MCQ Questions for Class 12 Maths Chapter 4 Determinants

Determinants Class 12 MCQs Questions with Answers

Determinants Class 12 MCQ Questions Question 1.
If A is a square matrix of order 3, such that A(adj A) = 10, then | adj A | is equal to
(A) 1
(B) 10
(C) 100
(D) 101
Answer:
(C) 100

Explanation:
Consider the equation
A(adjA) = |A|I
Here, A (adj A) =10 I
Then, |A| = 10
Since, | adj A| = |A|^n-1
Where n is order of matrix
Here, = |A|^3-1
= 10² = 100

MCQ On Determinants Class 12 Chapter 4 Question 2.
If A is a 3 x 3matrix Such that |A| = 8, then 3|A| equals
(A) 8
(B) 24
(C) 72
(D) 216
Answer:
(D) 216

Explanation:
Here |A| = 8
Then |3A| = 3³|A| = 27 x 8 = 216

MCQ Of Determinants Class 12 Chapter 4 Question 3
If A is skew symmetric matrix of order 3, then the value of |A| is
(A) 3
(B) 0
(C) 9
(D) 27
Answer:
(B) 0

Explanation:
Determinant value of skew I symmetric matrix is always ’O’.

if \(\left|\begin{array}{lll}
2 & 3 & 2 \\
x & x & x \\
4 & 9 & 1
\end{array}\right|\) + 3 = 0, then the value of x is
(A) 3
(B) 0
(C) -1
(D) 1
Answer:
(C) -1

Explanation:
\(\left|\begin{array}{lll}
2 & 3 & 2 \\
x & x & x \\
4 & 9 & 1
\end{array}\right|\) + 3 = 0
On expanding along R₁
2(x – 9x) – 3(x – 4x) + 2(9x – 4x) + 3 = 0
2(-8x)-3(-3x) + 2(5x) + 3 = 0
-16x + 9x + 10x + 3 = 0
3x + 3 = 0

3x = -3
x = \(\frac {3}{3}\)
x = -1

Determinants Class 12 MCQ Chapter 4 Question 5.
Let A = \(\frac {3}{3}\) and B = \(\frac {3}{3}\) then |AB| is equal to …………
(A) 460
(B) 2000
(C) 3000
(D) -7000
Answer:
(D) -7000

Explanation:
A = \(\left[\begin{array}{cc}
200 & 50 \\
10 & 2
\end{array}\right]\)
B = \(\left[\begin{array}{cc}
50 & 40 \\
2 & 3
\end{array}\right]\)
AB = \(\left[\begin{array}{cc}
200 & 50 \\
10 & 2
\end{array}\right]\)
= \(\left[\begin{array}{cc}
50 & 40 \\
2 & 3
\end{array}\right]\)
= \(\left[\begin{array}{cc}
10000+100 & 8000+150 \\
500+4 & 400+6
\end{array}\right]\)
AB = \(\left[\begin{array}{cc}
10100 & 8150 \\
504 & 406
\end{array}\right]\)
|AB| = (10100(406) – (504)(8150)
= 4100600 – 4107600
= -7000

Determinants MCQs With Answers Question 6.
If A = \(\left[\begin{array}{lll}
a & 0 & 0 \\
0 & a & 0 \\
0 & 0 & a
\end{array}\right]\), then det (adj A) equals

(A) a²⁷
(B) a⁹
(C) a⁶
(D) a²
Answer:
(C) a⁶

Explanation:
\(\left[\begin{array}{lll}
a & 0 & 0 \\
0 & a & 0 \\
0 & 0 & a
\end{array}\right]\)
Det (A) = a(a x a – 0 x 0) – 0 + 0 = a³
Ðet(adjA) = (a³)²
= a⁶

Determinants MCQ Maths Chapter 4 Question 7.
If A is any square matrix of order 3 x 3 such that
|A|= 3,then thevalueof adj |A| is?
(A) 3
(B) !
(C) 9
(D) 27
Answer:
(C) 9

Explanation:
|A| = 3,
n = 3
|adj A| = |A| = 3² = 9

Class 12 Maths Chapter 4 MCQ Question 8.
If = \(\left|\begin{array}{cc}
2 x & 5 \\
8 & x
\end{array}\right|=\left|\begin{array}{cc}
6 & -2 \\
7 & 3
\end{array}\right|\) then the value of x is
(A) 3
(B) ±3
(C) ±6
(D) 6
Answer:
(C) ±6

Explanation:
Given that
\(\left|\begin{array}{cc}
2 x & 5 \\
8 & x
\end{array}\right|=\left|\begin{array}{cc}
6 & -2 \\
7 & 3
\end{array}\right|\)
⇒ 2x² – 40 = 18 + 14
⇒ 2x² = 32 + 40
⇒ x² = \(\frac {72}{2}\)
x² = 36
∴ x = ±6

Determinant MCQ Maths Chapter 4 Question 9.
The value of determinant \(\left|\begin{array}{lll}
a-b & b+c & a \\
b-a & c+a & b \\
c-a & a+b & c
\end{array}\right|\) is
(A) a³ + b³ + c³
(B) 3bc
(C) a³ + b³ + c³ – 3abc
(D) None of these
Answer:
(D) None of these

Explanation:
We have

[∴ C₁ → C₁ + C₁ and C₂ → C₂ + C₃]
[a + b + c] \(\left|\begin{array}{lll}
a+c & 1 & a \\
b+c & 1 & b \\
c+b & 1 & c
\end{array}\right|\)
[Taking (a + b + c) common from C₂]
[∴R₂ → R₂ – R₃ and R₂ → R₂ – R₃]
[a + b + c] \(\left|\begin{array}{ccc}
a-b & 0 & a-c \\
0 & 0 & b-c \\
c+b & 1 & c
\end{array}\right|\)
[Expanding a long R₂]
= (a + b + c)(b – c)(a – b)
= (a + b + c)(b – c)(a – b)

Determinants MCQs Class 12 Question 10.
The area of a triangle with vertices (-3,0), (3, 0) and (0, k) is 9 sQuestion units. Then, the value of k will be
(A) 9
(B) 3
(C) – 9
(D) 6
Answer:
(B) 3

Explanation:
We know that, area of a triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃ y₃) is given by
∆ = \(\frac {1}{2}\)\(\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∆ = \(\frac {1}{2}\)\(\left|\begin{array}{ccc}
-3 & 0 & 1 \\
3 & 0 & 1 \\
0 & k & 1
\end{array}\right|\)
[Expanding along R₁]
9 = \(\frac {1}{2}\) [-3(-k)-0 + 1(3k)]
⇒ 18 = 3k + 3k
18 = 6k
∴ k = \(\frac {18}{6}\) = 3

Class 12 Determinants MCQ Maths Question 11.
The determinant \(\left|\begin{array}{lll}
b^{2}-a b & b-c & b c-a c \\
a b-a^{2} & a-b & b^{2}-a b \\
b c-a c & c-a & a b-a^{2}
\end{array}\right|\) is equal to
(A) abc(b – c)(c – a)(a – b)
(B) (b – c)(c – a)(a – b)
(C) (a + b + c)(b – c)(c – a)(a – b)
(D) None of these
Answer:
(D) None of these

Explanation:
We have
\(\left|\begin{array}{lll}
b^{2}-a b & b-c & b c-a c \\
a b-a^{2} & a-b & b^{2}-a b \\
b c-a c & c-a & a b-a^{2}
\end{array}\right|=\left|\begin{array}{lll}
b(b-a) & b-c & c(b-a) \\
a(b-a) & a-b & b(b-a) \\
c(b-a) & c-a & a(b-a)
\end{array}\right|\)

=(b – a)² \(\left|\begin{array}{lll}
b & b-c & c \\
a & a-b & b \\
c & c-a & a
\end{array}\right|\)
[On taking (b – a) common from C₁ and C₃ each]
= (b – a)²\(\left|\begin{array}{lll}
b-c & b-c & c \\
a-b & a-b & b \\
c-a & c-a & a
\end{array}\right|\)
[∵C₂ → C₁ – C₃]
= 0
[Since, two columns C₁ and C₂ are identical, so the value of determinant is zero.]

MCQ Questions On Determinants Class 12 Question 12.
If A = \(\left|\begin{array}{ccc}
2 & \lambda & -3 \\
0 & 2 & 5 \\
1 & 1 & 3
\end{array}\right|\) Then A^-1 exist if
(A) A = 2
(B) λ ≠2
(C) λ ≠ – 2
(D) None of these
Answer:
(D) None of these

Explanation:
Given that,
A = \(\left|\begin{array}{ccc}
2 & \lambda & -3 \\
0 & 2 & 5 \\
1 & 1 & 3
\end{array}\right|\)
Expanding along R₁,
|A| = 2(6 – 5) – (4 – 5) – 3(-2)
= 2 + 5λ + 6
We know that A^-1 exists, If A is non-singular
matrix, i.e., |A|≠ 0
∴ 2 + 5A + 6 ≠ 0
5 λ ≠ -8
∴λ ≠ \(\frac {-8}{5}\)
So,A^-1 exists if and only λ ≠ \(\frac {-8}{5}\)

MCQ Of Chapter 4 Maths Class 12 Question 13.
IfA and B are invertible matrices, then which of the following is not correct?
(A) adj A =|A|.A^-1
(B) det(A’) =[det(A)]^-1
(C) (AB)^-1 = B^-1 A^-1
(D) (A + B)^-1 = B^-1 + A^-1
Answer:
(D) (A + B)^-1 = B^-1 + A^-1

Explanation:
Since, A and B are invertible matrices, so, we can say that
(AB)^-1 = B^-1 A^-1 ………..(i)
Also, A^-1 = \(\frac{1}{|A|}\) (adj A) ………..(ii)
⇒ adj A = A^-1.|A|
Also, det (A)^-1 = [det (A)]^-1

= det(A).det(A)^-1 = 1 ……..(iii)
From equation (iii), we conclude that it is true
Again, (A + B)^-1 = \(\frac{1}{|(A+B)|}\) adj (A + B)
(A + B)^-1 = B^-1 + A^-1 ………..(iv)

MCQ On Determinants Maths Chapter 4 Question 14.
if \(\left|\begin{array}{cc}
x & 2 \\
18 & x
\end{array}\right|=\left|\begin{array}{cc}
6 & 2 \\
18 & 6
\end{array}\right|\) then xis equal to
(A) 6
(B) ±6
(C) -6
(D) 0
Answer:
(B) ±6

Explanation:
\(\left|\begin{array}{cc}
x & 2 \\
18 & x
\end{array}\right|=\left|\begin{array}{cc}
6 & 2 \\
18 & 6
\end{array}\right|\)
⇒ x² – 36 = 36 – 36
⇒ x² – 36 = 0
⇒ x = ±6

MCQ Determinants Class 12 Question 15.
Let A be a non-singular square matrix of order 3 x 3. Then |adi A| is equal to
(A) |A|
(B) |A|²
(C) |A|³
(D) 3 |A|
Answer:
(B) |A|²

Explanation:
We know that,
= (adj A) A = |A|I = \(\left[\begin{array}{ccc}
|A| & 0 & 0 \\
0 & |A| & 0 \\
0 & 0 & |A|
\end{array}\right]\)
= |(adj A) A | = \(\left|\begin{array}{ccc}
|A| & 0 & 0 \\
0 & |A| & 0 \\
0 & 0 & \mid A
\end{array}\right|\)
= | adj A | |A | = |A |³ \(\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\) = |A |³
|adj A | = |A |²

Determinants MCQs Maths Chapter 4 Question 16.
If A is an invertible matrix of order 2, then det (A^-1) is equal to
(A) det (A)
(B) \(\frac {1}{det (A)}\)
(C) 1
(D) O
Answer:
(B) \(\frac {1}{det (A)}\)

Explanation:
Given that A is an invertible matiix, A^-1 exists and A^-1 = \(\frac {1}{|A |}\) adj. A.
As matrix A is of order 2, let A = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Then, |A | = ad – bc and adj A = \(\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right]\)

Now

A^-1 = \(\frac {1}{|A |}\) adj.A = \(\left[\begin{array}{cc}
\frac{d}{|A|} & \frac{-b}{|A|} \\
\frac{-c}{|A|} & \frac{a}{|A|}
\end{array}\right]\)
∴\(\left|A^{-1}\right|\) = \(\left|\begin{array}{cc}
\frac{d}{|A|} & \frac{-b}{|A|} \\
\frac{-c}{|A|} & \frac{a}{|A|}
\end{array}\right|\)
= \(\frac{1}{|A|^{2}}\left|\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right|\)
\(\frac{1}{|A|^{2}} \cdot|A|\)
= \(\frac{1}{|A|}\)

Assertion And Reason Based MCQs (1 Mark each)

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is True

MCQs On Determinants Class 12 Question 1.
Let A be a 2 x 2 matrix.
Assertion (A): adj (adj A) = A
Reason (R): |adj A| = |A|
Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
adj (adjA) = |A|^n-2 A
Here
n = 2 ⇒ adj (adj A) = A
Hence A is true.
| adi A| = |A|^n-1
n = 2 = |adi A|= |A|
Hence R is true.
R is not the correct explanation for A

Ch 4 Maths Class 12 MCQ Determinants Question 2.
Assertion (A): If A = \(\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 4
\end{array}\right]\), then
A^-1 = \(\left[\begin{array}{ccc}
\frac{1}{2} & 0 & 0 \\
0 & \frac{1}{3} & 0 \\
0 & 0 & \frac{1}{4}
\end{array}\right]\)
Reason (R):
The inverse of an invertible diagonal matrix is a diagonal matrix.
Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
|A| = 24
Adj A = \(\left[\begin{array}{ccc}
12 & 0 & 0 \\
0 & 8 & 0 \\
0 & 0 & 6
\end{array}\right]\)

= \(\left[\begin{array}{ccc}
\frac{1}{2} & 0 & 0 \\
0 & \frac{1}{3} & 0 \\
0 & 0 & \frac{1}{4}
\end{array}\right]\)
Hence A is true.
A is a diagonal matrix and its inverse is also
diagonal matrix. Hence R is true.
But R is not the correct explanation of A.

MCQ On Determinants Class 12 Pdf Question 3.
Assertion (A): If every element of a third order determinant of value A is multiplied by 5, then the value of the new determinant is 125 ∆.
Reason (R): If k is a scalar and A is an n x n matrix, then
Answer:
Option (A) is correct.

Explanation:
If k is a scalar and A is an n x n
matrix, then |kA| = kⁿ|A|.
This is a property of the determinant. Hence R is true.
Using this property, |5A| = 5³ ∆ = 125 ∆
Hence A is true.
R is the correct explanation of A.

Class 12 Maths Ch 4 MCQ Determinants Question 4.
Assertion (A): If the matrix A = \(\left[\begin{array}{ccc}
1 & 3 & \lambda+2 \\
2 & 4 & 8 \\
3 & 5 & 10
\end{array}\right]\) is
singular, then λ = 4.
Reason (R): IfA isa singular matrix, then |A| = 0.
Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
A matrix is said to be singular if
|A| = 0.
Hence R is true.
\(\left[\begin{array}{ccc}
1 & 3 & \lambda+2 \\
2 & 4 & 8 \\
3 & 5 & 10
\end{array}\right]\) = 0
⇒ 1(40 – 40) – 3(20 – 24) = 0
0 + 12 + – 2λ – 4 = 0
λ = 4
Hence A is true.
R is the correct explanation for A.

Chapter 4 Maths Class 12 MCQ Question 5.
Given A = \(\left[\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right]\)
Assertion (A): 2A^-1 = 9I – A
Reason (R): A^-1 = \(\frac {1}{|A|}\) (adjA) is true
Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
A^-1 = \(\frac {1}{|A|}\) (adjA) is true.
Hence R is true
|A| = 2,
A^-1 = \(\frac {1}{2}\)\(\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right]\)
LHS = 2A^-1 = \(\frac {1}{2}\)\(\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right]\)
RHS = 9 \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right]\)
= \(\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right]\)
∴ 2A^-1 = 9I – A is true.
R is the correct explanation for A.

Question 6.
Assertion (A): If A = \(\left[\begin{array}{cc}
2 & 3 \\
5 & -2
\end{array}\right]\) and A^-1 = kA, then K = \(\frac {1}{9}\)
Reason (R): \(\left|A^{-1}\right|=\frac{1}{|A|}\)
Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
|A| = -4 – 15 = -19
A^-1 = \(\frac{-1}{19}\)\(\left[\begin{array}{cc}
-2 & -3 \\
-5 & 2
\end{array}\right]\)
= \(\frac{-1}{19}\left[\begin{array}{cc}
-2 & -3 \\
-5 & 2
\end{array}\right]=\left[\begin{array}{cc}
2 k & 3 k \\
5 k & -2 k
\end{array}\right]\)
= K = \(\frac {1}{9}\)
A is false
\(\left|A^{-1}\right|=\frac{1}{|A|}\) is true
R is true

Case-Based MCQs

Attempt any four sub-parts from each question.
Each sub-part carries 1 mark.
I. Read the following text and answer the following questions on the basis of the same:

Manjit wants to donate a rectangular plot o land for a school in his village. When he was asked to give dimensions of the plot, he told that if its length is decreased by 54) m and breadth is increased by 50 m, then its area will remain same, but if length is decreased by 10 m and breadth is decreased by 50 m, then its area will decrease by 5300 m²

Question 1.
The equations in terms of X and Y are
(A) x – y = 50, Zx – y = 550
(B) x – y = 50,2x + y = 550
(C) x + y = 50, 2x + y = 550
(D) x + y= 50,2x + y = 550
Answer:
(B) x – y = 50,2x + y = 550

Question 2.
Which of the following matrix equation is represented by the given information
(A) \(\left[\begin{array}{rr}
1 & -1 \\
2 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{r}
50 \\
550
\end{array}\right]\)
(B) \(\left[\begin{array}{ll}
1 & 1 \\
2 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{r}
50 \\
550
\end{array}\right]\)
(C) \(\left[\begin{array}{rr}
1 & 1 \\
2 & -1
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{r}
50 \\
550
\end{array}\right]\)
(D) \(\left[\begin{array}{ll}
1 & 1 \\
2 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{r}
-50 \\
-550
\end{array}\right]\)
Ans.
(A) \(\left[\begin{array}{rr}
1 & -1 \\
2 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{r}
50 \\
550
\end{array}\right]\)

Question 3.
The value of x (length of rectangular field) is
(A) 150m
(B) 400m
(C) 200m
(D) 320m
Answer:
(C) 200m

Explanation:
We have,
\(\left[\begin{array}{cc}
1 & -1 \\
2 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
50 \\
550
\end{array}\right]\)

Let A = \(\left[\begin{array}{cc}
1 & -1 \\
2 & 1
\end{array}\right]\)
B = \(\left[\begin{array}{c}
50 \\
550
\end{array}\right]\)
X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\)
Now AX = B
X = AB
Adj (A) = \(\left[\begin{array}{cc}
1 & 1 \\
-2 & 1
\end{array}\right]\)
= 1 + 2
= 3

= \(\frac{1}{3}\left[\begin{array}{cc}
1 & 1 \\
-2 & 1
\end{array}\right]\)
= \(\frac{1}{3}\left[\begin{array}{cc}
1 & 1 \\
-2 & 1
\end{array}\right]\left[\begin{array}{c}
50 \\
550
\end{array}\right]\)
= \(\left[\begin{array}{cc}
\frac{1}{3} & \frac{1}{3} \\
\frac{-2}{3} & \frac{1}{3}
\end{array}\right]\left[\begin{array}{c}
50 \\
550
\end{array}\right]\)
= \(\left[\begin{array}{c}
\frac{50}{3}+\frac{550}{3} \\
\frac{-100}{3}+\frac{550}{3}
\end{array}\right]\)
= \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
200 \\
150
\end{array}\right]\)
x = 200
y = 150

Question 4.
The value of y (breadth of rectangular field) is
(A) 150 m
(B) 200m
(C) 430m.
(D) 350m
Ans.
(A) 150 m

Question 5.

How much is the area of rectangular field?
(A) 60000 sq.m.
(B) 30000 sq.m.
(C) 30000m
(D) 3000m
Ans.
(B) 30000 sq.m.

Explanation:
Area of rectangular field
= xy
200 x 150
= 30000 sqm.

II. Read the following text and answer the following questions on the basis of the same:
The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to kept the colony neat and clean. The sum of all the, awardees is 12.

Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. The sum of the number of awardees for honesty and supervision is twice the number of awardees for helping.

Question 1.
x + y + z = ……………
(A) 3
(B) 5
(C) 7
(D) 12
Answer:
(D) 12

Explanation:
x + y + z = 12 ………….(i)
2x + 3y + 3z = 33 ………….(ii)
x – 2y + z = 0 ………..(iii)
A = \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
2 & 3 & 3 \\
1 & -2 & 1
\end{array}\right]\)
B = \(\left[\begin{array}{c}
12 \\
33 \\
0
\end{array}\right]\)
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
|A| =1(3 + 6) -1(2 – 3) + 1(-4-3)
= 9 + 1 – 7
= 3

= \(\frac{1}{3}\left[\begin{array}{ccc}
9 & -3 & 0 \\
1 & 0 & -1 \\
-7 & 3 & 1
\end{array}\right]\)
X = AB
= \(\frac{1}{3}\left[\begin{array}{ccc}
9 & -3 & 0 \\
1 & 0 & -1 \\
-7 & 3 & 1
\end{array}\right]\left[\begin{array}{c}
12 \\
33 \\
0
\end{array}\right]\)
= \(\frac{1}{3}\left[\begin{array}{c}
9 \\
12 \\
15
\end{array}\right]\)
= \(\left[\begin{array}{l}
3 \\
4 \\
5
\end{array}\right]\)
= x = 3, y = 4, z = 5
x + y + z = 12 [from(i)]

Question 2.
x – 2y = …………..
(A) z
(B) – z
(C) 2z
(D) -2z
Answer:
(B) – z

Explanation:
x — =— z [from (iii)I

Question 3.
The value of z is ………..
(A) 3
(B) 4
(C) 5
(D) 6
Answer:
(C) 5

Explanation:
z = 5

Question 4.
The value of x + 2y = …………..
(A) 9
(B) 10
(C) 11
(D) 12
Ans.
(C) 11

Explanation:
x + 2y = 3 + 8 = 11

Question 5.
The value of 2x + 3y + 5z = …………..
(A) 40
(B) 43
(C) 50
(D) 53
Answer:
(B) 43

Explanation:
2x + 3y + 5z = 6 + 12 + 25 = 43

III. Read the following text and answer the following questions. On the basis of the same:
Two schools Oxford and Navdeep want to award their selected students on the values of sincerity, truthfulness and helpfulness. Oxford wants to award ₹ x each, ₹ y each and ₹z each for the three respective values to 3,2 and 1 students respectively with a total award money of ₹ 1600. Navdeep wants to spend ₹ 2300 to award its 4, 1 and 3 students on the respective values (by giving the same amount to the three values as before). The total amount of the award for one prize on each is ₹ 900.

Question 1.
x + y + z = …………..
(A) 800
(B) 900
(C) 1000
(D) 12000
Answer:
(B) 900

Explanation:
From the above information, we have
3x + 2y + z = 1600 …(i)
4x + y + 3z = 2300 …(ii)
x + y + z = 900 …….(iii)
A = \(\left[\begin{array}{lll}
3 & 2 & 1 \\
4 & 1 & 3 \\
1 & 1 & 1
\end{array}\right]\)
B = \(\left[\begin{array}{c}
1600 \\
2300 \\
900
\end{array}\right]\)
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
|A| = 3(1 – 3) -2(4 – 3) +1(4 – 1) = -6 -2 + 3 = -5
A^-1 = \(\frac{1}{-5}\left[\begin{array}{ccc}
-2 & -1 & 5 \\
-1 & 2 & -5 \\
3 & -1 & -5
\end{array}\right]\)
X = A^-1B

= \(\frac{-1}{5}\left[\begin{array}{ccc}
-2 & -1 & 5 \\
-1 & 2 & -5 \\
3 & -1 & -5
\end{array}\right]\left[\begin{array}{c}
1600 \\
2300 \\
900
\end{array}\right]\)
= \(\frac{-1}{5}\left[\begin{array}{l}
-1000 \\
-1500 \\
-2000
\end{array}\right]\)
= \(\left[\begin{array}{l}
200 \\
300 \\
400
\end{array}\right]\)
x = 200, y = 300, z = 400
x + y + z = 900 [from (iii)]

Question 2.
4x + y + 3z = …………..
(A) 1600
(B) 2300
(C) 900
(D) 1200
Answer:
(B) 2300

Explanation:
4x + y + 3z = 2300

Question 3.
The value of y is .
(A) 200
(B) 250
(C) 300
(D) 350
Answer:
(C) 300

Explanation:
4x + y + 3z = 2300 [from (ii)] y = 300

Question 4.
The value of 1x + 3y is ………….
(A) 1000
(B) 1100
(C) 1200
(D) 1300
Answer:
(D) 1300

Explanation:
2x + 3y = 400 + 900 = 1300

Question 5.
y – x = …………
(A) 100
(B) 200
(C) 300
(D) 400
Ans.
(A) 100

Explanation:
y – x = 300 – 200 = 100

MCQ Questions for Class 12 Maths with Answers