RD Sharma Solutions Class 12 Mathematics Chapter 5 FBQ

Edited By Kuldeep Maurya | Updated on Jan 25, 2022 06:54 PM IST

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RD Sharma Class 12 Solutions Chapter 5 FBQ Determinants - Other Exercise

Determinants Excercise:FBQ

Determinants exercise Fill in the blanks question 1

Answer: $\left | A \right |=6$
Hint: Here, we use basic concept of diagonal and determinant of matrix
Given: $\begin{bmatrix} 1 &0 &0 \\ 0&2 &0 \\ 0&0 & 3 \end{bmatrix}$
Solution:
$det(A)=\left | A \right |=\begin{bmatrix} 1 &0 &0 \\ 0&2 &0 \\ 0&0 & 3 \end{bmatrix}$
$=1\begin{bmatrix} 2 & 0\\ 0&3 \end{bmatrix}-0\begin{bmatrix} 0 &0 \\ 0&3 \end{bmatrix}+0\begin{bmatrix} 0 &0 \\ 2&0 \end{bmatrix}$
$=1(6-0)-0(0-0)+0(0-0)$
$=1(6)-0+0$
$\left | A \right |=6$

Determinants exercise Fill in the blanks question 2

Answer: $\left | A \right |=0$
Hint: Here, we use basic concept of skew symmetric matrix and determinant of matrix
Given: A is $3\times 3$ order skew symmetric matrix
Solution: Here, $A_{\left [ 3\times 3 \right ]}$ and $A=-A^{T}$ ( Skew Symmetric )
So, $A=-A^{T}$
$\left | A \right |=-\left | A^{T} \right |$ [Taking determinant both side]
$\left | A \right |=-\left | A \right |\; \; \; \; \; \; \; \;$ $\left [ \left | A^{T} \right |=\left | A \right | \right ]$
$\left | A \right |+\left | A \right |=0$
$2\left | A \right |=0$
$\left | A \right |=0$
So, here determinant of 3 $3\times 3$ skew symmetric matrix always be zero.

Determinants exercise Fill in the blanks question 3

Answer: x = 4
Hint: Here, we use basic concept of singular matrix
Given: $A=\begin{bmatrix} 1 &3 &x+2 \\ 2& 4 & 8\\ 3& 5& 10 \end{bmatrix}$ is singular
Solution: Here, A is regular
So, $\left | A \right |$ is zero
$\left | A \right |=0$
So,
$\begin{bmatrix} 1 &3 &x+2 \\ 2& 4 & 8\\ 3& 5& 10 \end{bmatrix}=0$
$1\begin{bmatrix} 4 & 8\\ 5& 10 \end{bmatrix}-3\begin{bmatrix} 2 & 8\\ 3& 10 \end{bmatrix}+(x+2)\begin{bmatrix} 2 & 4\\ 3& 5 \end{bmatrix}=0$
$1(40-40)-3(20-24)+(x+2)(10-12)=0$
$0-3(-4)+(x+2)(-2)=0$
$12+\left [ -2x-4 \right ]=0$
$8-2x=0$
$8=2x$
$x=\frac{8}{2}=4$

Determinants exercise Fill in the blanks question 4

Answer: $K^{2}$
Hint: Here, we use basic concept of singular matrix
Given: A and B non singular square matrix of order ‘n’
Such that A = KB
Solution: Here, A and B are non singular
And both are n order of singular matrix
Here
$A=KB$
$\left | A \right |=\left | K \right |\left | B \right |$ (Let's take both side determinate)
$\left | A \right |=K^{n}\left | B \right |$
$\frac{\left | A \right |}{\left |B \right |}=K^{n}$ [A and B are $n\times n$ matrix]
So, $\frac{\left | A \right |}{\left |B \right |}=K^{n}$

Determinants exercise Fill in the blanks question 6

Answer: $\left | 5A \right |=125$
Hint: Here, we use basic concept $\left | KA \right |=K^{n}\left | A \right |$
Given: Here, A is $2\times 2$ matrix
So, n = 2
$\left | A \right |=5$
Solution: Here, $n=2$ and $\left | A \right |=5$
So, $\left |5 A \right |=?$
So, here 5 is just like k
$\left |K A \right |=K^{n}\left | A \right |$
$\left |5 A \right |=5^{2}\times 5\; \; \; \; \; \; \; \; \left [ K=5,\left | A \right |=5,n=2 \right ]$
$\left |5 A \right |=125$
$\left |5 A \right |$ is equal to 125.

Determinants exercise Fill in the blanks question 5

Answer: $R-\left | 1 \right |$
Hint: Here, we use basic concept of non-singular matrix
Given: $A=\begin{bmatrix} a & 2\\ 2&4 \end{bmatrix}$ is non-singular
Solution: Here, A is non-singular
So, $\left | A \right |\neq 0$
$\begin{bmatrix} a & 2\\ 2&4 \end{bmatrix}\neq 0$
$(a(4))-(2)(2)\neq 0$
$4a-4\neq 0$
$4a\neq 4$
$a\neq 1$
So, $a$ should be in $R-\left | 1 \right |$

Determinants exercise Fill in the blanks question 7

Answer: $\left | 3AB \right |=-81$
Hint: Here, we use basic concept of $\left | KA \right |=K^{n}\left | A \right |$
Given: A and B are square matrix of order 3. So, n = 3
$\left | A \right |=(-1), \; \; \left | B \right |=3$
Solution: Here, $\left | 3AB \right |$
Here 3 is just like K. So,
$\left | KAB \right |=K^{n}\left | A \right |\left | B \right |$
$\left | 3AB \right |=3^{3}\times -1(3)\; \; \; \; \; \; \; \left [ n=3,\left | A \right | =-1,\left | B \right |=3\right ]$
$\left | 3AB \right |=-81$
So, here $\left | 3AB \right |=-81$

Determinants exercise Fill in the blanks question 8

Answer: $\partial =\pm 3$
Hint: Here, we use basic concept of determinant of matrix
Given: $\left | A \right |^{3}=125$ , $A=\begin{bmatrix} \partial & 2\\ 2& \partial \end{bmatrix}$
Solution: Here, $A=\begin{bmatrix} \partial & 2\\ 2& \partial \end{bmatrix}$
So, $\left | A \right |=\partial \times \partial -2\times 2$
$\left | A \right |=\partial^{2} -4$ ........ $(1)$
And, $\left | A \right |^{3}=125$
$\left ( \left | A \right | \right )^{3}=125$
$\left ( \left | A \right | \right )^{3}=(5)^{3}$
$\left | A \right |=5$ ........... $(2)$
From equation (1) and equation (2)
$\partial ^{2}-4=5$
$\partial ^{2}=4+5$
$\partial ^{2}=9$
$\partial=\pm 3$

Determinants exercise Fill in the blanks question 9

Answer: $x=e^{2}$
Hint: Here, we use basic concept of determinant of matrix
Given: $A=\left[\begin{array}{cc} \log x & -1 \\ -\log x & 2 \end{array}\right]$
$det (A) = 2$
Solution: Here, $A=\left[\begin{array}{cc} \log x & -1 \\ -\log x & 2 \end{array}\right]$
$\begin{aligned} &|A|=\log x \times 2-(-1 \times-\log x) \\ &=2 \log x-\log x \\ &|A|=\log x \end{aligned}$
$\begin{aligned} &\text { And }|A|=2 \text { (Given) }\\ &\text { So, }|A|=\log x=2 \end{aligned}$
$\begin{aligned} &e^{\log x}=e^{2} \\ &x=e^{2}\left[e^{\log x}=x\right] \end{aligned}$

Determinants exercise Fill in the blanks question 10

Answer: 1
Hint: Here, we use basic concept of identity matrix and determinant of matrix
Given: Here I is order of 10 and identity matrix
Solution: Here, I is identity matrix and it’s order is 10.
Any order of identity matrix is determinant of matrix be 1
Let I is $2\times 2$ matrix
$\begin{aligned} &I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ &|I|=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=1-0=1 \\ &S o,\left|I_{10}\right|=1 \end{aligned}$

Determinants exercise Fill in the blanks question 11

Answer: $8\left | A \right |$
Hint: Here, we use basic concept of determinant of matrix $\left | KA \right |=K^{n}\left | A \right |$
Given: A is $3\times 3$ matrix. So, n = 3
Solution: Here A is $3\times 3$ matrix. So, n = 3
$\left | -2A \right |=-2A$ is constant. So, $K=(-2)$
Let’s put value of K in below, Formula,
$\begin{aligned} &|K A|=K^{n}|A| \\ &|-2 A|=(-2)^{n}|A| \end{aligned} \quad[K=-2]$
$\begin{aligned} &|-2 A|=(-2)^{3}|A| \quad[n=3] \\ &|-2 A|=-8|A| \end{aligned}$
So, here $8\left | A \right |$ is our answer.

Determinants exercise Fill in the blanks question 12

Answer: 5
Hint: Here, we use basic concept of determinant of matrix
Given: A is $3\times 3$ matrix. So, n = 3
$|A|=5 \; \; C_{i j}=\operatorname{cofactor} \text { of } a_{i j} . \text { So, let } C_{i j}=A_{i j}, C=A$
Solution: A is $\left | a_{ij} \right |$ be a square matrix
So, $A=\left(\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right) 1 \leq i \leq 3$
$\begin{aligned} &|A|=5 \\ &|A|=a_{11}\left[\begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right]-a_{12}\left[\begin{array}{cc} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array}\right]+a_{13}\left[\begin{array}{cc} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array}\right] \\ &|A|=a_{11}\left[\begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right]+a_{12}\left[\begin{array}{cc} a_{23} & a_{21} \\ a_{33} & a_{31} \end{array}\right]+a_{13}\left[\begin{array}{ll} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array}\right] \end{aligned}$
[Interchange column of a12]
$\begin{aligned} &\Rightarrow|A|=a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13} \\ &5=a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13} \\ &5=a_{11} C_{11}+a_{12} C_{12}+a_{13} C_{13} \quad[A=C] \end{aligned}$

Determinants exercise Fill in the blanks question 13

Answer: 0
Hint: Here we use basic concept of determinant of matrix
Given: A is $3\times 3$ matrix. So, n = 3
$\begin{aligned} &|A|=5\\ &C_{i y} \text { is cofactor of } C_{11 j} \end{aligned}$
Solution:
$\begin{aligned} &A_{i j}=\left(\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right) \\ &\text { and cofactor }\left(A_{i j}\right)=(-1)^{i+j}|\operatorname{mij}| \end{aligned}$
So, here
$\begin{aligned} &a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23} \\ &A_{21}=(-1)^{2+1}\left[a_{12} a_{33}-a_{13} a_{32}\right] \\ &A_{22}=(-1)^{2+2}\left[a_{11} a_{33}-a_{13} a_{31}\right] \\ &A_{23}=(-1)^{2+3}\left[a_{11} a_{32}-a_{12} a_{31}\right] \end{aligned}$
$\begin{aligned} &=a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23} \\ &=\left(-a_{11} a_{12} a_{33}\right)+a_{11} a_{13} a_{32}+a_{11} a_{12} a_{33}-a_{12} a_{13} a_{31}-a_{12} a_{13} a_{32}+a_{12} a_{13} a_{31} \\ &=\left[a_{11} a_{12} a_{33}-a_{11} a_{12} a_{33}\right]+\left[a_{11} a_{13} a_{32}-a_{11} a_{13} a_{32}\right]+\left[a_{12} a_{13} a_{31}-a_{12} a_{13} a_{31}\right] \\ &\text { So } 0+0+0 \\ &=0 \end{aligned}$

Determinants exercise Fill in the blanks question 14

Answer: 576
Hint: Here, we use basic concept of diagonal matrix and determinant of matrix
Given: $A= diag(2,34)$
Solution: $diag(2,34)$
$\text { So, } A=\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{array}\right]$
Let’s find determinant of A
$\begin{aligned} &|A|=2\left[\begin{array}{ll} 3 & 0 \\ 0 & 4 \end{array}\right]-0\left[\begin{array}{ll} 0 & 4 \\ 0 & 0 \end{array}\right]+0\left[\begin{array}{ll} 0 & 3 \\ 0 & 0 \end{array}\right] \\ &=2(12-0)-0(0-0)+0(0-0) \\ &=2(12)-0+0 \\ &|A|=24 \end{aligned}$
So,
$\begin{aligned} &\left|A^{2}\right|=|A|^{2}=(24)^{2} \\ &\left|A^{2}\right|=576 \end{aligned}$

Determinants exercise Fill in the blanks question 15

Answer: $|C|=|A|^{2}=2^{2}=4$
Hint: Here, we use basic concept of diagonal matrix and determinant of matrix
Given: $\begin{aligned} &A=\left[c_{i j}\right]_{3 \times 3}|A|=2 \text { and } c_{i j} \text { be cofactor of } a_{i j}\\ &C=\left[c_{i j}\right] \end{aligned}$
Solution: c is cofactor matrix
$\begin{aligned} &A d j A=(C)^{T}\\ &\text { But Adj } A=\text { Det. of } C \end{aligned}$
Because both are transpose of each other,
$|\operatorname{adj} A|=|C|=|A|^{n-1}$
Here n = 3 because A is $3\times 3$ matrix,
$\begin{aligned} &|C|=|A|^{3-1} \\ &|C|=|A|^{2} \end{aligned}$
Here,
$\begin{aligned} &|A|=2 \\ &|C|=(2)^{2}=4 \end{aligned}$

Determinants exercise Fill in the blanks question 16

Answer: 0
Hint: Here, we use basic concept of determinant of matrix.
Given: $\Delta=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 x & 6 x & 9 x \end{array}\right]$
Solution:
$\Delta=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 x & 6 x & 9 x \end{array}\right]$
$\begin{aligned} &=1\left[\begin{array}{cc} 5 & 6 \\ 6 x & 9 x \end{array}\right]-2\left[\begin{array}{cc} 4 & 6 \\ 3 x & 9 x \end{array}\right]+3\left[\begin{array}{cc} 4 & 5 \\ 3 x & 6 x \end{array}\right] \\ &=1(45 x-36 x)-2(36 x-18 x)+3(24 x-15 x) \\ &=9 x-36 x+27 x \\ &=0 \end{aligned}$

Determinants exercise Fill in the blanks question 17

Answer: 0
Hint: Here, we use basic concept of determinant of matrix
Given: $\Delta=\left[\begin{array}{ccc} \sec ^{2} \theta & \tan ^{2} \theta & 1 \\ \tan ^{2} \theta & \sec ^{2} \theta & -1 \\ 12 & 10 & 2 \end{array}\right]$
Solution: We know that
$\sec ^{2} \theta-\tan ^{2} \theta=1$ ..............(1)
Applying $C_{2}\rightarrow C_{2}-C_{1}$
$\begin{aligned} \Delta &=\left[\begin{array}{ccc} \sec ^{2} \theta & \sec ^{2} \theta-\tan ^{2} \theta & 1 \\ \tan ^{2} \theta & \tan ^{2} \theta-\sec ^{2} \theta & -1 \\ 12 & 12-10 & 2 \end{array}\right] \\ \Delta &=\left[\begin{array}{ccc} \sec ^{2} \theta & 1 & 1 \\ \tan ^{2} \theta & -1 & -1 \\ 12 & 2 & 2 \end{array}\right] \end{aligned}$ [From equation (1)]
$\Delta =0$ because according to the property of determinant, if any 2 columns are same then $\Delta =0$

Determinants exercise Fill in the blanks question 18

Answer: 0
Hint: Here, we use basic concept of determinant of matrix
Given: $\Delta=\left[\begin{array}{ccc} 0 & x-y & y-z \\ y-x & 0 & z-x \\ z-y & x-z & 0 \end{array}\right]$
Solution: Here $\Delta$ is skew symmetric matrix.
Here, clearly we can see that $\Delta$ is skew symmetric and its diagonal are 0
So, $\Delta=0$

Determinants exercise Fill in the blanks question 19

Answer: 120
Hint: Here, we use basic concept of determinant of matrix
Given:
$\begin{aligned} &b_{i 1}=2 a_{i 1} \\ &b_{i 2}=3 a_{i 2} \\ &b_{i 3}=4 a_{i 3} i, i=1,2,3 \end{aligned}$
Let $\left | A \right |=\left [ a_{ij} \right ]$ and $B=b_{ij}$ be a square matrix of order 3 and $\left | A \right |=5$
Solution:
$\begin{aligned} &|B|=\left(\begin{array}{lll} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array}\right) \\ &|B|=\left(\begin{array}{lll} 2 a_{11} & 3 a_{12} & 4 a_{13} \\ 2 a_{21} & 3 a_{22} & 4 a_{23} \\ 2 a_{31} & 3 a_{32} & 4 a_{33} \end{array}\right) \end{aligned}$
Let’s transfer $\left | B \right |=\left | B^{T} \right |$ So,
$\begin{aligned} &|B|=\left(\begin{array}{ccc} 2 a_{11} & 2 a_{12} & 2 a_{13} \\ 3 a_{21} & 3 a_{22} & 3 a_{23} \\ 4 a_{31} & 4 a_{32} & 4 a_{33} \end{array}\right) \\ &|B|=2 \times 3 \times 4\left(\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right) \end{aligned}$
$\begin{aligned} &=24 \times\left|A^{T}\right| \\ &=24 \times 5 \quad\left[|A|=\left|A^{T}\right|=5\right] \\ &=120 \end{aligned}$

Determinants exercise Fill in the blanks question 20

Answer: -64
Hint: Here, we use basic concept of determinant
$|K A B|=K^{n}|A||B|$
Given: $n = 3$
$\left | A \right |=-2$
$\left | B \right |=4$
Solution: A and B are $3\times 3$ order of matrix
So, n = 3
Here, $\left | 2AB \right |\; \; \; \; \; \; \; So,K=2$
So, now
$|K A B|=K^{n}|A||B|$
$|2 A B|=2^{3} \times-2 \times 4 \; \; \; \; \; \; \; \; \; \quad[n=3,|A|=-2,|B|=4]$
$\begin{aligned} &|2 A B|=8 \times-2 \times 4 \\ &|2 A B|=8 \times-8 \\ &|2 A B|=-64 \end{aligned}$

Determinants exercise Fill in the blanks question 21

Answer: y
Hint: Here, we use basic concept of determinant of matrix
Given: $\left[\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ -\cos x & \sin x & \sin y \end{array}\right]$
Solution: Formula expanding,
$\begin{aligned} &{\left[\begin{array}{ccc} \cos x \cos y-\sin x \sin y & -\sin x \cos y+\cos x \sin y & \cos ^{2} y-\sin ^{2} y \\ \sin x & \cos x & \sin y \\ -\cos x & \sin x & \sin y \end{array}\right]} \\ &R_{1} \rightarrow R_{1}+R_{2}(\sin y)+R_{3}(\cos y) & \end{aligned}$
$\left[\begin{array}{ccc} \cos x \cos y-\sin x \sin y+\sin y \sin x-\cos y \cos x & -\sin x \cos y-\cos x \sin y+\sin y \cos x+\cos y \sin x & \cos ^{2} y-\sin ^{2} y+\sin ^{2} y-\cos ^{2} y \\ \sin x & \cos x & \sin y \\ -\cos x & \sin x & -\sin y \end{array}\right]$ $\text { So }\left[\begin{array}{ccc} 0 & 0 & 0 \\ \sin x & \cos x & \sin x \\ -\cos x & \sin x & -\cos x \end{array}\right]=0$
Here, we get answer is zero which is obtained by multiplying 2nd row with sin y and 3rd row with cosy.
So, Value of determinant depends on y.

Determinants exercise Fill in the blanks question 22

Answer: zero
Hint: Here, we use basic concept of determinant of matrix
Given: $\left[\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}+2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}+3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}+4^{-x}\right)^{2} & 1 \end{array}\right]$
Solution: Here, clearly 2 column has same value
According to property of determinant, if two column has same value then determinant of matrix should be zero.

Determinants exercise Fill in the blanks question 23

Answer: zero
Hint: Here, we use basic concept of determinant.
Given: $\left[\begin{array}{lll} (1+x)^{17} & (1+x)^{19} & (1+x)^{23} \\ (1+x)^{23} & (1+x)^{24} & (1+x)^{34} \\ (1+x)^{41} & (1+x)^{43} & (1+x)^{47} \end{array}\right]$
$A+B x+c x^{2}+\ldots \ldots \ldots \text { then find } A=?$
Solution: Let’s take something common from 3 rows
$(1+x)^{17}\left(1+x^{23}\right)\left(1+x^{41}\right)\left[\begin{array}{ccc} 1 & (1+x)^{2} & (1+x)^{6} \\ 1 & (1+x)^{6} & (1+x)^{11} \\ 1 & (1+x)^{2} & (1+x)^{6} \end{array}\right]$
Here, we clearly see that 1^st and 3^rd row has same value
So, f(x) = 0
So, here x = 0 and
$\begin{aligned} &A+B x+C x^{2} \ldots \ldots \ldots .=0 \\ &S o, A=0 \end{aligned}$

Determinants exercise Fill in the blanks question 24

Answer: $\frac{1}{2}$
Hint: Here, we use basic concept of determinant of matrix.
Given: $\cos 2 \theta=0$
$\left[\begin{array}{ccc} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{array}\right]^{2}=?$
Solution:
$\cos 2 \theta=0$
$\begin{aligned} &\cos 2 \theta=\cos 90^{\circ} \quad\left(\cos 90^{\circ}=0\right) \\ &2 \theta=90^{\circ} \\ &\theta=45^{\circ} \end{aligned}$
Let’s put $\theta=45^{\circ}$ in matrix
$\left[\begin{array}{ccc} 0 & \cos 45^{\circ} & \sin 45^{\circ} \\ \cos 45^{\circ} & \sin 45^{\circ} & 0 \\ \sin 45^{\circ} & 0 & \cos 45^{\circ} \end{array}\right]^{2}$
$\left[\begin{array}{ccc} 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \end{array}\right]^{2} \quad\left[\sin 45^{\circ}=\frac{1}{\sqrt{2}}, \cos 45^{\circ}=\frac{1}{\sqrt{2}}\right]$
Let’s take $\frac{1}{\sqrt{2}}$ in common
$=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$
$\begin{aligned} &=\left[\frac{1}{2 \sqrt{2}}[-1 \times(1-0)+1(0-1)]\right]^{2} \\ &=\frac{1}{8} \times[-1-1]^{2} \\ &=\frac{1}{8} \times(-2)^{2} \\ &=\frac{1}{2} \end{aligned}$

Determinants exercise Fill in the blanks question 25

Answer: 9
Hint: Here, we use basic concept of minors in matrix
Given: Here A is $3\times 3$ matrix
Solution: In any matrix minors are equal to total number of elements
So, here total number of elements of $A=3\times 3=9$
So, total minors in A is 9

Determinants exercise Fill in the blanks question 26

Answer: 2,7
Hint: Here, we use basic concept of determinant of matrix
Given: $x=-y$ and $\left[\begin{array}{lll} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{array}\right]=0$
Solution:
$\left[\begin{array}{lll} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{array}\right]$
$\begin{aligned} &=x\left[\begin{array}{ll} x & 2 \\ 6 & x \end{array}\right]-3\left[\begin{array}{cc} 2 & 2 \\ 7 & x \end{array}\right]+7\left[\begin{array}{cc} 2 & x \\ 7 & 6 \end{array}\right] \\ &=x\left(x^{2}-12\right)-3(2 x-14)+7(12-7 x) \\ &f(x)=x^{3}-12 x-6 x+42+84-49 x=0 \\ &f(x)=x^{3}-67 x+126=0 \end{aligned}$
Let put x = 2 and x = 7
$\begin{aligned} &f(2)=(2)^{3}-67(2)+126 \\ &=8-134+126 \\ &=0 \\ &f(7)=(7)^{3}-67(7)+126 \\ &=49 \times 7-67+126 \\ &=0 \end{aligned}$
So, 2 and 7 are other roots.

Determinants exercise Fill in the blanks question 27

Answer: $(y-z)(z-x)[y-x+x y z]$
Hint: Here, we use basic concept of determinant of matrix
Given: $\left[\begin{array}{ccc} 0 & x y z & x-z \\ y-x & 0 & y-z \\ z-x & z-y & 0 \end{array}\right]$
Solution:
$0\left[\begin{array}{cc} 0 & y-z \\ z-y & 0 \end{array}\right]-x y z\left[\begin{array}{cc} y-x & y-z \\ z-x & 0 \end{array}\right]+x-z\left[\begin{array}{cc} y-x & 0 \\ z-x & z-y \end{array}\right]$
$\begin{aligned} &=0(0-(y-z)) \times(z-y)-x y z(0-(y-z)(z-x))+(x-z)((y-x)(z-y)-0) \\ &=x y z(y-z)(z-x)+(z-x)(y-z)(y-z) \\ &=(y-z)(z-x)[x y z+(y-x)] \\ &=(y-z)(z-x)[y-x+x y z] \end{aligned}$

Determinants exercise Fill in the blanks question 28

Answer: 675
Hint: Here, we use basic concept of determinant of matrix $|K A B|=K^{n}|A||B|$
Given: A is $3\times 3$ matrix. So, n = 3
$|A|=5,|B|=5$
Solution: $|3 A B|$
Here 3 is constant K. So,
$\begin{aligned} &|K A B|=K^{n}|A||B| \\ &|3 A B|=3^{n}|A||B| \quad(K=3) \\ &|3 A B|=3^{3} \times 5 \times 5[n=3,|A|=5,|B|=5] \\ &=27 \times 25 \\ &|3 A B|=675 \end{aligned}$

Determinants exercise Fill in the blanks question 29

Answer: AP
Hint: Here, we use basic concept of determinant of matrix.
Given: $\left[\begin{array}{ccc} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{array}\right]=0$
Solution:
So, Let’s
$\begin{aligned} &R_{2} \rightarrow R_{2}-R_{1} \\ &R_{3} \rightarrow R_{3}-R_{1} \\ &{\left[\begin{array}{ccc} x+1 & x+2 & x+a \\ 1 & 1 & b-a \\ 2 & 2 & c-a \end{array}\right]=0} \end{aligned}$
Let’s perform differentiate
$\begin{aligned} &=x+a[0+0]-1[(0-a)-2(b-a)]+(x+a) 0=0 \\ &c+a-2 b=0 \\ &2 b=a+c \end{aligned}$
So a,b,c are in A.P

Determinants exercise Fill in the blanks question 30

Answer: Zero
Hint: Here, we use basic concept of determinant of matrix.
Given: $\left[\begin{array}{ccc} \sin A & \cos A & \sin A+\cos B \\ \sin B & \cos A & \sin B+\cos B \\ \sin C & \cos A & \sin C+\cos B \end{array}\right]=0$
Solution: Let’s take sin A, sin B and sin C common from respectively 1^st , 2^nd and 3^rd row
$\sin A \sin B \sin C\left[\begin{array}{lll} 1 & \cos A & 1+\cos B \\ 1 & \cos A & 1+\cos B \\ 1 & \cos A & 1+\cos B \end{array}\right]$
Here, all rows are same
According to properties of determinants, if there are 2 or more rows are same then, value of determinants is zero.
So, Answer is zero.

Determinants exercise Fill in the blanks question 31

Answer: 8
Hint: Here, we use basic concept of determinant of matrix.
Given: $\left[\begin{array}{lll} x+a & p+u & l+f \\ y+b & q+v & m+g \\ z+c & r+w & n+h \end{array}\right] K=?$
Solution: Here, we use the property of determinant if elements of row and column of determinant are expressed as sum of two (or more) terms, then it can be expressed as sum of two ( or more) determinant.
Let’s take $\Delta$ of determinant
$\Delta=\left[\begin{array}{ccc} x+a & p+u & l+f \\ y+b & q+v & m+g \\ z+c & r+w & n+h \end{array}\right]$
Let’s use property for C1
$\begin{aligned} &\Delta=\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right] \\ &\Delta=\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right]+\left[\begin{array}{lll} a & p+u & l+f \\ b & q+v & m+g \\ c & r+w & n+h \end{array}\right] \end{aligned}$
Property using for C2
$\Delta=\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right]+\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right]+\left[\begin{array}{ccc} a & p+u & l+f \\ b & q+v & m+g \\ c & r+w & n+h \end{array}\right]$ $+\left[\begin{array}{ccc} a & p+u & l+f \\ b & q+v & m+g \\ c & r+w & n+h \end{array}\right]$
Property using for C3
$\Delta=\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right]+\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right]+\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right]$
$+\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right]+\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right]+\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right]$
$+\left[\begin{array}{ccc} a & p+u & l+f \\ b & q+v & m+g \\ c & r+w & n+h \end{array}\right]+\left[\begin{array}{ccc} a & p+u & l+f \\ b & q+v & m+g \\ c & r+w & n+h \end{array}\right]+\left[\begin{array}{ccc} a & p+u & l+f \\ b & q+v & m+g \\ c & r+w & n+h \end{array}\right]$
So, here we split in 8 parts.
So, k = 8

Determinants exercise Fill in the blanks question 32

Answer: $\frac{1}{2}$
Hint: Here we use basic concept of determinant of matrix.
Given: $\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1 & 1 & 1+\cot \theta \end{array}\right]$
Solution:
$\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1 & 1 & 1+\cot \theta \end{array}\right]$
$\begin{aligned} &C_{2} \rightarrow C_{2}-C_{1}, C_{3} \rightarrow C_{3}-1 \\ &\Delta=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 1 & \sin \theta & 0 \\ 1 & 0 & \cot \theta \end{array}\right] \\ &1\left[\begin{array}{cc} \sin \theta & 0 \\ 0 & \cot \theta \end{array}\right]-0\left[\begin{array}{cc} 1 & 0 \\ 1 & \cot \theta \end{array}\right]+0\left[\begin{array}{cc} 1 & \sin \theta \\ 1 & 0 \end{array}\right] \end{aligned}$
$\begin{aligned} &1(\sin \theta \times \cot \theta)-0(0)+0(0) \\ &\sin \theta \times \cot \theta-0+0 \\ &\sin \theta \times \frac{\cos \theta}{\sin \theta}=\cos \theta \\ &\Delta=\cos \theta \end{aligned}$
Let's $\Delta =0$
$\cos \; \theta =0=\frac{\pi }{2} \left [ So,maximum\; value\; is\; \frac{1}{2}\; of \; \cos \theta \right ]$

Determinants exercise Fill in the blanks question 34

Answer: 0
Hint: Here, we use basic concept of determinant of matrix
Given: $D=\left[\begin{array}{ccc} \sqrt{23}+\sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15}+\sqrt{46} & 5 & \sqrt{10} \\ 3+\sqrt{115} & \sqrt{15} & 5 \end{array}\right]$
Solution: Let’s use separation property of determinant
$\begin{aligned} &{\left[\begin{array}{ccc} \sqrt{23} & \sqrt{5} & \sqrt{5} \\ \sqrt{15} & 5 & \sqrt{10} \\ 3 & \sqrt{15} & 5 \end{array}\right]+\left[\begin{array}{ccc} \sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15} & 5 & \sqrt{10} \\ 3 & \sqrt{15} & 5 \end{array}\right]} \\ &{\left[\begin{array}{lll} \sqrt{23} & \sqrt{5} & \sqrt{5} \\ \sqrt{2 \times 23} & \sqrt{5 \times 5} & \sqrt{2 \times 5} \\ \sqrt{5 \times 23} & \sqrt{5 \times 3} & \sqrt{5 \times 5} \end{array}\right]+\left[\begin{array}{ccc} \sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{3 \times 5} & \sqrt{5 \times 5} & \sqrt{2 \times 5} \\ \sqrt{3 \times 3} & \sqrt{5 \times 3} & \sqrt{5 \times 5} \end{array}\right]} \end{aligned}$
$\begin{aligned} C_{1} & \rightarrow \frac{C_{1}}{\sqrt{3}} \\ C_{2} & \rightarrow \frac{C_{2}}{\sqrt{5}} \text { and } C_{3} \rightarrow \frac{C_{3}}{\sqrt{5}} \end{aligned}$
$\sqrt{23} \sqrt{5} \sqrt{5}\left[\begin{array}{ccc} 1 & 1 & 1 \\ \sqrt{2} & \sqrt{5} & \sqrt{2} \\ \sqrt{5} & \sqrt{3} & \sqrt{5} \end{array}\right]+\sqrt{3} \sqrt{5} \sqrt{5}\left[\begin{array}{ccc} 1 & 1 & 1 \\ \sqrt{5} & \sqrt{5} & \sqrt{2} \\ \sqrt{3} & \sqrt{3} & \sqrt{5} \end{array}\right]$
$C_{1} \rightarrow C_{1}-C_{3} \text { and } C_{1} \rightarrow C_{1}-C_{2}$
$5 \sqrt{23}\left[\begin{array}{ccc} 0 & 1 & 1 \\ 0 & \sqrt{5} & \sqrt{2} \\ 0 & \sqrt{3} & \sqrt{5} \end{array}\right]+5 \sqrt{23}\left[\begin{array}{ccc} 0 & 1 & 1 \\ 0 & \sqrt{5} & \sqrt{2} \\ 0 & \sqrt{3} & \sqrt{5} \end{array}\right]$
Here, both column has 0 value of each column
So, determinant value is zero.

Determinants exercise Fill in the blanks question 35

Answer: 0
Hint: Here, we use basic concept of determinant of matrix
Given: $\left[\begin{array}{ccc} \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} & \cos 180^{\circ} \\ -\sin ^{2} 67^{\circ} & -\sin ^{2} 23^{\circ} & \cos ^{2} 180^{\circ} \\ -1 & \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right]$
Solution:
$\left[\begin{array}{ccc} \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} & \cos 180^{\circ} \\ -\sin ^{2} 67^{\circ} & -\sin ^{2} 23^{\circ} & \cos ^{2} 180^{\circ} \\ -1 & \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right]$
$\begin{aligned} &{\left[\begin{array}{ccc} \sin ^{2} 23^{\circ} & \sin ^{2}\left(90^{\circ}-23^{\circ}\right) & -1 \\ -\sin ^{2}\left(90^{\circ}-23^{\circ}\right) & -\sin ^{2} 23^{\circ} & 1 \\ -1 & \sin ^{2} 23^{\circ} & \cos ^{2}\left(90^{\circ}-23^{\circ}\right) \end{array}\right]} \\ &{\left[\begin{array}{ccc} \sin ^{2} 23^{\circ} & \cos ^{2} 23 & -1 \\ -\cos ^{2} 23 & -\sin ^{2} 23^{\circ} & 1 \\ -1 & \sin ^{2} 23^{\circ} & \cos ^{2} 23^{\circ} \end{array}\right]} \end{aligned}$
Let's apply $C_{1} \rightarrow C_{1}+C_{2}$
$\begin{aligned} &{\left[\begin{array}{ccc} \sin ^{2} 23^{\circ}+\cos ^{2} 23^{\circ} & \cos ^{2} 23 & -1 \\ -\cos ^{2} 23-\sin ^{2} 23^{\circ} & -\sin ^{2} 23^{\circ} & 1 \\ -1+\sin ^{2} 23^{\circ} & \sin ^{2} 23^{\circ} & \cos ^{2} 23^{\circ} \end{array}\right]} \\ &{\left[\begin{array}{ccc} 1 & 1 & -1 \\ -1 & -\sin ^{2} 23^{\circ} & 1 \\ -\cos ^{2} 23^{\circ} & \sin ^{2} 23^{\circ} & \cos ^{2} 23^{\circ} \end{array}\right]} \\ &(-1)\left[\begin{array}{ccc} 1 & 1 & -1 \\ -1 & -\sin ^{2} 23^{\circ} & 1 \\ -\cos ^{2} 23^{\circ} & \sin ^{2} 23^{\circ} & \cos ^{2} 23^{\circ} \end{array}\right] \end{aligned}$
So, here two columns become same. So, determinant value becomes zero.
So, our answer is zero.

Determinants exercise Fill in the blanks question 36

Answer: 0,-6
Hint: Here we use basic determinant
Given: $\left[\begin{array}{ccc} 2-x & 2+x & 2+x \\ 2+x & 2-x & 2+x \\ 2+x & 2+x & 2-x \end{array}\right]=0$
Solution: Here, firstly
$2-x\left[\begin{array}{ll} 2-x & 2+x \\ 2+x & 2-x \end{array}\right]-2+x\left[\begin{array}{cc} 2+x & 2+x \\ 2+x & 2-x \end{array}\right]+2+x\left[\begin{array}{cc} 2+x & 2-x \\ 2+x & 2+x \end{array}\right]$
$\begin{aligned} &0=(2-x)\left[(2-x)^{2}-(2+x)^{2}\right]-[2+x]\left[2^{2}-x^{2}[2+x]^{2}\right]+ &(2+x)\left[(2+x)^{2}-\left[\left(2^{2}-x^{2}\right)\right]\right. \end{aligned}$
$\begin{aligned} &0=8 x^{2}+16 x+8 x+8 x^{2}+8 x^{2}+4 x^{3}\end{aligned}$
$\begin{aligned} &0=8 x^{2}+8 x^{2}+8 x^{2}+4 x^{3} \end{aligned}$
$\begin{aligned} &0=24 x^{2}+4 x^{3} \end{aligned}$
Let’s take 4x² common
$\begin{aligned} &4 x^{2}(6+x)=0 \\ &4 x^{2}=0 \text { and } 6+x=0 \\ &x^{2}=0 \text { and } x=-6 \end{aligned}$

Determinants exercise Fill in the blanks question 37

Answer: -4
Hint: Here, we use basic component of determinant
Given: A is square matrix of order 3
So, n = 3
$\left | A \right |=2$
Solution:
Here, $\left |-A A^{T} \right |$
So, $\left | -A \right |\times \left |A^{T} \right |$
Here, $\left |A^{T} \right |=\left | A \right |$ because both are transpose to each other
So, $\left | -A \right |\times \left | A \right |$
Let’s take -1 common
$\begin{aligned} &-1([A]) \times|A| \\ &-1(2) \times 2[|A|=2] \\ &S o,\left|-A A^{T}\right|=-4 \end{aligned}$

Determinants exercise Fill in the blanks question 33

Answer: zero
Hint: Here, we use basic concept of determinant of matrix
Given: $D=\left[\begin{array}{ccc} \sin ^{2} A & \cot A & 1 \\ \sin ^{2} B & \cot B & 1 \\ \sin ^{2} C & \cot C & 1 \end{array}\right]$
Solution: Let’s perform some row operations
$\begin{aligned} &R_{1} \rightarrow R_{1}-R_{3} \\ &R_{2} \rightarrow R_{2}-R_{3} \\ &{\left[\begin{array}{ccc} \sin ^{2} A-\sin ^{2} C & \cot A-\cot c & 0 \\ \sin ^{2} B-\sin ^{2} C & \cot B-\cot c & 0 \\ \sin ^{2} C & \cot C & 1 \end{array}\right]} \end{aligned}$
$=\sin (A-C) \times \sin (B-C)\left[\begin{array}{ccc} \sin B & \frac{1}{\sin A \sin C} & 0 \\ \sin A & \frac{1}{\sin B \sin C} & 0 \\ \sin ^{2} C & \cot c & 1 \end{array}\right]$
$\begin{aligned} &=\sin (A-C) \sin (B-C)\left[\frac{1}{\sin C}-\frac{1}{\sin C}\right] \\ &=\sin (A-C) \sin (B-C) \times 0 \\ &=0 \end{aligned}$

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