\ 2 \\ 3 \end{array} \right] \).Find AA’
Answer:

Question 3.
If A = [1 0 2] \({ B }=\left[ \begin{array}{l} 1 \\ 2 \\ 3 \end{array} \right] \). Find AB.
Answer:

Question 4.
If A \(\mathbf{A}=\left[\begin{array}{ll}i & \mathbf{0} \\\mathbf{0} & i \end{array}\right]\) find A²
Answer:

Question 5.
If [2 x 2] \(\left[\begin{array}{l}1 \\3 \\5\end{array}\right]\) = [9] S.T x = -1
Answer:

Question 6.
Find x such that \(\left[\begin{array}{ll}3 & x \\4 & 7\end{array}\right]\) is symmentric
Answer:
x = 4

Question 7.
Find x if [-1 x 4]\(\left[\begin{array}{c}1 \\2 \\-1\end{array}\right]\) = [3]
Answer:
– 1 +2x – 4 = 3
= 2x = 3 +5
x = 4

Question 8.
Give an example for scalar matrix
Answer:

Question 9.
If \(\mathbf{A}=\left[\begin{array}{l}4 \\5 \\6\end{array}\right] \mathbf{B}=\left[\begin{array}{l}3 \\4 \\5\end{array}\right]\) compute 2A +3B
Answer:

Question 10.
Find x if \(\left[\begin{array}{l}3 \\-1\end{array}\right]+\left[\begin{array}{l}2 x \\3
\end{array}\right]=\left[\begin{array}{l}1 \\2\end{array}\right]\)
Answer:
3x + 2x = 1 ⇒ 2x = -2  ⇒ x = -1

Question 11.
A company sold 22 scooters, 15 cars and 10 buses in June and 20,21 and 11 respectively in October. Represent the data in the matrix form.
Answer:
\(M=\left[\begin{array}{lll} 22 & 15 & 10 \\20 & 21 & 11\end{array}\right]\)

Question 12.
The following matrix shows the belongings of 3 friends Amar, Akbar and Antony

Write down the row matrix which represents the belongings of Anthony.
Answer:
Anthonys belongings = [6 8].

Question 13.
Find x if \(\left|\begin{array}{cc}x & -3 \\-3 & x\end{array}\right|=0\)
Answer:
x² – 9 = 0 ⇒ x = ± 3

Question 14.
Find x if  \(\left|\begin{array}{cc}2 x & -4 \\-2 & x\end{array}\right|=0\) then find x
Answer:
2x² -8 = 0 ⇒ x² = 4  ⇒ x = ± 2

Question 15.
Evalute \(\left|\begin{array}{cc}2 & 4 \\-5 & -1\end{array}\right|\)
Answer:
-2 + 20 = 18

Question 16.
\(\left|\begin{array}{ll}\cos \theta & \sin \theta \\\sin \theta & \cos \theta\end{array}\right|\)
Answer:
Cos²  θ + sin²θ = 1

Question 17.
Evaluate \(\left|\begin{array}{ccc}1 & 5 & 7 \\5 & 25 & 35 \\3 & -1 & 0\end{array}\right|\)
Answer:

= 5(0) = 0 (∴ Two rows are Indentical)

Question 18.
\(\mathbf{S} \mathbf{T}\left|\begin{array}{lll}\mathbf{1} & \mathbf{1 0} & \mathbf{2} \\\mathbf{2} & \mathbf{1 5} & \mathbf{2} \\\mathbf{3} & \mathbf{1 6} & \mathbf{6}\end{array}\right|=\mathbf{0}\) without expansion
Answer:

Question 19.
If \(\left|\begin{array}{lll}2 & x & 3 \\4 & 1 & 6 \\1 & 2 & 7\end{array}\right|\) = 0 with out expansion.
Answer:
2(7-12)-x (28 – 6) +3(8 – 1) = 0
– 10 – 22x + 21 = 0 = x = 1/2

Question 20.
Evaluate \(\left| \begin{matrix} 51 & 52 & 53 \\ 54 & 55 & 56 \\ 57 & 58 & 59 \end{matrix} \right| \) without expansion
Answer:

Question 21.
Evaluate \(\begin{vmatrix} 3003 & 3005 \\ 3006 & 3008 \end{vmatrix}\)
Answer:

Question 22.
If \( \mathbf{A}=\left[\begin{array}{lll}1 & 2 & 3 \\2 & x & 4 \\3 & 6 & 5
\end{array}\right]\)
Answer:

⇒ (5x-24)-2(10-12) + 3(12 -3x) = 0
⇒ 5x – 24 + 4 + 36 – 9x = 0
⇒- 4x = – 16
x = 4

Question 23.
If \(\left(\begin{array}{ll}3 & 2 \\x & 6\end{array}\right)\) is singular, find x
Answer:

= 2x – 18 = 0 ⇒ 2x = 18 ⇒ x = 9

Question 24.
Without expanding
\({ S.T }\left| \begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{array} \right| =0\)
Answer:

2nd PUC Basic Maths Matrices and Determinants Two Marks Questions and Answers

Question 1.
If \(\mathbf{A}=\left[\begin{array}{ll}3 & 2 \\4 & 1\end{array}\right]\) P.T. A2 – 4A – SI = 0. Where I is unit matrix and 0 is the null matrix of order 2 x 2
Answer:

Question 2.
If \(\mathbf{A}+\mathbf{B}=\left[\begin{array}{ll}7 & \mathbf{0} \\2 & 5
\end{array}\right] \& \mathbf{A}-\mathbf{B}=\left[\begin{array}{ll}3 & 0 \\0 & 3\end{array}\right]\)
Answer:

Question 3.
Find x and Y. If \(\left[\begin{array}{c}5 \\-3\end{array}\right]+\left[\begin{array}{ll}2 & x \\3 & y\end{array}\right]=\left[\begin{array}{l}1 \\6\end{array}\right]\)
Answer:
⇒ 5 + 2x = 1 and -3 +3y = 6
2x = 4 ⇒ x = 2
3y = 9 ⇒ y = 3

Question 4.
If 2A + \( \mathbf{B}=\left[\begin{array}{cc}-7 & 11 \\7 & 4\end{array}\right] \& \mathbf{A}+\mathbf{2 B}=\left[\begin{array}{ll}-8 & 13 \\14 & 5\end{array}\right]\)find A. Multity equation 1 by 2 we get
Answer:

Question 5.
Find AB if \(A=\left[\begin{array}{ll}3 & 2 \\5 & 1 \\1 & 1\end{array}\right] \quad B=\left[\begin{array}{l}1 \\3\end{array}\right]\)
Answer:

Question 6.
Sukhesh buys 3 kgs of dal, 2 kgs of Rice and 5 kgs of oil if the cost of each is Rs. 25, Rs. 20 and Rs. 75 respectively, Find the total cost by matrix method.
Answer:

Question 7.
A lady buys 8 kgs of Apples, 4 kgs of Bananas and 6kgs of oranges. Apples cost Rs. 30 per kg, Bananas cost Rs. 15 per kgs and oranges cost Rs. 24 per kg. Obtain the total cost using matrices.
Answer:

Question 8.
If \(\mathbf{A}=\left[\begin{array}{lll}6 & 2 & 4 \\3 & 2 & 0\end{array}\right] \text { and } B=\left[\begin{array}{cc}1 & 2 \\1 & -1 \\2 & 4\end{array}\right]\) find
2A’ – 3B
Answer:

Question 9.
If \(\mathbf{A}=\left[\begin{array}{lll}2 & 1 & -3 \\1 & 2 & 1\end{array}\right] \mathbf{B}=\left[\begin{array}{cc}-3 & 2 \\1 & 4 \\1 & 5\end{array}\right]\)
find (AB)^T
Answer:

Question 10.
A company is considering which of the 3 methods of production it should be use in producing 3 products A, B and C. The amount of each product produced by each method is as shown below.

Answer:
Further information relating to profit per unit as follows
Product        A    B    C
Profit/unit    10   4   6
Using matrix multiplication find which method maximizes the total profit

Question 11.
If A = \(\mathbf{A}=\left[\begin{array}{cc}1 & -1 \\2 & 0 \\1 & -3\end{array}\right] \& \mathbf{B}=\left[\begin{array}{cc}-2 & 4 \\3 & 2 \\1 & 0\end{array}\right]\). Find 2A + 3B.
Answer:

Question 12.
If \(\left[\begin{array}{ll}x^{2} & 3 \\1 & 2\end{array}\right]+\left[\begin{array}{ll}5 x & 1 \\4 & 3\end{array}\right]=\left[\begin{array}{ll}6 & 4 \\5 & 5\end{array}\right]\). find x.
Answer:
x² + 5x = 6
x² + 5x – 6 – 0 (x + 6)(x – 1) = 0
x = – 6 or 1
[By adding and equating the correspondy terms].

Question 13.
If \(\left[\begin{array}{ll}4 & 5 \\3 & 2\end{array}\right]+\left[\begin{array}{cc}2 & x-3 \\y-4 & 1\end{array}\right]=\left[\begin{array}{ll}6 & 1 \\2 & 3\end{array}\right]\) find x and y.
Answer:
5 + x – 3 = 1 ⇒ x = 1 – 2 = – 1
3 + y – 4 = 2 ⇒ y = 6 -3 = 3

Question 14.
If \(\mathbf{A}=\left[\begin{array}{ll}3 & -2 \\2 & 1\end{array}\right] \mathbf{B}=\left[\begin{array}{ll}1 & -1 \\6 & 2\end{array}\right]\)
Answer:

2nd PUC Basic Maths Matrices and Determinants Three Marks Questions and Answers

Question 1.
If \(\mathbf{A}=\left[\begin{array}{cc}2 & -1 \\-3 & 1 \\4 & 0\end{array}\right] \quad \mathbf{B}=\left[\begin{array}{lll}3 & 1 & -5 \\5 & 4 & -2\end{array}\right]\) find AB and BA.
Answer:

Question 2.
If \(A=\left[\begin{array}{ccc}2 & -1 & 3 \\-3 & 1 & 4 \\-2 & -1 & 5\end{array}\right]\) find A²
Answer:

Question 3.
If \(\mathbf{A}=\left[\begin{array}{cc}2 & 3 \\-4 & 1\end{array}\right] \& \mathbf{B}=\left[\begin{array}{cc}-1 & 5 \\6 & 2\end{array}\right]\) S.T. (AB)’ = B’ A’
Answer:

Question 4.
If \(\mathbf{A}=\left[\begin{array}{cc}2 & 3 \\1 & -1\end{array}\right] \& \mathbf{B}=\left[\begin{array}{l}-2 \\-1\end{array}\right]\) Find x such that Ax = B \(\text { Let } \mathbf{X}=\left[\begin{array}{l}x \\y\end{array}\right]=?\)
Answer:

Question 5.
If \(A=\left[\begin{array}{ll}1 & 2 \\4 & 2\end{array}\right]\) then S T
|2A| = 4|A|
Answer:

Question 6.
Find the inverse of the Matrix \(\text { (i) }\left[\begin{array}{ll}2 & 1 \\
3 & 2\end{array}\right] \text { (ii) }\left[\begin{array}{ll}2 & -1 \\3 & -2\end{array}\right]\)
Answer:

Question 7.
Solve the following equations by Cramer’s rule
(i) 5x + 7y – 3 = 0, 7x + 5y – 9=0
(ii) 4x – y = 5,3x + 2y =1
Answer:

Question 8.
If \(\mathbf{A}=\left[\begin{array}{ll}1 & 2 \\3 & 4\end{array}\right] \quad B=\left[\begin{array}{cc}-1 & 0 \\0 & 1\end{array}\right]\) Find adj (AB)
Answer:

Question 9.
\(\mathbf{P T}\left|\begin{array}{ccc}\mathbf{1} & \boldsymbol{a} & \boldsymbol{b}+\boldsymbol{c} \\\boldsymbol{1} & \boldsymbol{b} & \boldsymbol{c}+\boldsymbol{a} \\\boldsymbol{1} & \boldsymbol{c} & \boldsymbol{a}+\boldsymbol{b}\end{array}\right|=\boldsymbol{0}\)
Answer:

Question 10.
\(\text { S.T }\left|\begin{array}{lll} 2 & 2^{2} & 2^{3} \\2^{2} & 2^{3} & 2^{4} \\2^{4} & 2^{5} & 2^{6}\end{array}\right|=0\)
Answer:

Question 11.
\(\mathbf{P T}\left|\begin{array}{lll}x-y & y-z & z-x \\a-b & b-c & c-a \\p-q & q-r & r-p\end{array}\right|=0 \)
Answer:

2nd PUC Basic Maths Matrices and Determinants Four Marks Questions and Answers

Question 1.
P.T \(\left|\begin{array}{lll}1 & 1 & 1 \\x & y & z \\x^{2} & y^{2} & z^{2}
\end{array}\right|\) = (x -y) (y – z) (z – x)
Answer:

Question 2.
P.T \(\left|\begin{array}{lll}1 & x & y z \\1 & y & z x \\1 & z & x y\end{array}\right|\) = (x -y)(y – z)(z – x)
Answer:

Question 3.
S.T \(\left|\begin{array}{lll}x & y & y \\y & x & y \\y & y & x\end{array}\right|\)= (x+2y) (x – y)²
Answer:

Question 2.
P. T \(\left|\begin{array}{lll}-a^{2} & a b & a c \\a b & -b^{2} & b c \\a c & b c & -c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}\)
Answer:

Question 4.
S. T \(\left|\begin{array}{ccc}1+a & b & c \\a & 1+b & c \\a & b & 1+c \end{array}\right|\) = 1 + a + b + c
Answer:

Question 5.
P.T \(\left|\begin{array}{lll}x & p & q \\p & x & q \\p & q & x\end{array}\right|\)
Answer:

Question 6.
P.T \(\left|\begin{array}{ccc}1 & y+z & y^{2}+z^{2} \\1 & z+x & z^{2}+x^{2} \\
1 \cdot x+y & x^{2}+y^{2}\end{array}\right|=(x-y)(y-z)(z-x)\)
Answer:

Question 7.
If \(A=\left[\begin{array}{ll}1 & 3 \\4 & 5\end{array}\right]\) PT A adjA A =|A| I
Answer:

Question 8.
P. T\(\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\2 b & b-c-a & 2 b \\2 c & 2 c & c-a-b\end{array}\right|\) = (a + b + c )³ with out actnal expresian
Answer:

Question 9.
\({ ST }=\left| \begin{array}{ccc} { 1 }+{ a } & { 1 } & { 1 } \\ { 1 } & { 1+b } & { 1 } \\ { 1 } & { 1 } & { 1 }+{ c } \end{array} \right| ={ abc }({ 1+\frac { 1 }{ { a } } +\frac { { 1 } }{ { b } } +\frac { { 1 } }{ { c } } ) }\)
Answer:

Question 10.
Solve for \(x:\left|\begin{array}{ccc}x+1 & x+2 & 3 \\3 & x+2 & x+1 \\x+1 & 2 & x+1\end{array}\right|=0\)
Answer:

Question 11.
Solve \(\left|\begin{array}{ccc}3 x-8 & 3 & 3 \\3 & 3 x-8 & 3 \\3 & 3 & 3 x-8 \end{array}\right|\) = 0 without direct expansion
Answer:

Question 12.
If \(\mathbf{A}=\left[\begin{array}{ccc}2 & 1 & -1 \\0 & 1 & 3 \\4 & 0 & 5
\end{array}\right]\)find A^-1
Answer:

Question 13.
Solve using cramers Rule
(i) 5x – y – 4z = 4, x +4y +2z = 12, 3x – y -z = 3
Answer:

Question 14.
Solve the following by Matrix Method:- 2x -y + z = 3, x +3y -2z =1, x+y+z = 6 The given equations is Matrix Method is
Answer:

2nd PUC Basic Maths Matrices and Determinants Five or Six Marks Questions and Answers

Question 36.
A Salesman has the following record of sales during 3 months for 3 items A,B,& C which have different rates of commision.

Find out the reters of commission on iterm A,B & C₁¹
Answer:
Let x, y & z denote the raters of commission in Rupees / unit for A,B&C iterms respec­tively. Then the data given can be expressed as a system of linear equations.
100x +100y + 200z = 900
300x+200y+ 100z= 1000
100x + 200y + 30z = 1400
x+y +2 z = 9
3x + 2y +z = 10
x + 2y + 3z
Solve these equations by Matrix Method.

Question 37.
The Monthly expenditure an office for 3 months is given in the table

Assuming that the salary for the different Categories of the staft did net vary from months to month, calculate the salary for each type of staft per Month.
Answer:
Let the Salary of the clerks, peon & typists be x, y,& z
8x + 4y + 6z = 3750
9x +9y + 6z = 5000
12x + 9y + 12z = 8850
The given equation in Matrix form is

Question 38.
TheThe prices of 3 commodities X,Y,Z are x,y,z & respectively. A sells 4 units of Z & purchases 24 units of X & 5 units of Y. B Sells 5units of Y, & purchases 1 unit of X and 1 unit of Z.C sells 1 unit of X, 1 unit of Z and purchases 1 units of Y. In the process Rs 700, Rs 350 & 250 remain with A, B & C respective. Find the pice / unit of X
Answer:
The given data in Matrix form is

A x = B ⇒ x = A^-1B
where  x is the price of X
y is the price of Y z is the price of Z
Here purchase is indicated by α – ve Sign & Sales is indicated by α + ve sign.
By Solve the equation by Matrix Method we get x = 150, y = 200 & z = 500

Question 39.
The cost of 4 kg Ragi, 3kg rice & 2 kg wheat is ₹60, the cost of 2kg Ragi, 4kgs rice & 6kg wheat is ₹90. The cost of 6 kg Ragi, 2kg Rice & 3kg wheat is ₹70. Find the cost of each item per kg by Matrix Method.
Answer:
Let the prices (per kg) of Ragi, Rice & wheat be ₹x,& Xz, respectively then the data in the system of equations is given by
4x + 3y + 2z = 60
2x + 4y + 6z = 90
6x + 2y + 3z = 70
These equation in form of Matrix is

Question 40.
The Sum of three nembers 3 is 6. If we maltiply third number by 3 & add second num­ber to it, we get 11 By adding I and III numbers, we get doduble of the 2nd number Represent it algebraically and find the number 8 using matrix method.
Answer:
Let the I, II & III nos be donoted by x,y & z respectively. Then according to given conditions we have, x + y + z = 6
y + 32 = 11
x + z = 2y ⇒ x – 2y + z ≠ 0                                                    .
These equations can be put in Matrix form as

Question 41.
Food I has one unit of Vitamin A, 2 units of vitamin &units of Vitamin C Food II has 1,3 & 9 units respectively & Food III has 1, 2 & 1 unit respectively units of vitamin A, 17 units of vitamin in & 37 units of vitamin C are required. Find all possible amount 3 foods that will provide exactly these amounts of the vitamins.
Answer:
Let x, y & z be the amounts of the food I, II & III respectively. Then we have the following ret of simaltaneous linear equations.
x + y + z = 7
2x + 3y + 2z = 17
4x + 9y + z = 37
By using Matrix Method Solve the above equations The given equations in Matrix form

|A|= 1(3 – 18)-1(2 – 8)+ 1(18 – 12)
= -15 + 6 + 6 =-3 ≠ 0 ∴ A^-1 exists
By Solving the equation X = A^-1B we get
x = 2, y = 3, z = 2

Question 42.
PT \(\left|\begin{array}{lll}1 & 1 & 1 \\a & b & c \\a^{3} & b^{3} & c^{3} \end{array}\right|\) = (a – b) (b – c) (c – a)(a + b +c)
Answer:

= (a- b)(b – c) [(c – a)(c + a) +b(c – a)]
= (a – b)(b – c)(c – a)(a+b+c) = R.H.S

Question 43.
ST\(\left|\begin{array}{ccc}1 & 1 & 1 \\x^{2} & y^{2} & z^{2} \\x^{3} & y^{3} & z^{3} \end{array}\right|\)=(x – y)(y – z)(xy + yz + zx)
Answer:

x²y (x – y) (y – z)[(xz² – x²z) + (yz² – x²y) ]
(x – y)(y – z)[x = (z – x) + y(z² – x²)]
(x – y)(y – z)(z – x)[zx + y(z + x)]
(x – y)(y – z)(z – x)(xy + yz + zx) = R.H.S

Question 44.
If \(A=\left[\begin{array}{lll}1 & 2 & 3 \\1 & 3 & 4 \\1 & 4 & 3\end{array}\right]\) verify A adj A . A = |A|
Answer:
|A| = 1(9 – 16) -2(3 – 4) + 3(4 -3)
= (-7) + 2 + 3 = -2

Question 45.
If A = \(=\left[\begin{array}{lll}1 & 2 & 3 \\1 & 3 & 4 \\1 & 4 & 3\end{array}\right]\) then verify A adj A =|A|. I
Answer: