MCQ 1011 Mark
If $\text{a, b, c}$ are in $A.P$., then the determinant $\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+2\text{a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$
Answer$\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+2\text{a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$
$=\begin{vmatrix}0&0&2(\text{a}+\text{c}-2\text{b})\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$
$[$Applying $R_1 \rightarrow R_1+R_3-R_2, R_1 \rightarrow R_1-R_2]$
$=\begin{vmatrix}0&0&0\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}\ [\because \text{a, b, c}$ are in $A.P.]$
$=0$
View full question & answer→MCQ 1021 Mark
If $A$ is a singular matrix, then $A \text{(adj A)}$ is a.
AnswerGiven $A$ is a singular matrix.
$\Rightarrow ∣A∣ = 0$
$A(\text{adj A}) = |A| I = 0I = O$
$\therefore \text{A(adj A)}$ is a zero matrix.
View full question & answer→MCQ 1031 Mark
Let $\text{A}=\begin{vmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix},$ where $0\leq\theta\leq2\pi.$ Then :
- A
Det $(A)=0$
- B
Det $(A)\in(2,\infty)$
- C
Det $(A)\in(2,4)$
- ✓
Det $(A)\in[2,4]$
AnswerCorrect option: D. Det $(A)\in[2,4]$
$\begin{vmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix}$
$=\begin{vmatrix}1&\sin\theta&2\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix}\ [$Applying $C _3 \rightarrow C _3+ C _1]$
$=2\times\begin{vmatrix}-\sin\theta&1\\-1&-\sin\theta\end{vmatrix}\ [$Expanding along $C _3]$
$=2(\sin^2\theta+1)$
Given, $0\leq\theta\leq2\pi$
$-1\leq\sin\theta\leq1$
$0\leq\sin^2\theta\leq1$
$|\text{A}|=2(\sin^2\theta+1)$
$|\text{A}|=2\times1=2\ [\theta=0]$
$|\text{A}|=2\times2=4\ [\theta=2\pi]$
Det $(A)\in[2,4]$
View full question & answer→MCQ 1041 Mark
If $\text{A}=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix},$ then $A^5 =$
Answer$\text{A}=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$
$\Rightarrow\text{A}=2\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\Rightarrow\text{A}=2\text{I}$
$\Rightarrow\text{A}^5=(2\text{I})^5$
$\Rightarrow\text{A}^5=16\times2\text{I}$
$\Rightarrow\text{A}^5=16\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$
$\Rightarrow\text{A}^5=16\text{A}$
View full question & answer→MCQ 1051 Mark
If the determinant $\begin{vmatrix}\text{a}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}&\text{c}&2\text{b}\alpha+3\text{c}\\2\text{a}\alpha+ 3\text{b}&2\text{b}\alpha+3\text{c}&0\end{vmatrix}=0,$ then :
- A
$\text{a, b, c}$ are in $H.P.$
- ✓
$\alpha$ is a root of $4 a x^2+12 b x+9 c=0$ or $\text{a, b, c}$ are in $G.P.$
- C
$\text{a, b, c}$ are in $G.P$. only.
- D
$\text{a, b, c}$ are in $A.P.$
AnswerCorrect option: B. $\alpha$ is a root of $4 a x^2+12 b x+9 c=0$ or $\text{a, b, c}$ are in $G.P.$
Let $\triangle=\begin{vmatrix}\text{a}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}&\text{c}&2\text{b}\alpha+3\text{c}\\2\text{a}\alpha+3\text{b}&2\text{b}\alpha+3\text{c}&0\end{vmatrix}$
$=\begin{vmatrix}\text{a}-\text{b}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}-\text{c}& \text{c}&2\text{b}\alpha+3\text{c}\\2\text{a}\alpha+ 3\text{b}-2\text{b}\alpha-3&2\text{b}\alpha+3\text{c}&0\end{vmatrix}\ [$Applying $C _1 \rightarrow C _1- C _2]$
$=\begin{vmatrix}\text{a}-\text{b}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}-\text{c}& \text{c}&2\text{b}\alpha+3\text{c}\\2(\text{a}-\text{b})\alpha+3(\text{b}-\text{c})&2\text{b}\alpha+3\text{b}&0\end{vmatrix}$
$=2\alpha(2\text{a}\alpha+3\text{b})-3(2\text{b}\alpha+3\text{c})\begin{vmatrix}\text{a}-\text{b}&\text{b}\\\text{b}-\text{c}&\text{c}\end{vmatrix}\ [$Expanding along $R]$
$=-(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})(\text{ac}-\text{b}^2)$
But $\triangle=0\ [$Given$]$
$\Rightarrow-(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})(\text{ac}-\text{b}^2)=0$
$\Rightarrow(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})=0$
Or $(\text{ac}-\text{b}^2)=0$
$\Rightarrow \alpha$ is a root of $4 a x^2+12 b x+9 c=0$
Or $ac = b ^2$, i.e. $\text{a , b , c}$ are in $G.P.$
View full question & answer→MCQ 1061 Mark
If $\text{A}+\text{B}+\text{C}=\pi,$ then the value of $\begin{vmatrix}\sin(\text{A}+\text{B}+\text{C})&\sin(\text{A}+\text{C})&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\\cos(\text{A}+\text{B})&\tan(\text{B}+\text{C})&0\end{vmatrix}$ is equal to :
Answer$\text{A}+\text{B}+\text{C}=\pi$
$\text{A}+\text{C}=\pi-\text{B},\text{A}+\text{B}=\pi-\text{C}$ and $\text{B}+\text{C}=\pi-\text{A}$
Thus the determinant becomes
$\begin{vmatrix}\sin\pi&\sin(\pi-\text{B})&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\\cos(\pi-\text{C})&\tan(\pi-\text{A})&0\end{vmatrix}$
$=\begin{vmatrix}0&\sin\text{B}&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\-\cos\text{C}&-\tan\text{A}&0\end{vmatrix}$
$[\sin\pi=0,\sin(\pi-\text{B}),\cos(\pi-\text{C})=-\cos\text{C},\tan(\pi-\text{A})=-\tan\text{A}]$
It is a skew symmetric matrix of the odd order $3$.
Thus by property of determinants, we get
$|\triangle|=0$
$\begin{vmatrix}0&\sin\text{B}&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\-\cos\text{C}&-\tan\text{A}&0\end{vmatrix}$
View full question & answer→MCQ 1071 Mark
If $\text{A}=\begin{vmatrix} 10 & 2 \\ 30 & 6 \end{vmatrix}$ then $|\text{A}|$
Answer$\text{A}=\begin{vmatrix} 10 & 2 \\ 30 & 6 \end{vmatrix}$
$=10\times6-30\times2$
$=60-60$
$=0$
View full question & answer→MCQ 1081 Mark
If $A$ is any skew $-$ symmetric matrix of odd order then $∣A∣$ equals:
Answerif $A$ is skew symmetric matrix
then $\text{A} = -\text{A}^\text{T}$
$\therefore |\text{A}|=-|\text{A}^\text{T}|=-|\text{A}|$
$\Rightarrow 2|\text{A}|=0$
$\Rightarrow|\text{A}|=0$
View full question & answer→MCQ 1091 Mark
Evaluate $\begin{bmatrix}5&-4\\1&\sqrt{3}\end{bmatrix}$
- A
$4\sqrt{3}+4$
- B
$4\sqrt{3}+5$
- ✓
$5\sqrt{3}+4$
- D
$4\sqrt{3}-4$
AnswerCorrect option: C. $5\sqrt{3}+4$
Evaluating along $\text{R}_1$, we get
$\triangle5(\sqrt3)-(-4)^1$
$=5\sqrt{3}+4$
View full question & answer→MCQ 1101 Mark
Let $\text{f(x)}=\begin{vmatrix}\cos\text{x}&\text{x}&1\\2\sin\text{x}&\text{x}&2\text{x}\\\sin\text{x}&\text{x}&\text{x}\end{vmatrix},$ then $\lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}}{\text{x}^2}$ is equal to :
Answer$\text{f(x)}=\begin{vmatrix}\cos\text{x}&\text{x}&1\\2\sin\text{x}&\text{x}&2\text{x}\\\sin\text{x}&\text{x}&\text{x}\end{vmatrix}$
$=\begin{vmatrix}\cos\text{x}&\text{x}&1\\\sin\text{x}&0&\text{x}\\\sin\text{x}&\text{x}&\text{x}\end{vmatrix}\ [$Applying $R_2 \rightarrow R_2-R_3]$
$=\begin{vmatrix}\cos\text{x}&\text{x}&1\\\sin\text{x}&0&\text{x}\\\sin\text{x}-\cos\text{x}&0&\text{x}-1\end{vmatrix} \ [$Applying $\left.R _3 \rightarrow R_3-R_1\right]$
$=-\text{x}[\text{x}\sin\text{x}-\sin\text{x}-\text{x}\sin\text{x}+\text{x}\cos\text{x}]$
$=-\text{x}(\text{x}\cos\text{x}-\sin\text{x})$
$\therefore\ \lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}}{\text{x}^2}=\lim_\limits{\text{x}\rightarrow0}\frac{\text{x}(\sin\text{x}-\text{x}\cos\text{x})}{\text{x}^2}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}^2}-\lim_\limits{\text{x}\rightarrow0}\cos\text{x}$
$=1-1=0$
Hence, the correct option is $(a)$
View full question & answer→MCQ 1111 Mark
Find the minor of the element $1$ in the determinant $\triangle=\begin{bmatrix}1&5\\3&8\end{bmatrix}$ is:
AnswerThe minor of the element $1$ can be obtained by deleting the first row and the first column
$\therefore\text{ M}_{11}=8$
View full question & answer→MCQ 1121 Mark
The value of the determinant $\begin{vmatrix} 5 &\text{amp; } 1 \\ 3 &\text{amp; } 2 \end{vmatrix}$
Answer$5 \times 2 - 1 \times 3 = 7$
View full question & answer→MCQ 1131 Mark
Which of the following is not correct?
- A
$|\text{A}|=|\text{A}^{\text{T}}|,$ where $\text{A}=[\text{a}_{\text{ij}}]_{3\times3}$
- B
$|\text{kA}|=|\text{k}^3|,$ where $\text{A}=[\text{a}_{\text{ij}}]_{3\times3}$
- C
If a is a skew$-$symmetric of odd order, then $|A| = 0$
- ✓
$\begin{vmatrix}\text{a}&\text{c}\\\text{e}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{b}&\text{c}\\\text{f}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{a}&\text{d}\\\text{e}&\text{h}\end{vmatrix}+\begin{vmatrix}\text{b}&\text{d}\\\text{f}&\text{h}\end{vmatrix}$
AnswerCorrect option: D. $\begin{vmatrix}\text{a}&\text{c}\\\text{e}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{b}&\text{c}\\\text{f}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{a}&\text{d}\\\text{e}&\text{h}\end{vmatrix}+\begin{vmatrix}\text{b}&\text{d}\\\text{f}&\text{h}\end{vmatrix}$
$\begin{vmatrix}\text{a}+\text{b}&\text{c}+\text{d}\\\text{e}+\text{f}&\text{g}+\text{h} \end{vmatrix}=\begin{vmatrix}\text{a}+\text{b}&\text{c}\\\text{e}+\text{f}&\text{h} \end{vmatrix}+\begin{vmatrix}\text{a}+\text{b}&\text{d}\\\text{e}+\text{f}&\text{h}\end{vmatrix}$
$=\begin{vmatrix}\text{a}&\text{c}\\\text{e}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{b}&\text{c}\\\text{f}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{a}&\text{d}\\\text{e}&\text{h}\end{vmatrix}+\begin{vmatrix}\text{b}&\text{d}\\\text{f}&\text{h}\end{vmatrix}$
View full question & answer→MCQ 1141 Mark
Let $\text{P}$ and $\text{Q}$ be $3\times3$ matrices with $\text{P}\neq\text{Q}.$ If $\text{P}^3=\text{Q}^3$ and $\text{P}^2\text{Q}=\text{Q}^2\text{P}$ then determinant of $(\text{P}^2+\text{Q}^2)$ is equal to:
Answer$\text{P}^3=\text{Q}^3$
$\Rightarrow\text{P}^3- \text{P}^2\text{Q}=\text{Q}^3- \text{Q}^2\text{P}$
$\Rightarrow\text{P}^2(\text{P- Q})=\text{Q}^2(\text{Q- P})$
$\Rightarrow \text{P}^2(\text{P- Q})-\text{Q}^2(\text{Q- P})=0$
$\Rightarrow (\text{P}^2+\text{Q}^2)(\text{P}-\text{Q})=0$
$\Rightarrow|\text{P}^2+\text{Q}^2|=0$
View full question & answer→MCQ 1151 Mark
Evaluate $\begin{bmatrix}1&0&1\\0&0&1\\1&0&1\end{bmatrix}$ is:
Answer$\triangle=\begin{bmatrix}1&0&1\\0&0&1\\1&0&1\end{bmatrix}$
$\triangle=1\begin{bmatrix}0&1\\0&1\end{bmatrix}-0\begin{bmatrix}0&1\\1&1\end{bmatrix}+1\begin{bmatrix}0&0\\1&0\end{bmatrix}$
$\triangle=1(0-0)-0(0-1)+1(0-0)$
$\triangle=0-0+0=0.$
View full question & answer→MCQ 1161 Mark
For the system of equations : $ x + 2y + 3z = 1, 2x + y + 3z = 2, 5x + 5y + 9z = 4$
- ✓
There is only one solution.
- B
There exists infinitely many solution.
- C
- D
AnswerCorrect option: A. There is only one solution.
$x + 2y + 3z = 1$
$2x + y + 3z = 2$
$5x + 5y + 9z = 4$
The determinant of the coefficient matrix $\begin{bmatrix}1&2&3\\2&1&3\\5&5&9\end{bmatrix}$ is
$= -6 - 2(18 - 15) + 3(10 - 5)$
$= -6 - 6 + 15$
$=3\neq0$
The right hand side is also non zero.
The system has a unique solution.
View full question & answer→MCQ 1171 Mark
If the coordinates of the vertices of a triangle are $(0, 0), (0, 2)$ and $(3, 1),$ then area of the triangle is :
- ✓
$\text{ sq.units}$
- B
$-3 \text{ sq.units}$
- C
$2 \text{ sq.units}$
- D
$1 \text{ sq.units}$
AnswerCorrect option: A. $\text{ sq.units}$
Area of triangle $=\frac{1}{2} \begin{vmatrix} 0 &\text{amp; }0 &\text{amp; 1} \\ 0&\text{amp; 2} &\text{amp; 1} \\3 &\text{amp;1} &\text{amp; 1} \end{vmatrix}= \frac{1}{2}\times|-6|=3$
View full question & answer→MCQ 1181 Mark
If for the matrix $A, A^3=1$, than $A^{-1}=$
Answer$A^3=1$
$\Rightarrow A^{-1} A^3=A^{-1} \mid$
$\Rightarrow\left|A^2=A^{-1}\right|$
$\Rightarrow A^2=A^{-1}$
View full question & answer→MCQ 1191 Mark
If $\text{x, y, z}$ are different from zero and $\begin{vmatrix}1+\text{x}&1&1\\1&1+\text{y}&1\\1&1&1+\text{z}\end{vmatrix}=0,$ then the value $x^{-1}+y^{-1}+z^{-1}$ is:
- A
$xyz$
- B
$x^{-1}+y^{-1}+z^{-1}$
- C
$-x - y - z$
- ✓
$-1$
Answer$\begin{vmatrix}1+\text{x}&1&1\\1&1+\text{y}&1\\1&1&1+\text{z}\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}&0&-\text{z}\\0&\text{y}&-\text{z}\\1&1&1+\text{z}\end{vmatrix}=0$ $[$Applying $R_2 \rightarrow R_2-R_3$ and $R_1 \rightarrow R_1-R_3]$
$\Rightarrow\text{x}\big[\text{y}(1+\text{z})+\text{z}\big]+1(\text{yz})=0\ [$Expanding along first column$]$
$\Rightarrow\text{x}[\text{y}+\text{yz}+\text{z}]+\text{yz}=0$
$\Rightarrow\text{xy}+\text{xyz}+\text{xz}+\text{yz}=0$
$\Rightarrow\text{xy}+\text{yz}+\text{zx}=-\text{xyz}$
$\Rightarrow\frac{\text{xy}}{\text{xyz}}+\frac{\text{yz}}{\text{xyz}}+\frac{\text{zx}}{\text{xyz}}=-\frac{\text{xyz}}{\text{xyz}}$
$\Rightarrow\frac{1}{\text{z}}+\frac{1}{\text{x}}+\frac{1}{\text{y}}=-1$
$\Rightarrow\text{x}^{-1}+\text{y}^{-1}+\text{z}^{-1}=-1$
Hence, the correct option is $(d)$
View full question & answer→MCQ 1201 Mark
Choose the correct answer.
Which of the following is correct:
- A
Determinant is a square matrix.
- B
Determinants is a number associated to a matrix.
- ✓
Determinants is a number associated to a square matrix.
- D
AnswerCorrect option: C. Determinants is a number associated to a square matrix.
Since, Determinants is a number associated to a square matrix.
Therefore, option (c) is correct.
View full question & answer→MCQ 1211 Mark
Evaluate $\begin{bmatrix}5&4&3\\3&4&1\\5&6&1\end{bmatrix}$ is:
AnswerExpanding along the first row, we get
$\triangle=5\begin{bmatrix}4&1\\6&1\end{bmatrix}-4\begin{bmatrix}3&1\\5&1\end{bmatrix}+3\begin{bmatrix}3&4\\5&6\end{bmatrix}$
$=5(4-6)-4(3-5)+3(18-20)$
$=5(-2)-4(-2)+3(-2)$
$=-10+8-6$
$=-8.$
View full question & answer→MCQ 1221 Mark
Find the determinant of the matrix $\text{A}=\begin{bmatrix}-\cos\theta&\tan\theta\\\cot\theta&\cos\theta\end{bmatrix}$ is:
- ✓
$\sin^2\theta$
- B
$\sin\theta$
- C
$-\sin\theta$
- D
$-\sin^2\theta$
AnswerCorrect option: A. $\sin^2\theta$
Given that, $\text{A}=\begin{bmatrix}-\cos\theta&-\tan\theta\\\cot\theta&\cos\theta\end{bmatrix}$
$|\text{A}|=\begin{bmatrix}-\cos\theta&-\tan\theta\\\cot\theta&\cos\theta\end{bmatrix}$
$|\text{A}|=-\cos\theta (\cos\theta )-\cot\theta(-\tan\theta) $
$|\text{A}|=-\cos^2\theta+1=\sin^2\theta.$
View full question & answer→MCQ 1231 Mark
If $x = – 4$ is a root of $\triangle=\begin{bmatrix}\text{x}&2&3\\1&\text{x}&1\\3&2&\text{x}\end{bmatrix}=0,$ then the other roots are:
- ✓
$1, 3$
- B
$0, 2$
- C
$-1, 1$
- D
$2, 4$
AnswerCorrect option: A. $1, 3$
View full question & answer→MCQ 1241 Mark
Evaluate $\begin{bmatrix}\sqrt{3}&\sqrt{2}\\-1&2\sqrt{3}\end{bmatrix}$ is:
- A
$6-3\sqrt{2}$
- B
$6-\sqrt{2}$
- C
$6+3\sqrt{2}$
- ✓
$6+\sqrt{2}$
AnswerCorrect option: D. $6+\sqrt{2}$
$\triangle=\begin{bmatrix}\sqrt{3}&\sqrt{2}\\-1&2\sqrt{3}\end{bmatrix}$
$\triangle=(\sqrt{3}\times2\sqrt{3})+\sqrt{2}$
$\triangle=6+\sqrt{2}$
View full question & answer→MCQ 1251 Mark
The system of equation $x + y + z = 2, 3x − y + 2z = 6$ and $3x + y + z = −18$ has:
- ✓
- B
- C
An infinite number of solutions.
- D
Zero solution as the only solution.
View full question & answer→MCQ 1261 Mark
The maximum value of $\triangle=\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1+\cos\theta&1&1\end{vmatrix}$ is $(\theta$ is real$):$
- ✓
$\frac{1}{2}$
- B
$\frac{\sqrt{3}}{2}$
- C
$\sqrt{2}$
- D
$-\frac{\sqrt{3}}{2}$
AnswerCorrect option: A. $\frac{1}{2}$
View full question & answer→MCQ 1271 Mark
Evaluate $\begin{bmatrix}8\text{x}+1&2\text{x}-2\\\text{x}^2-1&3\text{x}+5\end{bmatrix}$ is:
- A
$-2 x^3-26 x^2+45 x+3$
- ✓
$-2 x^3+26 x^2+45 x+3$
- C
$-2 x^3+26 x^2+45 x-3$
- D
$-2 x^3-26 x^2-45 x+3$
AnswerCorrect option: B. $-2 x^3+26 x^2+45 x+3$
View full question & answer→MCQ 1281 Mark
The value of the determinant $\begin{vmatrix}\text{x}&\text{x}+\text{y}&\text{x}+2\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\\text{x}+\text{y}&\text{x}+2\text{y}&\text{x}\end{vmatrix}$ is:
- A
$9 x^2(x+y)$
- ✓
$9 y^2(x+y)$
- C
$3 y^2(x+y)$
- D
$7 x^2(x+y)$
AnswerCorrect option: B. $9 y^2(x+y)$
View full question & answer→MCQ 1291 Mark
The value of the determinant $\begin{vmatrix}\text{a}^2&\text{a}&1\\\cos\text{nx}&\cos(\text{n}+1)\text{x}&\cos(\text{n}+2)\text{x}\\\sin\text{nx}&\sin(\text{n}+1)\text{x}&\sin(\text{n}+2)\text{x}\end{vmatrix}$ is independent of:
View full question & answer→MCQ 1301 Mark
Evaluate $\begin{bmatrix}5&0&5\\1&4&3\\0&8&6\end{bmatrix}$ is:
Answer$\triangle=\begin{bmatrix}5&0&5\\1&4&3\\0&8&6\end{bmatrix}$
Expanding along $\text{R}_1,$ we get:
$\triangle=5\begin{bmatrix}4&3\\8&6\end{bmatrix}-0\begin{bmatrix}1&3\\0&6\end{bmatrix}+5\begin{bmatrix}1&4\\0&8\end{bmatrix}$
$\triangle=5(24-24)-0+5(8-0)$
$\triangle=0-0+40=40.$
View full question & answer→MCQ 1311 Mark
If $\begin{bmatrix}\text{x} &\text{amp; } 1 &\text{amp; 1}\\ 2 &\text{amp; 3} &\text{amp; 4}\\ 1 &\text{amp; 1} &\text{amp; 1}\end{bmatrix}$ has no inverse, then $\text{x}=$
AnswerWe know that, If Dett $= 0$ there is no inverse
$\Rightarrow D = x(3 - 4) - 1(2 - 4) + 1(2 - 3) = 0$
$\Rightarrow -x + 2 - 1 = 0$
$\Rightarrow x = 1$
View full question & answer→MCQ 1321 Mark
For any $2 \times 2$ matrix, if $\text{A(adj A)}=\begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix},$ then $|A|$ is equal to :
Answer$\text{A(adj A)}\begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix}$
By definition, we have
$\text{A(adj A) = |A|I = (adj A)A}\ ($Where $I$ is the identity matrix$)$
$\Rightarrow |A|I = A(\text{adj A})$
$\Rightarrow|\text{A}|\text{I}=10\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$\Rightarrow|\text{A}|=10$
View full question & answer→MCQ 1331 Mark
If $A, B$ are two $n \times n$ non $-$ singular matrices, then
- ✓
$AB$ is non $-$ singular.
- B
$AB$ is singular.
- C
$(A B)^{-1} A^{-1} B^{-1}$
- D
$(A B)^{-1}$ does not exist.
AnswerCorrect option: A. $AB$ is non $-$ singular.
$A$ and $B$ are non-singular matrices of order $n \times n.$
$\therefore|\text{A}|\neq0$ and $|\text{B}|\neq0\ .....(\text{i})$
$A$ and $B$ are of the same order, so $AB$ is defined and is of the same order.
Thus,
$|AB| = |A\|B|$
$\Rightarrow|\text{AB}|\neq0\ \big[$Using $(1)\big]$
Thus, $Ab$ is non $-$ singular.
View full question & answer→MCQ 1341 Mark
If $A^2-A+I=0$, then the inverse of $A$ is:
- A
$A^{-2}$
- B
$A + I$
- ✓
$I - A$
- D
$A - I$
AnswerCorrect option: C. $I - A$
View full question & answer→MCQ 1351 Mark
How much is the area of rectangular field:
- A
$60000 \ \text{sq.m}$
- ✓
$30000 \ \text{sq.m}$
- C
$30000m$
- D
$3000m$
AnswerCorrect option: B. $30000 \ \text{sq.m}$
View full question & answer→MCQ 1361 Mark
Evaluate $\begin{bmatrix}\text{i}&-1\\-1&\text{i}\end{bmatrix}$
View full question & answer→MCQ 1371 Mark
The number of line segments possible with three collinear points is $...........$
View full question & answer→MCQ 1381 Mark
Let $\begin{vmatrix}\text{x}&2&\text{x}\\\text{x}^2&\text{x}&6\\\text{x}&\text{x}&6\end{vmatrix}=\text{ax}^4+\text{bx}^3+\text{cx}^2+\text{dx}+\text{e}.$ Then, the value of $5a + 4b + 3c + 2d + e$ is equal to:
Answer$\triangle=\begin{vmatrix}\text{x}&2&\text{x}\\\text{x}^2&\text{x}&6\\\text{x}&\text{x}&6\end{vmatrix}$
$=\text{x}\begin{vmatrix}\text{x}&6\\\text{x}&6\end{vmatrix}-\text{x}^2\begin{vmatrix}2&\text{x}\\\text{x}&6\end{vmatrix}+\text{x}\begin{vmatrix}2&\text{x}\\\text{x}&6\end{vmatrix}$
$=0-x^2\left(12-x^2\right)+x\left(12-x^2\right)$
$=x^4-12 x^2+12 x-x^3$
$=a x^4+b x^3+c x^2+d x+e$
$\Rightarrow x^4-12 x^2+12 x-x^3=a x^4+b x^3+c x^2+d x+e$
$\Rightarrow a=1, b=-1, c=-12, d=12, e=0$
Thus,
$5 a+4 b+3 c+2 d+e=5-4-36+24+0=-11$
View full question & answer→MCQ 1391 Mark
Find the minor of the element $2$ in the determinant $\triangle=\begin{bmatrix}1&9\\2&3\end{bmatrix}$?
AnswerThe minor of the element $2$ can be obtained by deleting the first row and the first column
$\therefore\text{M}_{11}=9$
View full question & answer→MCQ 1401 Mark
If $\triangle=\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}$ and $A_{ji}$ is cofactors of $a_{ji}$, then value of $\triangle$ is given by:
- A
$a_{11} A_{31}+a_{12} A_{32}+a_{13} A_{33}$
- B
$a_{11} A_{11}+a_{12} A_{21}+a_{13} A_{31}$
- C
$a_{21} A_{11}+a_{22} A_{12}+a_{23} A_{13}$
- ✓
$a_{11} A_{11}+a_{21} A_{21}+a_{31} A_{31}$
AnswerCorrect option: D. $a_{11} A_{11}+a_{21} A_{21}+a_{31} A_{31}$
We know that: $\triangle$ = Sum of the product of the elements of a column (or a row) with their corresponding cofactors $\therefore\triangle = \text{a}_{11}\text{A}_{11} +\text{a}_{21}\text{A}_{21} + \text{a}_{31}\text{A}_{31}$ Hence, the value of $\triangle$ is given by the expression given in alternative d. the correct answer is d.
View full question & answer→MCQ 1411 Mark
The matrix $\begin{bmatrix} 5 & 10 & 3 \\ -2 & -4 & 6 \\ -1 & -2 & \text{b} \end{bmatrix}$ is a singular matrix, if the value of $b$ is:
- A
$-3$
- B
$3$
- C
$0$
- ✓
Now$-$existent
AnswerCorrect option: D. Now$-$existent
$\begin{bmatrix} 5 & 10 & 3 \\ -2 & -4 & 6 \\ -1 & -2 & \text{b} \end{bmatrix}$ is singular matrix.
$\Rightarrow\begin{vmatrix} 5 & 10 & 3 \\ -2 & -4 & 6 \\ -1 & -2 & \text{b} \end{vmatrix}=0$
$\Rightarrow 5(-4b + 12) - 10(-2b + 6) + 3(4 - 4) = 0$
$\Rightarrow -20b + 60 + 20b - 60 = 0$
$b$ does not exist.
View full question & answer→MCQ 1421 Mark
If $\text{A}=\begin{vmatrix}\text{a}_{11}&\text{a}_{12}&\text{a}_{13}\\\text{a}_{21}&\text{a}_{22}&\text{a}_{23}\\\text{a}_{31}&\text{a}_{32}&\text{a}_{33}\end{vmatrix}$ and $C_\text{ij}$ is cofactor of $a_\text{ij}$ in $a$, then value of $|A|$ is given by:
- A
$a_{11} C_{31}+a_{12} C_{32}+a_{13} C_{33}$
- B
$a_{11} C_{11}+a_{12} C_{21}+a_{13} C_{31}$
- C
$a_{21} C_{11}+a_{22} C_{12}+a_{23} C_{13}$
- ✓
$a_{11} C_{11}+a_{21} C_{21}+a_{13} C_{31}$
AnswerCorrect option: D. $a_{11} C_{11}+a_{21} C_{21}+a_{13} C_{31}$
Properties of determinants state that if a is a square matrix of the order $n,$ then Det $(A)$ is the sum of products of elements of a row $($or a column$)$ with the
corresponding cofactor of that element.
View full question & answer→MCQ 1431 Mark
If x, y, z are nonzero real numbers, then the inverse of matrix $\text{A}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$ is
- ✓
$\begin{bmatrix}\text{x}^{-1}&0&0\\0&\text{y}^{-1}&0\\0&0&\text{z}^{-1}\end{bmatrix}$
- B
$\text{xyz}\begin{bmatrix}\text{x}^{-1}&0&0\\0&\text{y}^{-1}&0\\0&0&\text{z}^{-1}\end{bmatrix}$
- C
$\frac{1}{\text{xyz}}\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$
- D
$\frac{1}{\text{xyz}}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
AnswerCorrect option: A. $\begin{bmatrix}\text{x}^{-1}&0&0\\0&\text{y}^{-1}&0\\0&0&\text{z}^{-1}\end{bmatrix}$
$\text{A}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$ $\therefore$
$|A|=x(y z-0)=x y z \neq 0$
Now, $A_{11}=y z, A_{12}=0, A_{13}=0$
$A_{21}=0, A_{22}=x z, A_{23}=0$
$A_{31}=0, A_{32}=0, A_{33}=x y a$
$\therefore\text{adj. A}=\begin{bmatrix}\text{yz}&0&0\\0&\text{xz}&0\\0&0&\text{xy}\end{bmatrix}$ $\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj. A}$ $=\frac{1}{\text{xyz}}\begin{bmatrix}\text{yz}&0&0\\0&\text{xz}&0\\0&0&\text{xy}\end{bmatrix}$ $=\begin{bmatrix}\frac{\text{yz}}{\text{xyz}}&0&0\\0&\frac{\text{xz}}{\text{xyz}}&0\\0&0&\frac{\text{xy}}{\text{xyz}}\end{bmatrix}$ $=\begin{bmatrix}\frac{1}{\text{x}}&0&0\\0&\frac{1}{\text{y}}&0\\0&0&\frac{1}{\text{z}}\end{bmatrix}=\begin{bmatrix}\text{x}^{-1}&0&0\\0&\text{y}^{-1}&0\\0&0&\text{z}^{-1}\end{bmatrix}$ The correct answer is a.
View full question & answer→MCQ 1441 Mark
$ \begin{bmatrix}1 & \text{x} & \text{x}^2 \\1 & \text{y} & \text{y}^2 \\1 & \text{z} & \text{z}^2\end{bmatrix}$
AnswerCorrect option: D. $(x - y) (y - z) (z - x)$
View full question & answer→MCQ 1451 Mark
If $\displaystyle \begin{vmatrix} 2 &\text{amp;} -4 \\ 9 &\text{amp; d}-3 \end{vmatrix}=4$ then $\text{d}=$
AnswerGiven, determinant of the matrix $ \displaystyle \begin{vmatrix} 2 &\text{amp;} -4 \\ 9 &\text{amp; d}-3 \end{vmatrix}=4$
$\Rightarrow2(\text{d}−3)−(9)(−4)=4$
$\Rightarrow2\text{d}-6+36 = 4$
$\Rightarrow 2\text{d}=-26$
$\Rightarrow\text{d} = -13$
View full question & answer→MCQ 1461 Mark
Choose the correct answer from given four options in each of the Exercise : Let $\text{f}(\text{t})=\begin{vmatrix}\cos\text{t}&\text{t}&1\\2\sin\text{t}&\text{t}&2\text{t}\\\sin\text{t}&\text{t}&\text{t}\end{vmatrix} ,$ then $\lim\limits_{\text{t}\rightarrow0}\frac{\text{f(t)}}{\text{t}^2}$ is equal to:
AnswerWe have, $\text{f}(\text{t})=\begin{vmatrix}\cos\text{t}&\text{t}&1\\2\sin\text{t}&\text{t}&2\text{t}\\\sin\text{t}&\text{t}&\text{t}\end{vmatrix} $
$\Rightarrow\ \frac{\text{f(x)}}{\text{t}^2}=\frac{1}{\text{t}^2}\begin{vmatrix}\cos\text{t}&\text{t}&1\\2\sin\text{t}&\text{t}&2\text{t}\\\sin\text{t}&\text{t}&\text{t}\end{vmatrix} $
$=\begin{vmatrix}\cos\text{t}&\text{t}&1\\\frac{2\sin\text{t}}{\text{t}}&1&2\\\frac{\sin\text{t}}{\text{t}}&1&1\end{vmatrix} $
$\big[$Dividing ${R}_2$ and ${R}_3$ by $\ '\text{t}'\big]$
$\Rightarrow\ \lim\limits_{\text{t}\rightarrow0}\frac{\text{f(t)}}{\text{t}^2}=\begin{vmatrix} \lim\limits_{\text{t}\rightarrow0}\text{t}\cos\text{t}&\lim\limits_{\text{t}\rightarrow0}\text{t}&\lim\limits_{\text{t}\rightarrow0}1\\\lim\limits_{\text{t}\rightarrow0}\text{t}\frac{2\sin}{\text{t}}&\lim\limits_{\text{t}\rightarrow0}1&\lim\limits_{\text{t}\rightarrow0}2\\\lim\limits_{\text{t}\rightarrow0}\text{t}\frac{\sin\text{t}}{\text{t}}&\lim\limits_{\text{t}\rightarrow0}1&\lim\limits_{\text{t}\rightarrow0}1\end{vmatrix}$
$=\begin{vmatrix}1&0&1\\2&1&2\\1&1&1\end{vmatrix}$ $\bigg(\because\lim\limits_{\text{t}\rightarrow 0}\frac{\sin\text{t}}{\text{t}}=1\bigg)$
$=1(1-2)-0+1(2-1)$
$=0$
View full question & answer→MCQ 1471 Mark
$\text{Let A}=\begin{bmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{bmatrix},$ where $0\leq\theta\leq2\pi.$ Then
AnswerCorrect option: D. Det $(A) \in [2, 4]$
$\text{A}=\begin{bmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{bmatrix}$ $\therefore|\text{A}|=1(1+\sin^2\theta)-\sin\theta(-\sin\theta+\sin\theta)+1(\sin^2\theta+1)$ $=1+\sin^2\theta+\sin^2\theta+1$
$=2+2 \sin ^2 \theta$
$=2\left(1+\sin ^2 \theta\right)$
Now, $0 \leq \theta \leq 2 \pi$
$\Rightarrow 0 \leq \sin \theta \leq 1$
$\Rightarrow 0 \leq 1+\sin ^2 \theta \leq 2$
$\Rightarrow 2 \leq 2\left(1+\sin ^2 \theta\right) \leq 4$
$\therefore \operatorname{Det}(A) \in[2,4]$
The correct answer is d.
View full question & answer→MCQ 1481 Mark
If $\text{A}=\begin{bmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{bmatrix},$ then $\text{A}^{-1}$ exists if:
- A
$\lambda=2$
- B
$\lambda\neq2$
- C
$\lambda\neq-2$
- ✓
Answer$\text{A}=\begin{bmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{bmatrix}$
The inverse of a matrix exists if its determinant is not equal to $0.$
Consider,
$|\text{A}|=\begin{bmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{bmatrix}\neq0$
$\Rightarrow|\text{A}| = 2 (6 – 5) – \lambda (0 – 5) + (-3) (0 – 2)\neq0$
$\Rightarrow2 + 5\lambda + 6 \neq 0$
$\Rightarrow5\lambda + 8 \neq 0$
$\Rightarrow5\lambda \neq -8$
$\Rightarrow\lambda\neq\frac{-8}{5}$
View full question & answer→MCQ 1491 Mark
Let $A$ be a nonsingular square matrix of order $3 \times 3$. Then $|adj. A|$ is equal to:
- A
$|A|$
- ✓
$|A|^2$
- C
$|A|^3$
- D
$3|A|$
AnswerCorrect option: B. $|A|^2$
If $A$ is a non-singular matrix of order $n \times n$, then $\mid$ adj. $\left.A|=| A\right|^{n-1}$. $\therefore$ Putting $n=3,|\operatorname{adj} . A|=|A|^2$
Therefore, option $(b)$ is correct.
View full question & answer→MCQ 1501 Mark
If $A$ satisfies the equation $\text{x}^2-5\text{x}^2+4\text{x}+\lambda=0$ then $A^{-1}$ exists if :
- A
$\lambda\neq1$
- B
$\lambda\neq2$
- C
$\lambda\neq-1$
- ✓
$\lambda\neq0$
AnswerCorrect option: D. $\lambda\neq0$
A satisfies $\text{x}^3-5\text{x}^2+4\text{x}+\lambda=0$
$\Rightarrow\text{A}^3-5\text{A}^2+4\text{A}=-\lambda$
Assuming $A^{-1}$ exists, we get
$\text{A}^{-1}(\text{A}^3-5\text{A}^2+4\text{A})=-\lambda\text{A}^{-1}$
$\Rightarrow\text{A}^2-5\text{A}+4=-\text{A}^{-1}\lambda$
$\Rightarrow\text{A}-1=\frac{-(\text{A}^2-5\text{A}+4)}{\lambda}$
Thus, $A^{-1}$ exists if $\lambda\neq0$.
View full question & answer→