MCQ 11 Mark
यदि $\left[\begin{array}{cc}x & 2 \\ 18 & x\end{array}\right]=\left[\begin{array}{cc}6 & 2 \\ 18 & 6\end{array}\right]$ तो x बराबर है-
AnswerCorrect option: B. $\pm 6$
(B) $\pm 6$
$x^2-36=36-36$
$\therefore$ $x^2=36=0$
या $x^2=36$
$x=\sqrt{36}= \pm 6$
View full question & answer→MCQ 21 Mark
यदि $\left|\begin{array}{ll}3 & 3 \\ x & 1\end{array}\right|=\left|\begin{array}{cc}-3 & x \\ 1 & 1\end{array}\right|$, तो x का मान है-
Answer(B) 3
$\left|\begin{array}{ll}3 & 3 \\ x & 1\end{array}\right|=\left|\begin{array}{cc}-3 & x \\ 1 & 1\end{array}\right|$
$\Rightarrow 3-3 x=-3-x$
$\Rightarrow 3+3=3 x-x$
$\Rightarrow 6=2 x \quad \Rightarrow x=3$
View full question & answer→MCQ 31 Mark
यदि $A=\left[\begin{array}{ccc}-2 & 0 & 0 \\ 1 & 2 & 3 \\ 5 & 1 & -1\end{array}\right]$ है, तो $| A (\operatorname{adj} . A )|$ का मान है-
- A
$100 I$
- B
$10 I$
- C
$10$
- ✓
$1000$
AnswerCorrect option: D. $1000$
View full question & answer→MCQ 41 Mark
यदि $A=\left[\begin{array}{cc}2 & 1 \\ -4 & -2\end{array}\right]$ है, तो $I - A + A ^2- A ^3+\ldots$ है-
- A
$\left[\begin{array}{cc}-1 & -1 \\ 4 & 3\end{array}\right]$
- B
$\left[\begin{array}{cc}3 & 1 \\ -4 & -1\end{array}\right]$
- C
$\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
- ✓
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
AnswerCorrect option: D. $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
(D) सही विकल्प $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ है।
View full question & answer→MCQ 51 Mark
दिया है कि $A ^{-1}=\frac{1}{7}\left[\begin{array}{cc}2 & 1 \\ -3 & 2\end{array}\right]$ है, तो आव्यूह A है-
- ✓
$7\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]$
- B
$\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]$
- C
$\frac{1}{7}\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]$
- D
$\frac{1}{49}\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]$
AnswerCorrect option: A. $7\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]$
(A) अतः सही विकल्प $7\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]$ है।
View full question & answer→MCQ 61 Mark
माना A एक ऐसा $3 \times 3$ आव्यूह है कि $|\operatorname{adj} A|=64$ है। तो $|A|$ बराबर है-
View full question & answer→MCQ 71 Mark
आव्यूह $X=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]$ का प्रतिलोम है-
- A
$\frac{1}{24}\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
- B
$24\left[\begin{array}{ccc}1 / 2 & 0 & 0 \\ 0 & 1 / 3 & 0 \\ 0 & 0 & 1 / 4\end{array}\right]$
- C
$\frac{1}{24}\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]$
- ✓
$\left[\begin{array}{ccc}1 / 2 & 0 & 0 \\ 0 & 1 / 3 & 0 \\ 0 & 0 & 1 / 4\end{array}\right]$
AnswerCorrect option: D. $\left[\begin{array}{ccc}1 / 2 & 0 & 0 \\ 0 & 1 / 3 & 0 \\ 0 & 0 & 1 / 4\end{array}\right]$
(D) $\left[\begin{array}{ccc}1 / 2 & 0 & 0 \\ 0 & 1 / 3 & 0 \\ 0 & 0 & 1 / 4\end{array}\right]$
यहाँ $|X|=\left|\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right|=2(3 \times 4)=24 \neq 0$
और
$\begin{array}{lll} C _{11}=12, & C _{12}=0, & C _{13}=0 \\ C _{21}=0, & C _{22}=8, & C _{23}=0 \\ C _{31}=0, & C _{31}=0, & C _{33}=6\end{array}$
$\therefore \quad \operatorname{adj} X=\left[\begin{array}{ccc}12 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 6\end{array}\right]$
$\therefore \quad X ^{-1}=\frac{1}{24}\left[\begin{array}{ccc}12 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 6\end{array}\right]=$ $\left[\begin{array}{ccc}1 / 2 & 0 & 0 \\ 0 & 1 / 3 & 0 \\ 0 & 0 & 1 / 4\end{array}\right]$
View full question & answer→MCQ 81 Mark
यदि $x=-1,\left|\begin{array}{lll}x & 2 & 3 \\ 1 & x & 1 \\ 3 & 2 & x\end{array}\right|=0$ का मूल है, तब इस समीकरण के अन्य दो मूलों का योग है-
Answerमाना $P ( X )=(x+4)(x-a)(x-b)$ एक त्रिघातीय बहुपद है।
$\therefore \alpha=-4, \beta=a, \gamma=b$, तीन मूल हैं।
यहाँ, $P ( X )=\left|\begin{array}{lll}x & 2 & 3 \\ 1 & x & 1 \\ 3 & 2 & x\end{array}\right|$
$=x\left(x^2-2\right)-2(x-3)+3(2-3 x)$
$=x^3-2 x-2 x+6+6-9 x$
$=x^2-13 x+12$
$\therefore$ मूलों का योग $=\alpha+\beta+\gamma=\frac{-b}{a}$
$\therefore-4+a+b=0 $
$\Rightarrow a+b=4$
View full question & answer→MCQ 91 Mark
माना $X =\left[x_{i j}\right]$ एक आव्यूह है, जहाँ $X=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3\end{array}\right]$ तब आव्यूह $Y =\left[m_{i j}\right]$, जहाँ $m_{i j}=x_{i j}$ का उपसारिणक है $-$
- A
$\left[\begin{array}{ccc}7 & -5 & -3 \\ 19 & 1 & -11 \\ -11 & 1 & 7\end{array}\right]$
- B
$\left[\begin{array}{ccc}7 & -19 & -11 \\ 5 & -1 & -1 \\ 3 & 11 & 7\end{array}\right]$
- C
$\left[\begin{array}{ccc}7 & 19 & -11 \\ -3 & 11 & 7 \\ -5 & -1 & -1\end{array}\right]$
- ✓
$\left[\begin{array}{ccc}7 & 19 & -11 \\ -1 & -1 & 1 \\ -3 & -11 & 7\end{array}\right]$
AnswerCorrect option: D. $\left[\begin{array}{ccc}7 & 19 & -11 \\ -1 & -1 & 1 \\ -3 & -11 & 7\end{array}\right]$
$\left[\begin{array}{ccc}7 & 19 & -11 \\ -1 & -1 & 1 \\ -3 & -11 & 7\end{array}\right]$
यहाँ $X =\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3\end{array}\right]$
$\therefore m_{11} =\left|\begin{array}{cc}4 & -5 \\ -1 & 3\end{array}\right|=12-5=7 $
$m_{12} =\left|\begin{array}{cc}3 & -5 \\ 2 & 3\end{array}\right|=9+10=19 $
$ m_{13} =\left|\begin{array}{cc}3 & 4 \\ 2 & -1\end{array}\right|=-3-8=-11$
इसी प्रकार,
$\begin{array}{l}m_{21}=-1, m_{22}=-1, m_{23}=1 \\ m_{31}=-3, m_{32}=-11, m_{33}=7\end{array}$
$Y =\left[m_{i j}\right]=\left[\begin{array}{lll}m_{11} & m_{12} & m_{13} \\ m_{21} & m_{22} & m_{23} \\ m_{31} & m_{32} & m_{33}\end{array}\right]=\left[\begin{array}{ccc}7 & 19 & -11 \\ -1 & -1 & 1 \\ -3 & -11 & 7\end{array}\right]$
View full question & answer→MCQ 101 Mark
यदि किसी आव्यूह A के लिए $A=\left[\begin{array}{cc}\alpha & -2 \\ -2 & \alpha\end{array}\right],\left|A^3\right|=125$ तब $\alpha$ का मान है$-$
- ✓
$\pm 3$
- B
$-3$
- C
$\pm 1$
- D
$1$
AnswerCorrect option: A. $\pm 3$
यहाँ $A=\left[\begin{array}{cc}\alpha & -2 \\ -2 & \alpha\end{array}\right]$ और $\left| A ^3\right|=125$
$\left| A ^n\right|=| A |^n$
$\Rightarrow\left|A^3\right|=| A |^3$
$\Rightarrow 125=|A|^3$
$\Rightarrow 125=\left(\alpha^2-4\right)^3$
$\Rightarrow \alpha^2-4=5$
$\Rightarrow \alpha^2=9$
$\Rightarrow \alpha= \pm 3$
View full question & answer→MCQ 111 Mark
यदि A एक 3 क्रम का वर्ग आव्यूह है तथा $|A| = 5$, तब $|adj A|$ है-
- A
$125$
- B
$-25$
- ✓
$25$
- D
$\pm 25$
Answer(C) $25$
दिया है- $|A|=-5$ तथा $n=3$
$|\operatorname{adj} A |=| A |^{n-1}=(-5)^{3-1}=(-5)^2=25$
View full question & answer→MCQ 121 Mark
रैखिक समीकरण निकाय $5 x+k y=5,3 x+3 y=5$ संगत है यदि-
- A
$k \neq-3$
- B
$k=-5$
- C
$k=5$
- ✓
$k \neq-5$
AnswerCorrect option: D. $k \neq-5$
(D) $k \neq-5$
रैखिक समीकरण निकाय संगत है यदि $\left|\begin{array}{ll}5 & k \\ 3 & 3\end{array}\right| \neq 0$
$\begin{aligned} 15-3 k & \neq 0 \\ 3 k & \neq 15 \\ k & \neq 5\end{aligned}$
View full question & answer→MCQ 131 Mark
यदि मैट्रिक्स $P=\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & 2 & 4\end{array}\right]$ के अवयव $p_{i j}$ के लिए सहखण्डज $C_{i j}$ से व्यक्त हो तो $C _{31} \cdot C _{23}$ का मान है-
Answer(A) 5
$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}-1 & 2 \\ 2 & -3\end{array}\right|=3-4=-1$
$C_{23}=(-1)^{2+3}\left|\begin{array}{cc}1 & -1 \\ 3 & 2\end{array}\right|=-(2+3)=-5$
अत: $C _{31} \cdot C _{23}=(-1)(-5)=5$
View full question & answer→MCQ 141 Mark
तीन बिन्दु $P (2 x, x+3), Q (0, x)$ तथा $R (x+3, x+6)$ संरेखीय हैं तब x का मान होगा-
Answer(D) P, Q तथा R संरेखीय हैं
$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}2 x & x+3 & 1 \\ 0 & x & 1 \\ x+3 & x+6 & 1\end{array}\right|=0$
$\begin{aligned} \Rightarrow \frac{1}{2}[2 x(x-x-6)-(x+3)(0-x-3) & \left.+(1)\left(0-x^2-3 x\right)\right]=0\end{aligned}$
$\Rightarrow \frac{1}{2}\left[-12 x+(x+3)^2-x^2-3 x\right]=0$
$\Rightarrow \quad-12 x+x^2+6 x+9-x^2-3 x=0$
$\begin{array}{ll}\Rightarrow & -9 x+9=0 \\ \Rightarrow & \underline{x=1}\end{array}$
View full question & answer→MCQ 151 Mark
यदि $A =\left[\begin{array}{ccc}3 & -1 & 2 \\ 2 & 1 & 3 \\ 1 & -3 & K\end{array}\right]$ एक अव्युत्क्रमणीय आव्यूह हो, तो K का मान होगा-
Answer(B) $-2$
एक अव्युत्क्रमणीय आव्यूह के लिए
$| A |=0 \therefore\left|\begin{array}{ccc}3 & -1 & 2 \\ 2 & 1 & 3 \\ 1 & -3 & K\end{array}\right|=0$
$3(K+9)+1(2 K-3)+2(-6-1)=0$
$\Rightarrow 3 K+27+2 K-3-14=0$
$\Rightarrow 5 K+10=0 \Rightarrow K=-2$
अतः सही विकल्प (B) है।
View full question & answer→MCQ 161 Mark
यदि $A=\left[\begin{array}{cc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right]$ तथा A (adj A) = KI तो K का मान होगा-
- A
$3$
- B
$2$
- ✓
$1$
- D
$\sin x \cdot \cos x$
Answer(C) $1$
$\operatorname{adj} A =\left[\begin{array}{ll} A _{11} & A_{12} \\ A_{21} & A_{22}\end{array}\right]^{\prime}=\left[\begin{array}{ll} A _{11} & A_{21} \\ A_{12} & A_{22}\end{array}\right]$
$\therefore \operatorname{adj} A =\left[\begin{array}{cc}\cos x & -\sin x \\ \sin x & \cos x\end{array}\right]$
$\therefore A (\operatorname{adj} A )=\left[\begin{array}{cc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right]\left[\begin{array}{cc}\cos x & -\sin x \\ \sin x & \cos x\end{array}\right]= KI$
$\Rightarrow\left[\begin{array}{cc}\cos ^2 x+\sin ^2 x & -\cos x \sin x+\sin x \cos x \\ -\sin x \cos x+\cos x \sin x & \sin ^2 x+\cos ^2 x\end{array}\right]$$= K \left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]= K \left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]= KI$
$\Rightarrow \quad I = KI \therefore K =1$
अतः सही विकल्प (C) है।
View full question & answer→MCQ 171 Mark
समान कोटि के वर्ग आव्यूह A, B, C के लिए $AB = AC\Rightarrow B = C$ हो, तो A होगा-
Answer(D) व्युत्क्रमणीय आव्यूह
चूँकि व्युत्क्रमणीय आव्यूह में $|A| \neq 0$
View full question & answer→MCQ 181 Mark
यदि $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$ हो, तो $A ^{-1}$ का मान होगा-
- A
$\left[\begin{array}{ll}4 & 2 \\ 3 & 1\end{array}\right]$
- ✓
$\left[\begin{array}{cc}-2 & 1 \\ \frac{3}{2} & -1 / 2\end{array}\right]$
- C
$\left[\begin{array}{cc}4 & -3 \\ 2 & 1\end{array}\right]$
- D
$\left[\begin{array}{ll}-4 & 3 \\ -2 & 1\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{cc}-2 & 1 \\ \frac{3}{2} & -1 / 2\end{array}\right]$
(B) $\left[\begin{array}{cc}-2 & 1 \\ \frac{3}{2} & -1 / 2\end{array}\right]$
$-A_{11}=4, A_{12}=-3, A_{21}=-2, A_{22}=1$
$\operatorname{adj} A =\left[\begin{array}{cc}4 & -3 \\ -2 & 1\end{array}\right]^{\prime}=\left[\begin{array}{cc}4 & -2 \\ -3 & 1\end{array}\right]$
$|A|=4-6=-2 \neq 0$
$A ^{-1}=\frac{\operatorname{adj} A }{| A |}=\frac{-1}{2}\left[\begin{array}{cc}4 & -2 \\ -3 & 1\end{array}\right]=\left[\begin{array}{cc}-2 & 1 \\ \frac{3}{2} & -1 / 2\end{array}\right]$
अतः सही विकल्प (B) है।
View full question & answer→MCQ 191 Mark
यदि $A =\left[\begin{array}{ll}0 & 1 \\ 1 & 2\end{array}\right]$ हो, तो $adj \ A$ होगा $-$
- ✓
$\left[\begin{array}{cc}2 & -1 \\ -1 & 0\end{array}\right]$
- B
$\left[\begin{array}{cc}2 & -1 \\ -2 & 0\end{array}\right]$
- C
$\left[\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right]$
- D
$\left[\begin{array}{cc}-2 & -1 \\ -1 & 0\end{array}\right]$
AnswerCorrect option: A. $\left[\begin{array}{cc}2 & -1 \\ -1 & 0\end{array}\right]$
$\left[\begin{array}{cc}2 & -1 \\ -1 & 0\end{array}\right]$
$ A _{11}=2 A_{12}=-1 A_{21}=-1 A_{22}=0$
$\operatorname{adj} A=\left[\begin{array}{ll} A _{11} & A_{12} \\ A_{21} & A_{22}\end{array}\right]^{\prime}=\left[\begin{array}{cc}2 & -1 \\ -1 & 0\end{array}\right]^{\prime}$
$=\left[\begin{array}{cc}2 & -1 \\ -1 & 0\end{array}\right]$
अतः सही विकल्प $(A)$ है।
View full question & answer→MCQ 201 Mark
सारणिक $\left|\begin{array}{cc}1 & \log _b a \\ \log _a b & 1\end{array}\right|=$
- A
$1$
- ✓
$0$
- C
$\log _a b$
- D
$\log _b a$
Answer(B) $0$
$1-\log _b a \times \log _a b=1-\log _a a=1-1=0$
$\because\log _a a=1$
अतः सही विकल्प (B) है।
View full question & answer→MCQ 211 Mark
सारणिक $\left|\begin{array}{ll}\cos 50^{\circ} & \sin 10^{\circ} \\ \sin 50^{\circ} & \cos 10^{\circ}\end{array}\right|$ का मान है$-$
- A
$0$
- B
$1$
- ✓
$\frac{1}{2}$
- D
$-\frac{1}{2}$
AnswerCorrect option: C. $\frac{1}{2}$
$\cos 50^{\circ} \cos 10^{\circ}-\sin 10^{\circ} \sin 50^{\circ}$
$=\cos \left(50^{\circ}+10^{\circ}\right)$
$=\cos 60^{\circ}$
$=\frac{1}{2}$
अतः सही विकल्प $(C)$ है।
View full question & answer→