Question Set C

cbse 12th Standard Maths Determinants Model Questions Papers Study Material - QB365

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Questions:

Generic 2

5 x 2 = 10

If \(\left[\begin{array}{cc}x & x-y \\ 2 x+y & 7\end{array}\right]=\left[\begin{array}{ll}3 & 1 \\ 8 & 7\end{array}\right]\), then find the value of y.

If a matrix has 5 elements, then write all possible orders it can have.

For what value of x, matrix \(\left[\begin{array}{ll} 6-x & 4 \\ 3-x & 1 \end{array}\right]\) is a singular matrix?

If there are two values of a which makes determinant,\(\begin{equation} \Delta=\left|\begin{array}{rrr} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2 a \end{array}\right|=86 \end{equation}\) then find the sum of these numbers.

If \(\left[ \begin{matrix} x & +3y & y \\ 7 & -x & 4 \end{matrix} \right] \)=\(\begin{bmatrix} 4 & -1 \\ 0 & 4 \end{bmatrix}\), find the values of x and y.

Generic 5

4 x 5 = 20

Using properties of determinants, prove the following:
\(\left| \begin{matrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{matrix} \right| \)= 4abc

Prove the following:
\(\left| \begin{matrix} a^{ 2 } & bc & ac+c^{ 2 } \\ a^{ 2 }+ab & b^{ 2 } & ac \\ ab & b^{ 2 }+bc & c^{ 2 } \end{matrix} \right| =4a^{ 2 }b^{ 2 }c^{ 2 }\)

Find the matrix X such that,\(\left[ \begin{matrix} 2 & -1 \\ 0 & 1 \\ -2 & 4 \end{matrix} \right] X=\left[ \begin{matrix} -1 & -8 & -10 \\ 3 & 4 & 0 \\ 10 & 20 & 10 \end{matrix} \right] \)

If \(A=\left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] \), verify that \({ A }^{ 2 }-4A-5l=0\)

Generic 3

6 x 3 = 18

10.

If A is an invertible matrix of order 2, then det (A^–1) is equal to
(i) det (A)
(ii) \(\frac{1}{det(A)}\)
(iii) 1
(iv) 0

11.

If the area of a triangle with vertices (-3,0), (3,0) and (0, k) is 9 sq units, then find the value of k.

12.

If \(A=\left[\begin{array}{rrr} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]\) then find A² - 5A +41 and hence find a matrix X, such that A² - 5A + 41 +X = 0.

13.

Using elementary transformations, find the inverse of matrix \(\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}.\)

14.

If \(A=diag.\left[ \begin{matrix} 3, & -5, & 7 \end{matrix} \right] \)then \(B=\left[ \begin{matrix} -1, & 2, & 4 \end{matrix} \right] ,\) then find \((2A+3B)\) .

15.

If x = -4 is root of \(\Delta=\left|\begin{matrix}x&2&3\\ 1&x&1\\3&2&x\end{matrix}\right|=0\)then find the other two roots.

Multiple Choice Question

15 x 1 = 15

16.

If the area of the tríangle with vertices (-3,0), (3, 0) and (0, k) is 9 sq units, then the value's of k will be

(a)

(b)

土3

(c)

-9

(d)

17.

If A is a square matrix such that A²=A, then (I + A)³ – 7A is

(a)

(b)

I + A

(c)

I - A

(d)

18.

If \(A=\left[a_{i j}\right]\) is a square matrix of order 2 such that \(a_{i j}=\left\{\begin{array}{ll}1, & \text { when } i \neq j \\ 0, & \text { when } i=j\end{array}\right.\), then \(A^2\) is

(a)

\(\left[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right]\)

(b)

\(\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]\)

(c)

\(\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]\)

(d)

\(\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \)

19.

A and B are square matrices of same order. If (A+B)²= A² + B², then

(a)

AB = BA

(b)

AB =-BA

(c)

AB = 0

(d)

BA = 0

20.

If A is a square matrix of order 3 such that the value of | adj A | = 8, then the value of |4^T | is

(a)

\(\sqrt{2}\)

(b)

-\(\sqrt{2}\)

(c)

(d)

2\(\sqrt{2}\)

21.

If \(A=\left[\begin{array}{lll} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right]\) then A⁶ is equal to

(a)

zero matrix

(b)

(c)

(d)

none of these

22.

Asquare matrix A is said to be non-singular, if

(a)

\(|A|=0\)

(b)

\(|A| \neq 0\)

(c)

\(|A|=-1\)

(d)

\(|A|=1\)

23.

For the system of equations 5x + 2y = 4; 7x +3y = 5 the values of x and yare respectively.

(a)

x = 2, y = -3

(b)

x = 2, y = 3

(c)

x = -2, y = -3

(d)

x = -2, y = 3

24.

If \(A=\left[\begin{array}{cc} 2 & 3 \\ -4 & -6 \end{array}\right]\) then which of the following is true?

(a)

\(A(\operatorname{adj} A) \neq|A| I\)

(b)

\(A(\operatorname{adj} A) \neq(\operatorname{adj} A) A\)

(c)

\(A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]\)

(d)

None of the above

25.

Let \(\Delta=\left|\begin{array}{lll} A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1 \end{array}\right| \text { and } \Delta_{1}=\left|\begin{array}{ccc} A & B & C \\ x & y & z \\ z y & z x & x y \end{array}\right|\) then

(a)

\(\Delta_{1}=-\Delta\)

(b)

\(\Delta \neq \Delta_{1}\)

(c)

\(\Delta^{2}-\Delta_{1}=0\)

(d)

None of these

26.

The value of x such that
\(\left[\begin{array}{lll} 1 & 2 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{array}\right]\left[\begin{array}{l} 0 \\ 2 \\ x \end{array}\right]=O, \mathrm{i}\)

(a)

(b)

(c)

-1

(d)

27.

\(A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]\), then if the value of \(\alpha\) is

(a)

\(\frac{\pi}{6}\)

(b)

\(\frac{\pi}{3}\)

(c)

\(\frac{3 \pi}{2}\)

(d)

\(\pi\)

28.

If A and B are square matrices of the sameorder, then (A + B) (A - B) is equal to

(a)

A²-B²

(b)

A² - BA - AB - B²

(c)

A² - B² + BA - AB

(d)

A² - BA + B² + AB

29.

If \(\left[\begin{array}{rr}1 & 2 \\ -2 & -b\end{array}\right]+\left[\begin{array}{ll}a & 4 \\ 3 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 0\end{array}\right]\), then \(a^{2}+b^{2}\) is equal to

(a)

(b)

(c)

(d)

30.

If a matrix has 8 elements, then which of the following will not be a possible order of the matrix?

(a)

1x 8

(b)

2 x 4

(c)

4x2

(d)

4 x 4

Case Study Questions

3 x 4 = 12

31.

Area of a triangle whose vertices are (x₁, y₁), (x_2' y₂) and (x_3, y₃) is given by thedeterminant
\(\begin{equation} \Delta=\frac{1}{2}\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right| \end{equation}\)
Since, area is a positive quantity, so we always take the absolute value of the determinant Δ. Also, the area of the triangle formed by three collinear points is zero.
Based on the above information, answer the following questions
(i) Find the area of the triangle whose vertices are (-2, 6), (3, -6) and (1, 5).

(a) 30 sq. units

(b) 35 sq. units

(c) 40 sq. units

(d) 15.5 sq. units

(ii) If the points (2, -3), (k, -1) and (0, 4) are collinear, then find the value of 4k

(a) 4

(b) \(\begin{equation} \frac{7}{140} \end{equation}\)

(c) 4

(d) \(\begin{equation} \frac{40}{7} \end{equation}\)

(iii) If the area of a triangle ABC, with vertices A(1, 3), B(O, 0) and C(k, 0) is 3 sq. units, then a value of k is

(a) 2

(b) 3

(c) 4

(d) 5

(iv) Using determinants, find the equation of the line joining the points A(1, 2) and B(3, 6).

(a) y = 2x

(b) x = 3y

(c) y = x

(d) 4x-y = 5

(v) If A = (11, 7), B = (5, 5) and C = (-1, 3), then

(a) \(\begin{equation} \Delta A B C \end{equation}\) is scalene triangle	(b) \(\begin{equation} \Delta A B C \end{equation}\) is equilateral triangle
(c) A, B and C are collinear	(d) None of these

32.

If there is a statement involving the natural number n such that
(i) The statement is true for n = 1
(ii) When the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1.
Then, the statement is true for all natural numbers n.
Also, if A is a square matrix of order n, then A² is defined as AA. In general, A^m = AA .... A (m times), where m is any positive integer.
Based on the above information, answer the following questions.
(i) If \(A=\left[\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right]\),then for any positive integer n,

(a) \(A^{n}=\left[\begin{array}{cc} 3 n & -4 n \\ n & -n \end{array}\right]\)

(b) \(A^{n}=\left[\begin{array}{cc} 1+2 n & -4 n \\ n & 1-2 n \end{array}\right]\)

(c) \(A^{n}=\left[\begin{array}{cc} 3 n & -8 n \\ 1 & -n \end{array}\right]\)

(d) \(A^{n}=\left[\begin{array}{cc} 1+3 n & -4 n \\ n & 1-3 n \end{array}\right]\)

(ii) If \(A=\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]\),then |Aⁿ| where \(n \in N\), is equal to

(a) 2ⁿ

(b) 3ⁿ

(c) n

(d) 1

(iii) If \(A=\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right]\), and \(I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\) then which of the following holds for all natural numbers \(n \geq 1\) ?

(a) Aⁿ= nA-(n-1}I

(b) Aⁿ= 2^n-1A-(n-1}I

(c) An = nA+(n-1)I

(d) Aⁿ= 2^n-1A+(n-1}I

(iv) Let \(A=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right]\) , and \(A^{n}=\left[a_{i j}\right]_{3 \times 3}\) for some positive integer n, then the cofactor of a₁₃ is

(a) aⁿ

(b) -aⁿ

(c) 2aⁿ

(d) 0

(v) If A is a square matrix such that IAI = 2, then for any positive integer n, IAⁿl is equal to

(a) 0

(b) 2n

(c) 2n

(d) n²

33.

Three schools A, Band C organized a mela for collecting funds for helping the rehabilitation of flood victims. They sold hand made fans, mats and plates from recycled material at a cost of Rs. 25, Rs.100 and Rs.50 each. The number of articles sold by school A, B, C are given below.

Artilcle\School	A	B	C
Fans	40	25	35
Mats	50	40	50
Plates	20	30	40

Based on above information, answer the following questions.
(i) If P be a 3 x 3 matrix represent the sale of handmade fans, mats and plates by three schools A, Band C, then

(ii) If Q be a 3 x 1 matrix represent the sale prices (in Rs) of given products per unit, then

(iii) The funds collected by school A by selling the given articles is

(a) Rs. 7000

(b) Rs. 6125

(c) Rs. 7875

(d) Rs. 8000

(iv) The funds collected by school B by selling the given articles is

(a) Rs. 5125

(b) Rs. 6125

(c) Rs. 7125

(d) Rs. 8125

(v) The total funds collected for the required purpose is

(a) Rs. 20000

(b) Rs. 21000

(c) Rs. 30000

(d) Rs. 35000

Assertion and reason

5 x 1 = 5

34.

Assertion adj(adj A) = |A|^n-2. A
Reason |adj A| = |A|^n-1
(a) Both A and R are correct; R is the correct explanation of A
(b) Both A and R are correct; R is not the correct explanation of A
(c) A is correct; R is incorrect
(d) R is correct; A is incorrect

35.

Assertion: If A = \(\begin{bmatrix} 2& 3\\ 1& 2\\ \end{bmatrix}\)and B = \(\begin{bmatrix} 2& -3\\ -1& 2\\ \end{bmatrix}\), then B is the inverse of A.
Reason: If A is a square matrix of order m and if there exists another square matrix B of the same order m, such that AB = BA = I, then B is called the inverse of A.
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct.

36.

For any square matrix A with real number entries, consider the following statements.
Assertion: A + A' is a symmetric matrix.
Reason: A - A' is a skew-symmetric matrix.
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct.

37.

Consider the system
2x + 3y + 6z = 8
x + 2y + 3z = 5
x + y + 3z = 4
Assertion: The above system of equation has no solution.
Reason: detA = 0 and (adj A A)B = 0, where
\(A=\begin{bmatrix} 2& 3& 6\\ 1& 2& 3\\ 1& 1& 3\\ \end{bmatrix}\)and \(B=\begin{bmatrix} 8 \\ 5 \\ 4 \\ \end{bmatrix}\)
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct.

38.

Consider the system
x + y + z = 1
2x + 2y + 2z = 2
4x + 4y + 4z = 3
Assertion: The above system has infinitely many solutions.
Reason: For the above system det A = 0 and (adj A) B = 0, where
\(A=\begin{bmatrix} 1& 1& 1\\ 2& 2& 2\\ 4& 4& 4\\ \end{bmatrix}\) and \(B=\begin{bmatrix} 1 \\ 2 \\ 3 \\ \end{bmatrix}\)
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct.

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Answers Key:

Generic 2

5 x 2 = 10

Given \(\left[\begin{array}{cc}x & x-y \\ 2 x+y & 7\end{array}\right]=\left[\begin{array}{ll}3 & 1 \\ 8 & 7\end{array}\right]\)
On equating the corresponding elements, we get
x = 3 and x - y = 1 ⇒ y = x - 1 = 3 -1 = 2

Given, a matrix has 5 elements. So, possible order of this matrix are 5 x 1 and 1 x 5.

x = 2

We have,\(\begin{equation} \Delta=\left|\begin{array}{rrr} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2 a \end{array}\right|=86 \end{equation}\)
\(\Rightarrow\) 1(2a² +4)- 2(-4a- 20)+0=8
[expanding along first column]
\(\Rightarrow 2 a^{2}+4+8 a+40=86\)
\(\Rightarrow\ 2 a^{2}+8 a+44-86=0\)
\(\Rightarrow a^{2}+4 a-21=0\)
\(\Rightarrow a^{2}+7 a-3 a-21=0\)
\(\Rightarrow (a+7)(a-3)=0\)
\(\Rightarrow a=-7\ and\ 3\)
\(\therefore\) Required sum = -7 + 3 = - 4

\(\left[ \begin{matrix} x & +3y & y \\ 7 & -x & 4 \end{matrix} \right] =\begin{bmatrix} 4 & -1 \\ 0 & 4 \end{bmatrix}\)
\(\Rightarrow x+3y=4;y=-1;7-x=0 \)
\(\Rightarrow x=7,y=-1\)

Generic 5

4 x 5 = 20

\(\Delta =\left| \begin{matrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{matrix} \right| \)
Using R₁--->R₁-R₂-R₃
= \(\left| \begin{matrix} 0 & -2c & -2b \\ b & c+a & b \\ c & c & a+b \end{matrix} \right| \)
R₂ --> cR₂, R₃--> bR₃ and taking common (-2) from R₁
= \(-\frac { 2 }{ bc } \left| \begin{matrix} 0 & c & b \\ bc & { c }^{ 2 }+ac & bc \\ bc & bc & ab+{ a }^{ 2 } \end{matrix} \right| \)
Taking common bc from C₁, c from C₂ & b from C₃
= \(2bc\left| \begin{matrix} 0 & 1 & 1 \\ 1 & c+a & c \\ 1 & b & a+b \end{matrix} \right| \)
= \(-2bc\left| \begin{matrix} 0 & 0 & 1 \\ 1 & a & c \\ 1 & -a & a+b \end{matrix} \right| \)
Expanding along R₁, we get
\(\Delta \)= -2bc(-2a)
\(\Delta \)= 4abc

\(\Delta =\left| \begin{matrix} a^{ 2 } & bc & ac+c^{ 2 } \\ a^{ 2 }+ab & b^{ 2 } & ac \\ ab & b^{ 2 }+bc & c^{ 2 } \end{matrix} \right| \)
Taking a,b ad c common from C₁, C₂ and C₃ respectively
\(=abc\left| \begin{matrix} a & c & a+c \\ a+b & b & a \\ b & b+c & c \end{matrix} \right| \)
Using \(C_{ 1 }\rightarrow C_{ 1 }+C_{ 2 }-C_{ 3 }\)
\(=abc\left| \begin{matrix} 0 & c & a+c \\ 2b & b & a \\ 2b & b+c & c \end{matrix} \right| \)
Using \(R_{ 2 }\rightarrow R_{ 2 }-R_{ 3 }\)
\(=abc\left| \begin{matrix} 0 & c & a+c \\ 0 & -c & a-c \\ 2b & b+c & c \end{matrix} \right| \)
Expanding along C₁,
\(\Delta =abc\left[ 0-0+2b(ca-c^{ 2 }+ca+c^{ 2 } \right] \)
\(\Rightarrow \Delta =abc(4abc)\Rightarrow \Delta =4a^{ 2 }b^{ 2 }c^{ 2 }\)

\(\text { Let } X=\left[\begin{array}{lll} a & b & c \\ d & e & f \end{array}\right] \)
\({\left[\begin{array}{rr} 2 & -1 \\ 0 & 1 \\ -2 & 4 \end{array}\right]\left[\begin{array}{lll} a & b & c \\ d & e & f \end{array}\right]=\left[\begin{array}{rrr} -1 & -8 & -10 \\ 3 & 4 & 0 \\ 10 & 20 & 10 \end{array}\right]} \)
\(\Rightarrow\left[\begin{array}{ccc} 2 a-d & 2 b-e & 2 c-f \\ d & e & f \\ -2 a+4 d & -2 b+4 e & -2 c+4 f \end{array}\right]\)
\(X=\left[ \begin{matrix} 1 & -2 & -5 \\ 3 & 4 & 0 \end{matrix} \right] \)

\(A^{2}-4 A-5 I\)
\(=\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]-\left[\begin{array}{lll} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{array}\right]-\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]\)
\(=\left[\begin{array}{ccc} 1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1 \end{array}\right]\)
\(-\left[\begin{array}{lll} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{array}\right]-\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]\)
\(=\left[\begin{array}{lll} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{array}\right]-\left[\begin{array}{lll} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{array}\right]-\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]\)
\(=\left[\begin{array}{lll} 9-4-5 & 8-8-0 & 8-8-0 \\ 8-8-0 & 9-4-5 & 8-8-0 \\ 8-8-0 & 8-8-0 & 9-4-5 \end{array}\right]\)
\(=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
\(\text { Hence, } A^{2}-4 A-5 I=O\)

Generic 3

6 x 3 = 18

10.

\(\text { Since } A \text { is an invertible matrix, }A^{-1} \text { exists and } A^{-1}=\frac{1}{|A|} \text { adjA. }\)
\(\text { As matrix } A \text { is of order } 2, \text { let } A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \text { . }\)
\(\text { Then, }|A|=a d-b c \text { and } a d j A=\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right] \text { . }\)
\(\text { Now, }A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\left[\begin{array}{ll} \frac{d}{|A|} & \frac{-b}{|A|} \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{array}\right]\)
\(\therefore \operatorname{det}\left(A^{-1}\right)=\frac{1}{\operatorname{det}(A)}\)
\(\text { Hence, the correct answer is } B \text { . }\)

11.

Given area of a triangle with vertices (-3, 0), (3, 0) and (0, k) is 9 sq units.
We have, \(\frac{1}{2}\left|\begin{array}{ccc} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{array}\right|=\pm 9 \Rightarrow\left|\begin{array}{ccc} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{array}\right|=\pm 18\)
\( \Rightarrow -3(0-k)-0+1(3 k-0)=\pm 18 \)
\(\Rightarrow 3 k+3 k=\pm 18 \Rightarrow 6 k=\pm 18 \)
\(\Rightarrow k=\pm 3 \)

12.

We have, \(A=\left[\begin{array}{rrr} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]\)
\(\therefore A^{2}=\left[\begin{array}{rrr} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]\left[\begin{array}{rrr} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]\)
\(=\left[\begin{array}{lll} 4+0+1 & 0+0-1 & 2+0+0 \\ 4+2+3 & 0+1-3 & 2+3+0 \\ 2-2+0 & 0-1+0 & 1-3+0 \end{array}\right]=\left[\begin{array}{llr} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{array}\right]\)
[multiplying rows by columns]
\(=\left[\begin{array}{rrr} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{array}\right]-5\left[\begin{array}{rrr} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]+4\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \)
\(=\left[\begin{array}{lll} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{array}\right]-\left[\begin{array}{ccc} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{array}\right]+\left[\begin{array}{ccc} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right] \)
\(=\left[\begin{array}{ccc} -5 & -1 & -3 \\ -1 & -7 & -10 \\ -5 & 4 & -2 \end{array}\right]+\left[\begin{array}{ccc} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right]=\left[\begin{array}{ccc} -1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2 \end{array}\right] \)
,Since, A² - SA + 41 + X = 0
ஃA² - 5A + 4I + X - X = O - X
[subtracting matrix X from both sides]
A² - 5A + 41 + 0 = - X
⇒ X = - (A² - SA + 4I)
\(\Rightarrow X=-\left[\begin{array}{rrr}-1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2\end{array}\right]=\left[\begin{array}{rrr}1 & 1 & 3 \\ 1 & 3 & 10 \\ 5 & -4 & -2\end{array}\right]\)

13.

\(We\quad know\quad that\quad A={ I }_{ 2 }A\)
\(i.e.\quad \begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}A\)
\( \Rightarrow \begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}A \quad [Applying\quad { R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 }]\)
\(\Rightarrow \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ -2 & 3 \end{bmatrix}A\quad [Applying\quad { R }_{ 2 }\rightarrow { R }_{ 2 }-{ 2R }_{ 1 }]\)
\( \Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 7 & -10 \\ -2 & 3 \end{bmatrix}A. \quad [Applying\quad { R }_{ 1 }\rightarrow { R }_{ 1 }-{ 3R }_{ 2 }]\\ \)
\(Hence, { A }^{ -1 }=\begin{bmatrix} 7 & -10 \\ -2 & 3 \end{bmatrix}.\)

14.

We have: \(A=\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7 \end{matrix} \right] \) and\(B=\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix} \right] .\)
\(\therefore \ 2A+3B=2\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7 \end{matrix} \right] +3\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 6 & 0 & 0 \\ 0 & -10 & 0 \\ 0 & 0 & 14 \end{matrix} \right] +\left[ \begin{matrix} -3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 12 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 6-3 & 0+0 & 0+0 \\ 0+0 & -10+6 & 0+0 \\ 0+0 & 0+0 & 14+12 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 26 \end{matrix} \right] \) =diag.\(\left[ \begin{matrix} 3, & -4, & 26 \end{matrix} \right] .\)

15.

\(\Delta=\begin{vmatrix}x+4&x+4&x+4\\1&x&1\\3&2&x \end{vmatrix}\)
\((x+4)\begin{vmatrix} 1&1&1\\1&x&1\\3&2&x\end{vmatrix}\)
\(=(x+4)\begin{vmatrix}1&0&0\\1&x-1&0\\3&-1&x-3 \end{vmatrix}\)
= (x+4) (1)[(x-1) (x-3)+0]
= (x+4) (x-1) (x-3)
Thus \(\Delta=0\Rightarrow x=-4,1,3.\)
Hence the other two roots are 1 and 3.

Multiple Choice Question

15 x 1 = 15

16.

(b)
土3

17.

(d)
I

18.

(d)
\(\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \)

19.

(b)
AB =-BA

20.

(d)
2\(\sqrt{2}\)

21.

22.

(b)
\(|A| \neq 0\)

23.

From the option, we can see only option (a) satisfy both the equations

24.

25.

\( \Delta_{1} =\left|\begin{array}{ccc} A & B & C \\ x & y & z \\ z y & z x & x y \end{array}\right| =\left|\begin{array}{ccc} A & x & z y \\ B & y & z x \\ C & z & x y \end{array}\right| \\ =\frac{1}{x y z}\left|\begin{array}{lll} A x & x^{2} & x y z \\ B y & y^{2} & x y z \\ C z & z^{2} & x y z \end{array}\right|=\frac{x y z}{x y z}\left|\begin{array}{ccc} A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1 \end{array}\right|=\Delta \)

26.

27.

Hint \(A+A^{\prime}=I\)I
\(\Rightarrow\left[\begin{array}{cc}2 \cos \alpha & 0 \\ 0 & 2 \cos \alpha\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\)
\(\Rightarrow\)\(2 \cos \alpha=1\)
\(\Rightarrow \cos \alpha=\frac{1}{2}\)
\(\Rightarrow \alpha=\frac{\pi}{3}\)

28.

(A + B) (A - B) = A(A - B) + B(A - B)
= A² - AB + BA - B²

29.

We have, \(\left[\begin{array}{cc}1 & 2 \\ -2 & -b\end{array}\right]+\left[\begin{array}{ll}a & 4 \\ 3 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 0\end{array}\right]\)
= \(\left[\begin{array}{cc}a+1 & 6 \\ 1 & 2-b\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 0\end{array}\right]\)
\(\Rightarrow\)a + 1 = 5, 2 - b = 0
\(\Rightarrow\)a = 4, b = 2
\(\Rightarrow\)\(a^{2}+b^{2}=20\)

30.

We know that if a matrix is of order m x n, then it has mn elements. Thus, to find all possible orders of a matrix with 8 elements, we will find all ordered pairs of natural numbers, whose product is 8. Thus, all possible ordered pair are (1,8), (8, I), (2, 4), (4, 2).

Case Study Questions

3 x 4 = 12

31.

(i) (d) : Let be the area of the triangle then,

\(\begin{equation} =\frac{1}{2}|-2(-6-5)-6(3-1)+1(15+6)| \end{equation}\) [Expanding along R_1]
\(\begin{equation} \Rightarrow \quad \Delta=\frac{1}{2}|43-12|=15.5 \mathrm{sq} . \text { units } \end{equation}\)
(ii) (d) : The given points are collinear.
\(\begin{equation} \therefore \frac{1}{2}\left|\begin{array}{ccc} 2 & -3 & 1 \\ k & -1 & 1 \\ 0 & 4 & 1 \end{array}\right|=0 \end{equation}\)
Expanding along R₁',we get
\(\begin{equation} 2(-1-4)+3(k)+1(4 k)=0 \end{equation}\)
\(\begin{equation} \Rightarrow 7 k-10=0 \Rightarrow k=\frac{10}{7} \Rightarrow 4 k=\frac{40}{7} \end{equation}\)
(iii) (a) : Area of \(\begin{equation} \Delta A B C=3 \text { sq. units } \end{equation}\) [Given]
\(\begin{equation} \Rightarrow \frac{1}{2}\left|\begin{array}{lll} 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 \end{array}\right|=\pm 3 \Rightarrow\left|\begin{array}{lll} 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 \end{array}\right|=\pm 6 \end{equation}\)
\(\begin{equation} \Rightarrow 1(0-0)-3(0-k)+1(0-0)=\pm 6 \end{equation}\)
\(\begin{equation} \Rightarrow 3 k=\pm 6 \Rightarrow k=\pm 2 . \end{equation}\)
(iv) (a) : Let Q(x, y) be any point on the line joining
A(1, 2) and B(3, 6). Then, area of \(\begin{equation} \Delta A B Q=0 \end{equation}\)
\(\begin{equation} \Rightarrow \frac{1}{2}\left|\begin{array}{lll} 1 & 2 & 1 \\ 3 & 6 & 1 \\ x & y & 1 \end{array}\right|=0 \end{equation}\)
\(\begin{equation} \Rightarrow 1(6-y)-2(3-x)+1(3 y-6 x)=0 \end{equation}\)
\(\begin{equation} \Rightarrow 6-y-6+2 x+3 y-6 x=0 \end{equation}\)
\(\begin{equation} \Rightarrow-4 x=-2 y \Rightarrow 2 x=y \end{equation}\)
(v) (c) : Area of ΔABC is given by
\(\begin{equation} \frac{1}{2}\left|\begin{array}{rrr} 11 & 7 & 1 \\ 5 & 5 & 1 \\ -1 & 3 & 1 \end{array}\right|=\frac{1}{2}[11(5-3)-7(5+1)+1(15+5)] \end{equation}\)
\(\begin{equation} =\frac{1}{2}[22-42+20]=0 \end{equation}\)
\(\therefore\) Points are collinear

32.

(i) (b) : we have \(A=\left[\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right]\)
\(\therefore A^{2}=\left[\begin{array}{rr} 3 & -4 \\ 1 & -1 \end{array}\right]\left[\begin{array}{rr} 3 & -4 \\ 1 & -1 \end{array}\right]=\left[\begin{array}{cc} 5 & -8 \\ 2 & -3 \end{array}\right]\) ,
which can be obtained from \(A^{n}=\left[\begin{array}{cc} 1+2 n & -4 n \\ n & 1-2 n \end{array}\right]\) for n = 2.
(ii) (d) : We have, \(A=\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]\)
\(\therefore|A|=\left|\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right|=1-0=1\)
Also, IAⁿl = IA .A A(n times)| = lAiⁿ = 1ⁿ = 1
(iii) (a) : For n = 1, all options are true
\(A^{2}=A \cdot A=\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right]\)
and \(A^{3}=A^{2} \cdot A=\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 3 & 1 \end{array}\right]\)
Putting n = 3, in (a), we get A³ = 3A - 2I
\(=3\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right]-\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]\)
\(=\left[\begin{array}{ll} 3 & 0 \\ 3 & 3 \end{array}\right]-\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 3 & 1 \end{array}\right]\) ,which is true
All other options are different' from A³ = 3A - 21 for n = 3.
(iv) (d) : we have \(A=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right]\)
\(\therefore A^{2}=A \cdot A=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right]\left[\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right]\)
\(=\left[\begin{array}{ccc} a^{2} & 0 & 0 \\ 0 & a^{2} & 0 \\ 0 & 0 & a^{2} \end{array}\right]\)
Similarly, \(A^{n}=\left[\begin{array}{ccc} a^{n} & 0 & 0 \\ 0 & a^{n} & 0 \\ 0 & 0 & a^{n} \end{array}\right]\)
Now, cofactor of \(a_{13}=(-1)^{1+3}\left|\begin{array}{cc} 0 & a^{n} \\ 0 & 0 \end{array}\right|=0\)
(v) (c) : We have, IAI = 2
and IAⁿl = IA . A A(n-times)|
= IAIIAI IAI(n-times) = lA|ⁿ = 2n

33.

(i) (a) : Clearly,

(ii) (d) : Since Q is a 3 x 1 matrix, therefore

(iii) (a) : Clearly, total funds collected by each school is given by the matrix.
\(P Q=\left[\begin{array}{ccc} 40 & 50 & 20 \\ 25 & 40 & 30 \\ 35 & 50 & 40 \end{array}\right]\left[\begin{array}{c} 25 \\ 100 \\ 50 \end{array}\right]\)
\(=\left[\begin{array}{c} 1000+5000+1000 \\ 625+4000+1500 \\ 875+5000+2000 \end{array}\right]=\left[\begin{array}{l} 7000 \\ 6125 \\ 7875 \end{array}\right]\)
\(\therefore\) Funds collected by school A is Rs. 7000
Funds collected by school B is Rs. 6125
Pimds collected by school C is Rs. 7875
(iv) (b)
(v) (b) : Total funds collected for the required purpose
= Rs. (7000 + 6125 + 7875) = Rs. 21000