Question Set C

CBSE 12th Standard Maths Determinants HOT Questions Study Material - QB365

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Questions:

Generic 2

5 x 2 = 10

Area of a triangle with vertices (k, 0), (1, 1) and (0, 3) is 5 sq units. Find the value(s) of k.

The sale figure of three car dealers during March 2015 showed that dealer A solds 6 deluxe, 4 premium and 5 standard cars, dealer B solds 8 deluxe, 3 premium and 4 standard cars and : dealer Csolds 4 deluxe, 2 premium and 3 standard cars. Write 3 x 3 matrices summarising sales data for March.

If \(A=\left[ \begin{matrix} 1 & 4 \\ 3 & 2 \\ 2 & 1 \end{matrix} \right] B=\left[ \begin{matrix} 5 & 2 \\ -1 & 0 \\ 1 & 1 \end{matrix} \right] \), then find the matrix X for which A + B - X = 0.

(Uniqueness of inverse) Inverse of a square matrix, if it exists, is unique

If A is a \(3\times 3\) matrix, whose elements are given by \({ a }_{ ij }=\frac { 1 }{ 3 } \left| -3i+j \right| \), then write the value \({ a }_{ 23 }\).

Generic 3

5 x 3 = 15

Find the minors and cofactors of the elements of first row of determinant \(\left|\begin{array}{rrr} 1 & 2 & 0 \\ 3 & 5 & -1 \\ 4 & 7 & 8 \end{array}\right|\)

Let A \(A=\left[\begin{array}{ll} 0 & 1 \\ 1 & 2 \end{array}\right] \text { and } f(x)=x^{2}+x-1\) then findf (A).

For the matrix \(A=\begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}\), verify that:
(i) \(A+A\prime \)is a symmetric matrix
(ii) \(A-A\prime \)is a skew-symmetric matrix.

Find the co - factors of the elements of the determinant: \(\left| \begin{matrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{matrix} \right| \) and verify that a₁₁ A₃₁+ a₁₂A₃₂ + a₁₃A₃₃ = 0.

10.

If \(A'=\begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}\) and \(B=\begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}\) then find (A+2B)'

Generic 5

1 x 5 = 5

11.

Determine the product \(\left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] ,\) and use it to solve the system of equations: x - y + z = 4, x - 2y - 2z = 9, 2x + y + 3z = 1.

Multiple Choice Question

9 x 1 = 9

12.

Let A be a 3 x 3 matrix such that \(|\operatorname{adj} A|=64\). Then, |A| is equal to

(a)

8 only

(b)

-8 only

(c)

(d)

8 or -8

13.

If \(\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|\), then the possible value(s) of x is/are

(a)

(b)

\(\sqrt{3}\)

(c)

\(-\sqrt{3}\)

(d)

\(\sqrt{3},-\sqrt{3}\)

14.

Let A be a skew- symmetric matrix of order 3 . If |A| = x, then (2023)^x is equal to

(a)

2023

(b)

\(\frac{1}{2023}\)

(c)

\((2023)^2\)

(d)

15.

If \(A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right], B=\left[\begin{array}{ll}x & 0 \\ 1 & 1\end{array}\right]\) and \(A=B^2\), then x equals

(a)

\(\pm 1\)

(b)

-1

(c)

(d)

16.

lf A is a square matrix of order 2 and |A| = -2, then value of |5A'| is

(a)

-50

(b)

-10

(c)

(d)

17.

The adjoint of the matrix \(A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\) is

(a)

\(\left[\begin{array}{ll} 4 & 2 \\ 3 & 1 \end{array}\right]\)

(b)

\(\left[\begin{array}{rr} -4 & 2 \\ 3 & -1 \end{array}\right]\)

(c)

\(\left[\begin{array}{rr} 4 & -2 \\ -3 & 1 \end{array}\right]\)

(d)

\(\left[\begin{array}{rr} 1 & -2 \\ -3 & 4 \end{array}\right]\)

18.

If \(\Delta=\left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right|\) and A_ij is cofactor of a_ij, then value of Δ is given by

(a)

a₁₁A₃₁+a₁₂+A₃₂+a₁₃A₃₃

(b)

a₁₁A₁₁ + a₁₂A₂₁+ a₁₃A₃₁

(c)

a₂₁A₁₁ + a₂₂A₁₂ + a₂A₁₃

(d)

a₁₁A₁₁ + a₂₁A₂₁+ a₃₁A₃₁

19.

If \(\Delta=\left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|\) then the cofactor A₂₁ is

(a)

-(he + fg)

(b)

fg -hc

(c)

fg + hc

(d)

hc-fg

20.

The value \(\left| \begin{matrix} 6 & 0 & -1 \\ 2 & 1 & 4 \\ 1 & 1 & 3 \end{matrix} \right| \) is

(a)

-7

(b)

(c)

(d)

Case Study Questions

1 x 4 = 4

21.

In a city there are two factories A and B. Each factory produces sports clothes for boys and girls. There are three types of clothes produced in both the factories, type I, II and III. For boys the number of units of types I, II and III respectively are 80, 70'and 65 in factory A and 85, 65 and 72 are in factory B. For girls the number of units of types I, II and III respectively are 80, 75, 90 in factory A and 50, 55, 80 are in factory B.

Based on the above information, answer the following questions.
(i) If P represents the matrix of number of units of each type produced by factory A for both boys and girls, then P is given by

(ii) If Q represents the matrix of number of units of each type produced by factory B for both boys and girls, then Q is given by

(iii) The total- production of sports clothes of each type for boys is given by the matrix

(iv) The total production of sports clothes of each type for girls is given by the matrix

(v) Let R be a 3 x 2 matrix that represent the total production of sports clothes of each type for boys and girls, then transpose of R is(iv) The total production of sports clothes of each type for girls is given by the matrix

Assertion and reason

2 x 1 = 2

22.

Assertion: The order of the matrix A is 3 x 5 and that of B is 2 x 3.Then the matrix AB is not possible.
Reason: No. of columns in A is not equal to no. of rows in B.
(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.

23.

Assertion: The possible dimensions of a matrix containing 32 elements is 6.
Reason: The No. of ways of expressing 32 as a product of two positive integers is 6.
(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

\(k = -\frac{7}{2},\frac{13}{2}\)

\(\left.\begin{array}{l} \text { Deluxe } & \text { Premium } & \text { Standard } \\ A & 6 & 4 & 5 \\ B & 8 & 3 & 4 \\ \text { C } & 4 & 2 & 3 \end{array}\right] \text { . } \)

We have A + B - X = 0
By adding X on both the sides,
A + B - X + X = 0 + X
\(\Rightarrow\) A + B = X
\(\Rightarrow X=\left[ \begin{matrix} 1 & 4 \\ 3 & 2 \\ 2 & 1 \end{matrix} \right] +\left[ \begin{matrix} 5 & 2 \\ -1 & 0 \\ 1 & 1 \end{matrix} \right] \)
\(\Rightarrow X=\left[ \begin{matrix} 6 & 6 \\ 2 & 2 \\ 3 & 2 \end{matrix} \right] \)

Let A = [a_ij] be a square matrix of order m. If possible, let B and C be two
inverses of A. We shall show that B = C.
Since B is the inverse of A
AB = BA = I ... (1)
Since C is also the inverse of A
AC = CA = I ... (2)
Thus B = BI = B (AC) = (BA) C = IC = C

Given \({ a }_{ ij }=\frac { 1 }{ 3 } \left| -3i+j \right| \)
\(\therefore \ { a }_{ 23 }=\frac { 1 }{ 3 } \left| -6+3 \right| =1\)

Generic 3

5 x 3 = 15

The minors and cofactors of the elements of first row of determinat \(\left|\begin{array}{rrr} 1 & 2 & 0 \\ 3 & 5 & -1 \\ 4 & 7 & 8 \end{array}\right|\)
\(M_{11}=\left|\begin{array}{rr} 5 & -1 \\ 7 & 8 \end{array}\right|=40+7=47\) [minor of an element a₁₁]
\(M_{12}=\left|\begin{array}{rr} 3 & -1 \\ 4 & 8 \end{array}\right|=24+4=28\) [minor of an element a₁₂]
\(M_{13}=\left|\begin{array}{ll} 3 & 5 \\ 4 & 7 \end{array}\right|=21-20=1\) [minor of an element a₁₃]
\(C_{11}=(-1)^{1+1} M_{11}=(-1)^{2} \times 47=47\) [cofactor of an element a₁₁]
\(C_{12}=(-1)^{1+2} M_{12}=(-1)^{3} \times 28=-28\) [cofactor of an element a₁₂]
\(C_{13}=(-1)^{1+3} M_{13}=(-1)^{4} \times 1=1\) [cofactor of an element a₁₃]

We have, \(A=\left[\begin{array}{ll} 0 & 1 \\ 1 & 2 \end{array}\right]\)
\(\therefore A^{2}=A \cdot A=\left[\begin{array}{ll} 0 & 1 \\ 1 & 2 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{ll} 0+1 & 0+2 \\ 0+2 & 1+4 \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ 2 & 5 \end{array}\right]\)
[multiplying rows by columns]
Now,
\(f(A)=A^{2}+A-I=\left[\begin{array}{ll} 1 & 2 \\ 2 & 5 \end{array}\right]+\left[\begin{array}{ll} 0 & 1 \\ 1 & 2 \end{array}\right]-\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\)
\(=\left[\begin{array}{ll} 1+0-1 & 2+1-0 \\ 2+1-0 & 5+2-1 \end{array}\right]=\left[\begin{array}{ll} 0 & 3 \\ 3 & 6 \end{array}\right]\)

We have: \(A=\begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}\)
\(\therefore\ A\prime =\begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}\)
\((i)\ A+A\prime =\begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}+\begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}\)
\(=\begin{bmatrix} 1+1 & 5+6 \\ 6+5 & 7+7 \end{bmatrix}\)
\(=\begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}\)
which is a symmetric matrix.
\((ii)\ A-A\prime =\begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}-\begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}\)
\(=\begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}+\begin{bmatrix} -1 & -6 \\ -5 & -7 \end{bmatrix}\)
\(=\begin{bmatrix} 1-1 & 5-6 \\ 6-5 & 7-7 \end{bmatrix}\)
\(=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}.\)

\(M_{11}=\begin{vmatrix} 0&4\\5&-7\end{vmatrix}=-0-20=-20\)
\(A_{11}=(-1)^{1+1}M_{11}=(-1)^2(-20)=-20\)
\(M_{12}=\begin{vmatrix}6&4\\1&-7 \end{vmatrix}=-42-4=-46\)
\(A_{12}=(-1)^{1+2}M_{12}=(-1)^3(-46)=(-1)(-46)=46\)
\(M_{13}=\begin{vmatrix}6&0\\1&5 \end{vmatrix}=30-0=30\)
\(A_{13}=(-1)^{1+3}M_{13}=(-1)^4(30)=30\)
\(M_{21}=\begin{vmatrix} -3&5\\5&-7\end{vmatrix}=21-25=-4\)
\(A_{21}=(-1)^{ 2+1}M_{21}=(-1)^3(-4=(-1 )(-4)=4)\)
\(M_{22}=\begin{vmatrix} 2&5\\1&-7\end{vmatrix}=-14-15=-19\)
\(A_{22}=(-1)^{2+2}M_{22}=(1)^4(-19)=-19\)
\(M_{23}=\begin{vmatrix} 2&-3\\1&5\end{vmatrix}=10+3=13\)
\(A_{23}=(-1)^{2+3}M_{23}=(-1)^513=-13\)
\(M_{31}=\begin{vmatrix}-3&5\\0&4 \end{vmatrix}=-12-0=-12\)
\(A_{31}=(-1)^{3+1}M_{31}=(-1)^4(-12)=-12\)
\(M_{32}=\begin{vmatrix} 2&5\\6&4\end{vmatrix}=8-30=-22\)
\(A_{32}=(-1)^{3+2}M_{32}=(-1)^5(-22)=(-1)(-22)=22\)
\(M_{33}=\begin{vmatrix} 2&-3\\6&0\end{vmatrix}=0+18=18\)
\(A_{33}=(-1)^{3+3}M_{33}=(-1)^6(18)=18\)
(ii)\(a_{11}A_{31}+a_{12}A_{32}+a_{13}A_{32}\)
\(=(2)(-12)+(-3)(22)+(5)(18)=-24-66+90=0\)

10.

\(A^{\prime}=\left[\begin{array}{rr} -2 & 3 \\ 1 & 2 \end{array}\right], B=\left[\begin{array}{rr} -1 & 0 \\ 1 & 2 \end{array}\right]\)
\(A^{\prime}=\left[\begin{array}{rr} -2 & 3 \\ 1 & 2 \end{array}\right] \Rightarrow A=\left[\begin{array}{rr} -2 & 1 \\ 3 & 2 \end{array}\right] \)
\(B=\left[\begin{array}{rr} -1 & 0 \\ 1 & 2 \end{array}\right] \Rightarrow 2 B=\left[\begin{array}{rr} -2 & 0 \\ 2 & 4 \end{array}\right]\)
\(\text { On adding }(i) \text { and }(i i), \text { we get }\)
\((A+2 B) =\left[\begin{array}{rr} -2 & 1 \\ 3 & 2 \end{array}\right]+\left[\begin{array}{rr} -2 & 0 \\ 2 & 4 \end{array}\right] \)
\(=\left[\begin{array}{rr} -4 & 1 \\ 5 & 6 \end{array}\right] \)
\((A+2 B)^{\prime} =\left[\begin{array}{rr} -4 & 5 \\ 1 & 6 \end{array}\right]\)

Generic 5

1 x 5 = 5

11.

\(\left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} -4+4+8 & 4-8+4 & -4-8+12 \\ -7+1+6 & 7-2+3 & -7-2+9 \\ 5-3-2 & -5+6-1 & 5+6-3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{matrix} \right] =8I\)
where I is the identity matrix
Let \(A=\left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \)
and \(B=\left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \)
\(\Rightarrow AB=8I\)
Post Multiplyingbith sides by B ^-1, we get
\(AB{ B }^{ -1 }=8I{ B }^{ -1 }\)
\(\Rightarrow A=8{ B }^{ -1 }\)
\(\Rightarrow { B }^{ -1 }=\frac { A }{ 8 } \)
Given Equations are:
x - y + z = 4
x - 2y - 2z = 9
and 2x + y + 3z = 1
\(\left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \)
\(\Rightarrow BX=C\)
where \(X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \ and\ C=\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \)
\(\Rightarrow X={ B }^{ -1 }C\)
Using \({ B }^{ -1 }=\frac { A }{ 8 } \)
\(\Rightarrow X=\frac { A }{ 8 } C\)
\(=\frac { 1 }{ 8 } \left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \)
\(=\frac { 1 }{ 8 } \left[ \begin{matrix} -16+36+4 \\ -28+9+3 \\ 20-27-1 \end{matrix} \right] \)
\(=\frac { 1 }{ 8 } \left[ \begin{matrix} 24 \\ -16 \\ -8 \end{matrix} \right] =\left[ \begin{matrix} 3 \\ -2 \\ -1 \end{matrix} \right] \)
(x, y, z) = (3, -2, -1)

Multiple Choice Question

9 x 1 = 9

12.

(d)
8 or -8

13.

(d)
\(\sqrt{3},-\sqrt{3}\)

14.

(d)
1

15.

16.

(a)
-50

17.

18.

△ Sum of product of elements of any row (or colunm) with their corresponding cofactors

19.

\(A_{21}=(-1)^{2+1} M_{21}=-M_{21}=-\left|\begin{array}{ll} h & g \\ f & c \end{array}\right|\)

20.

Δ = 6(-1)- 1(1) = -7

Case Study Questions

1 x 4 = 4

21.

(I) (d) : In factory A, number of units of types I, II and III for boys are 80, 70, 65 respectively and for girls number of units of types I, II and III are 80, 75, 90 respectively.

(ii) (a) : In factory B, number of units of types I, II and III for boys are 85, 65, 72 respectively and for girls number of units of types I, II and III are 50, 55, 80 respectively.

(iii) (c) : Let X be the matrix that represent the number of units of each type produced by factory A for boys, and Y be the matrix that represent the number of units of each type produced by factory B for boys.
Then, X = \(\begin{array}{ccc} \text { I } & \text { II } & \text { III } \\ {[170} & 130 & 130] \end{array}\) and Y = \(\begin{array}{ccc} \text { I } & \text { II } & \text { III } \\ {[85} & 65 & 72] \end{array}\)
Now, required matrix = X + Y = [80 70 65] + [85 65 72]
= [165 135 137]
(iv) (a): Required matrix = [80 75 90] + [50 55 80]
= [130 130 170]
(v) (a) : Clearly,R = P+Q
\(=\left[\begin{array}{ll} 80 & 80 \\ 70 & 75 \\ 65 & 90 \end{array}\right]+\left[\begin{array}{ll} 85 & 50 \\ 65 & 55 \\ 72 & 80 \end{array}\right]=\left[\begin{array}{ll} 165 & 130 \\ 135 & 130 \\ 137 & 170 \end{array}\right]\)
\(\therefore \quad R^{\prime}=\left[\begin{array}{lll} 165 & 135 & 137 \\ 130 & 130 & 170 \end{array}\right]\)