M.C.Q (1 Marks)

MCQ 1011 Mark

The value of $y ($breadth of rectangular field$)$ is:

✓
$150m$
B
$200m$
C
$430m$
D
$350m$

Answer

Correct option: A.

$150m$

MCQ 1021 Mark

$\begin{bmatrix}2\text{x}&5\\8&\text{x}\end{bmatrix}=\begin{bmatrix}6\text{x}&-2\\7&3\end{bmatrix},$ then the value of $\text{x}$ is:

A
$3$
B
$\pm3$
✓
$\pm6$
D
$6$

Answer

Correct option: C.

$\pm6$

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MCQ 1031 Mark

If $a, b, c$ are in $A.P.$, then the determinant $\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+2\text{a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$

✓
$0$
B
$1$
C
$x$
D
$2x$

Answer

Correct option: A.

$0$

$\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+2\text{a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$
$=\begin{vmatrix}0&0&2(\text{a}+\text{c}-2\text{b})\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$
$[$Applying $R_1 \rightarrow R_1 + R_3- R_2, R_1 \rightarrow R_1 - R_2]$
$=\begin{vmatrix}0&0&0\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$ $[\because a, b, c$ are in $A.P.]$
$=0$

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MCQ 1041 Mark

If A is a singular matrix, then A (adj A) is a.

A
scalar matrix
✓
zero matrix
C
identity matrix
D
orthogonal matrix

Answer

Correct option: B.

zero matrix

Given A is a singular matrix.

⇒ ∣A∣ = 0

A(adjA) = ∣A∣ I = 0I = O

∴ A(adjA) is a zero matrix.

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MCQ 1051 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
$\text{Det (A)}=0$
B
$\text{Det (A)}\in(2,\infty)$
C
$\text{Det (A)}\in(2,4)$
✓
$\text{Det (A)}\in[2,4]$

Answer

Correct option: D.

$\text{Det (A)}\in[2,4]$

$\begin{vmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix}$
$=\begin{vmatrix}1&\sin\theta&2\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix} [$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, $0\leq\theta\leq2\pi$
$-1\leq\sin\theta\leq1$
$0\leq\sin^2\theta\leq1$
$|\text{A}|=2(\sin^2\theta+1)$
$|\text{A}|=2\times1=2$ $[\theta=0]$
$|\text{A}|=2\times2=4$ $[\theta=2\pi]$
$\text{Det (A)}\in[2,4]$

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MCQ 1061 Mark

If $\text{A}=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix},$ then $A^5 =$

A
$5A$
B
$10A$
✓
$16A$
D
$32A$

Answer

Correct option: C.

$16A$

$\text{A}=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$
$\Rightarrow\text{A}=2\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\Rightarrow\text{A}=2\text{I}$
$\Rightarrow\text{A}^5=(2\text{I})^5$
$\Rightarrow\text{A}^5=16\times2\text{I}$
$\Rightarrow\text{A}^5=16\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$
$\Rightarrow\text{A}^5=16\text{A}$

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MCQ 1071 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
$a, b, c$ are in $H.P.$
✓
$\alpha$ is a root of $4ax^2 + 12bx + 9c = 0$ or $a, b, c$ are in $G.P$.
C
$a, b, c$ are in $G.P$. only.
D
$a, b, c$ are in $A.P$.

Answer

Correct option: B.

$\alpha$ is a root of $4ax^2 + 12bx + 9c = 0$ or $a, b, c$ 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 $R_3$]
$=-(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})(\text{ac}-\text{b}^2)$
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 $4ax^2 + 12bx + 9c = 0$ or $ac = b^2$, i.e. $a, b, c$ are in $G.P.$

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MCQ 1081 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:

✓
0
B
1
C
$2\sin\text{B}\tan\text{A}\cos\text{C}$
D
None of these.

Answer

Correct option: A.

$\text{A}+\text{B}+\text{C}=\pi$

$\text{A}+\text{C}=\pi-\text{B},\text{A}+\text{B}=\pi-\text{C}$ 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

$|\triangle|=0$

$\begin{vmatrix}0&\sin\text{B}&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\-\cos\text{C}&-\tan\text{A}&0\end{vmatrix}$

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MCQ 1091 Mark

If $\text{A}=\begin{vmatrix} 10 & 2 \\ 30 & 6 \end{vmatrix}$ then $|\text{A}|$

✓
0
B
10
C
12
D
60

Answer

Correct option: A.

$\text{A}=\begin{vmatrix} 10 & 2 \\ 30 & 6 \end{vmatrix}$
$=10\times6-30\times2=60-60=0$

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MCQ 1101 Mark

If A is any skew - symmetric matrix of odd order then ∣A∣ equals:

A
−1
✓
0
C
1
D
None of these

Answer

Correct option: B.

if A is skew symmetric matrix

then $\text{A} = \text{-A}^\text{T}$

$\therefore |\text{A}|=-|\text{A}^\text{T}|=-|\text{A}|$

$\Rightarrow 2|\text{A}|=0$

$\Rightarrow|\text{A}|=0$

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MCQ 1111 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
$4\sqrt{3}-4$

Answer

Correct option: C.

$5\sqrt{3}+4$

Evaluating along $\text{R}_1$,we get
$\triangle5(\sqrt3)-(-4)^1=5\sqrt{3}+4$

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MCQ 1121 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:

✓
$0$
B
$-1$
C
$2$
D
$3$

Answer

Correct option: A.

$0$

$\text{f(x)}=\begin{vmatrix}\cos\text{x}&\text{x}&1\\2\sin\text{x}&\text{x}&2\text{x}\\\sin\text{x}&\text{x}&\text{x}\end{vmatrix}$
$=\begin{vmatrix}\cos\text{x}&\text{x}&1\\\sin\text{x}&0&\text{x}\\\sin\text{x}&\text{x}&\text{x}\end{vmatrix} [$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 $R_3 \rightarrow R_3 - R_1]$
$=-\text{x}[\text{x}\sin\text{x}-\sin\text{x}-\text{x}\sin\text{x}+\text{x}\cos\text{x}]$
$=-\text{x}(\text{x}\cos\text{x}-\sin\text{x})$
$\therefore\ \lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}}{\text{x}^2}=\lim_\limits{\text{x}\rightarrow0}\frac{\text{x}(\sin\text{x}-\text{x}\cos\text{x})}{\text{x}^2}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}^2}-\lim_\limits{\text{x}\rightarrow0}\cos\text{x}$
$=1-1=0$
Hence, the correct option is $(a)$

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MCQ 1131 Mark

Find the minor of the element 1 in the determinant $\triangle=\begin{bmatrix}1&5\\3&8\end{bmatrix}$ is:

A
5
B
1
✓
8
D
3

Answer

Correct option: C.

The minor of the element 1 can be obtained by deleting the first row and the first column

$\therefore\text{ M}_{11}=8$

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MCQ 1141 Mark

The value of the determinant $\begin{vmatrix} 5 &\text{amp; } 1 \\ 3 &\text{amp; } 2 \end{vmatrix}$

A
4
B
5
C
6
✓
7

Answer

Correct option: D.

5 × 2 - 1 × 3 = 7

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MCQ 1151 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 $|A| = 0$
✓
$\begin{vmatrix}\text{a}&\text{c}\\\text{e}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{b}&\text{c}\\\text{f}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{a}&\text{d}\\\text{e}&\text{h}\end{vmatrix}+\begin{vmatrix}\text{b}&\text{d}\\\text{f}&\text{h}\end{vmatrix}$

Answer

Correct option: D.

$\begin{vmatrix}\text{a}&\text{c}\\\text{e}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{b}&\text{c}\\\text{f}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{a}&\text{d}\\\text{e}&\text{h}\end{vmatrix}+\begin{vmatrix}\text{b}&\text{d}\\\text{f}&\text{h}\end{vmatrix}$

$\begin{vmatrix}\text{a}+\text{b}&\text{c}+\text{d}\\\text{e}+\text{f}&\text{g}+\text{h} \end{vmatrix}=\begin{vmatrix}\text{a}+\text{b}&\text{c}\\\text{e}+\text{f}&\text{h} \end{vmatrix}+\begin{vmatrix}\text{a}+\text{b}&\text{d}\\\text{e}+\text{f}&\text{h}\end{vmatrix}$
$=\begin{vmatrix}\text{a}&\text{c}\\\text{e}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{b}&\text{c}\\\text{f}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{a}&\text{d}\\\text{e}&\text{h}\end{vmatrix}+\begin{vmatrix}\text{b}&\text{d}\\\text{f}&\text{h}\end{vmatrix}$

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MCQ 1161 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:

A
-2
B
1
✓
0
D
-1

Answer

Correct option: C.

$\text{P}^3=\text{Q}^3$
$\Rightarrow\text{P}^3- \text{P}^2\text{Q}=\text{Q}^3- \text{Q}^2\text{P}$

$\Rightarrow\text{P}^2(\text{P- Q})=\text{Q}^2(\text{Q- P})$

$\Rightarrow \text{P}^2(\text{P- Q})-\text{Q}^2(\text{Q- P})=0$

$\Rightarrow (\text{P}^2+\text{Q}^2)(\text{P}-\text{Q})=0\Rightarrow|\text{P}^2+\text{Q}^2|=0$

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MCQ 1171 Mark

Evaluate $\begin{bmatrix}1&0&1\\0&0&1\\1&0&1\end{bmatrix}$ is:

✓
2
B
0
C
1
D
-1

Answer

Correct option: A.

$\triangle=\begin{bmatrix}1&0&1\\0&0&1\\1&0&1\end{bmatrix}$

$\triangle=1\begin{bmatrix}0&1\\0&1\end{bmatrix}-0\begin{bmatrix}0&1\\1&1\end{bmatrix}+1\begin{bmatrix}0&0\\1&0\end{bmatrix}$

$\triangle=1(0-0)-0(0-1)+1(0-0)$

$\triangle=0-0+0=0.$

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MCQ 1181 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
There is no solution.
D
None of these.

Answer

Correct 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.

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MCQ 1191 Mark

If the coordinates of the vertices of a triangle are (0, 0), (0, 2) and(3, 1), then area of the triangle is:

✓
sq.units
B
-3 sq.units
C
2 sq.units
D
1 sq.units

Answer

Correct option: A.

sq.units

Area of triangle $=\frac{1}{2} \begin{vmatrix} 0 &\text{amp; }0 &\text{amp; 1} \\ 0&\text{amp; 2} &\text{amp; 1} \\3 &\text{amp;1} &\text{amp; 1} \end{vmatrix}= \frac{1}{2}\times|-6|=3$

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MCQ 1201 Mark

If for the matrix $A, A^3 = I,$ than $A^{-1} =$

✓
$A^2$
B
$A^3$
C
$A$
D
None of these.

Answer

Correct option: A.

$A^2$

$A^3 = Ia$
$\Rightarrow A^{-1}A^3 = A^{-1}I$
$\Rightarrow IA^2 = A^{-1}I$
$\Rightarrow A^2 = A^{-1}$

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MCQ 1211 Mark

If $x, y, z$ 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
$\text{xyz}$
B
$x^{-1} + y^{-1} + z^{-1}$
C
$-x - y - z$
✓
$-1$

Answer

Correct option: D.

$-1$

$\begin{vmatrix}1+\text{x}&1&1\\1&1+\text{y}&1\\1&1&1+\text{z}\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}&0&-\text{z}\\0&\text{y}&-\text{z}\\1&1&1+\text{z}\end{vmatrix}=0 [$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 $(d)$

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MCQ 1221 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
None of these.

Answer

Correct 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.

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MCQ 1231 Mark

Evaluate $\begin{bmatrix}5&4&3\\3&4&1\\5&6&1\end{bmatrix}$is:

A
4
B
-24
✓
-8
D
8

Answer

Correct option: C.

-8

Expanding along the first row, we get
$\triangle=5\begin{bmatrix}4&1\\6&1\end{bmatrix}-4\begin{bmatrix}3&1\\5&1\end{bmatrix}+3\begin{bmatrix}3&4\\5&6\end{bmatrix}$

$=5(4-6)-4(3-5)+3(18-20)$

$=5(-2)-4(-2)+3(-2)=-10+8-6=-8.$

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MCQ 1241 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
$-\sin^2\theta$

Answer

Correct 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) $

$|\text{A}|=-\cos^2\theta⁡+1=\sin^2⁡\theta.$

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MCQ 1251 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
$2, 4$

Answer

Correct option: A.

$1, 3$

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MCQ 1261 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
$6+3\sqrt{2}$
✓
$6+\sqrt{2}$

Answer

Correct option: D.

$6+\sqrt{2}$

$\triangle=\begin{bmatrix}\sqrt{3}&\sqrt{2}\\-1&2\sqrt{3}\end{bmatrix}$

$\triangle=(\sqrt{3}\times2\sqrt{3})+\sqrt{2}$

$\triangle=6+\sqrt{2}$

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MCQ 1271 Mark

The system of equation x + y + z = 2, 3x − y + 2z = 6 and 3x + y + z = −18 has:

✓
A unique solution.
B
No solution.
C
An infinite number of solutions.
D
Zero solution as the only solution.

Answer

Correct option: A.

A unique solution.

The given system of equations can be written in matrix form as follows:

$\begin{bmatrix}1&1&1\\3&-1&2\\3&1&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}2\\6\\-18\end{bmatrix}$

$\text{AX}=\text{B}$

Here,

$\text{A}=\begin{bmatrix}1&1&1\\3&-1&2\\3&1&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}2\\6\\-18\end{bmatrix}$

$|\text{A}|=1(-1-2)-1(3-6)+1(3+3)$

$=-3+3+6$

$=6\neq0$

So, the given system of equations has a unique solution.

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MCQ 1281 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
$-\frac{\sqrt{3}}{2}$

Answer

Correct option: A.

$\frac{1}{2}$

$\triangle=\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1+\cos\theta&1&1\end{vmatrix}$
$=\begin{vmatrix}1&1&1\\0&\sin\theta&0\\\cos\theta&0&0\end{vmatrix} [$Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1]$
$=-\sin\theta\cos\theta$
$=-\frac{\sin2\theta}{2}$
Now, maximum and minimum value of $\sin\theta$ is $1$ and $-1.$
So, the maximum value of $-\sin\theta$ is $1.$
So, the maximum value of $-\sin2\theta$ is $1.$
Therefore, the maximum value of $-\frac{\sin2\theta}{2}$ is $\frac{1}{2}$
Hence, the correct option is $(a)$

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MCQ 1291 Mark

Evaluate $\begin{bmatrix}8\text{x}+1&2\text{x}-2\\\text{x}^2-1&3\text{x}+5\end{bmatrix}$ is:

A
$-2x^3- 26x^2+ 45x + 3$
✓
$-2x^3+ 26x^2+ 45x + 3$
C
$-2x^3+ 26x^2+ 45x - 3$
D
$-2x^3- 26x^2- 45x + 3$

Answer

Correct option: B.

$-2x^3+ 26x^2+ 45x + 3$

Expanding along the first row, we get
$\triangle=8\text{x}+1(3\text{x}+5)-(2\text{x}-2)(\text{x}^2-1)$
$=(24\text{x}^2+43\text{x}+5)-(2\text{x}^3-2\text{x}^2-2\text{x}+2)$
$=-2\text{x}^3+26\text{x}^2+45\text{x}+3.$

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MCQ 1301 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
$9x^2(x + y)$
✓
$9y^2(x + y)$
C
$3y^2(x + y)$
D
$7x^2(x + y)$

Answer

Correct option: B.

$9y^2(x + y)$

$\begin{vmatrix}\text{x}&\text{x}+\text{y}&\text{x}+2\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\\text{x}+\text{y}&\text{x}+2\text{y}&\text{x}\end{vmatrix}$
$=\begin{vmatrix}-2\text{y}&\text{y}&\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\-\text{y}&2\text{y}&-\text{y}\end{vmatrix} \ [$Applying $R_1 \rightarrow R_1 - R_2$ and $R_3\rightarrow R_3 - R_2]$
$=\text{y}^2\begin{vmatrix}-2&1&1\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\-1&2&-1\end{vmatrix}\ [$Taking $(y)$ common from $R_1$ and from $R_3]$
$=\text{y}^2\begin{vmatrix}-2&-3&3\\\text{x}+2\text{y}&3\text{x}+4\text{y}&-\text{y}\\-1&0&0\end{vmatrix} \ [$Applying $C_2 \rightarrow C_2 + 2C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=\text{y}^2\big[-1(3\text{y}-9\text{x}-12\text{y})\big]$
$=\text{y}^2[9\text{y}+9\text{x}]$
$=9\text{y}^2(\text{y}+\text{x})$
Hence, the correct option is $(b)$

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MCQ 1311 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:

✓
n
B
a
C
x
D
None of these.

Answer

Correct option: A.

Let A = nx, B = (n - 1)x, C = (n + 2)x

⇒ C - B = x, B - A = x, C - A = 2x

Thus, the given determinant is

$\begin{vmatrix}\text{a}^2&\text{a}&1\\\cos\text{A}&\cos\text{B}&\cos\text{C}\\\sin\text{A}&\sin\text{B}&\sin\text{C}\end{vmatrix}$

$=\text{a}^2(\cos\text{B}\sin\text{C}-\cos\text{C}\sin\text{B})-\text{a}\times(\cos\text{A}\sin\text{C}-\cos\text{C}\sin\text{A})\\+1\times(\cos\text{A}\sin\text{B}-\sin\text{A}\cos\text{B})$

$=\text{a}^2\sin(\text{C}-\text{B})-\text{a}\sin(\text{C}-\text{A})+\sin(\text{B}-\text{A})$

$=\text{a}^2\sin{\text{x}}-\text{a}\sin2\text{x}+\sin\text{x}$ [Independent of n]

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MCQ 1321 Mark

Evaluate $\begin{bmatrix}5&0&5\\1&4&3\\0&8&6\end{bmatrix}$ is:

A
20
✓
0
C
-40
D
40

Answer

Correct option: B.

$\triangle=\begin{bmatrix}5&0&5\\1&4&3\\0&8&6\end{bmatrix}$
Expanding along $\text{R}_1,$ we get:

$\triangle=5\begin{bmatrix}4&3\\8&6\end{bmatrix}-0\begin{bmatrix}1&3\\0&6\end{bmatrix}+5\begin{bmatrix}1&4\\0&8\end{bmatrix}$

$\triangle=5(24-24)-0+5(8-0)$

$\triangle=0-0+40=40.$

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MCQ 1331 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 $\text{x}=$

A
-4
B
-2
✓
1
D
-3

Answer

Correct option: C.

We know that, If Dett = 0 there is no inverse

⇒ D = x(3 - 4) - 1(2 - 4) + 1(2 - 3) = 0

⇒ -x + 2 - 1 = 0

⇒ x = 1

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MCQ 1341 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:

A
$20$
B
$100$
✓
$10$
D
$0$

Answer

Correct option: C.

$10$

$\text{A(adj A)}\begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix}$
By definition, we have
$\text{A(adj A) = |A|I = (adj A)A}\ ($Where $I$ is the identity matrix$)$
$\Rightarrow \text{|A|I = A(adj A)}$
$\Rightarrow|\text{A}|\text{I}=10\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$\Rightarrow|\text{A}|=10$

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MCQ 1351 Mark

If $A,\ B$ are two $n \times n$ non $-$ singular matrices, then

✓
$AB$ is non $-$ singular.
B
$AB$ is singular.
C
$(AB)^{-1} A^{-1} B^{-1}.$
D
$(AB)^{-1}$ does not exist.

Answer

Correct 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,
$|AB| = |A||B|$
$\Rightarrow|\text{AB}|\neq0\ \big[\text{Using (1)}\big]$
Thus $, Ab$ is non $-$ singular.

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MCQ 1361 Mark

If $A^2 - A + I = 0,$ then the inverse of $A$ is:

A
$A^{-2}$
B
$A + I$
✓
$I - A$
D
$A - I$

Answer

Correct option: C.

$I - A$

$A^2 - A + I = 0$
$\Rightarrow A^{-1}A^2 - A^{-1}A + A^{-1}I = A^{-1}0$
$\Rightarrow A - I + A^{-1} = 0$
$\Rightarrow A^{-1} = I - A$

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MCQ 1371 Mark

Which of the following is correct?

✓
Determinant is a square matrix.
B
Determinant is a number associated with a matrix.
C
Determinant is a number associated with a square matrix.
D
None of these.

Answer

Correct option: A.

Determinant is a square matrix.

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MCQ 1381 Mark

How much is the area of rectangular field:

A
$60000 \ \text{sq.m}$
✓
$30000 \ \text{sq.m}$
C
$30000m$
D
$3000m$

Answer

Correct option: B.

$30000 \ \text{sq.m}$

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MCQ 1391 Mark

Evaluate $\begin{bmatrix}\text{i}&-1\\-1&\text{i}\end{bmatrix}$

✓
4
B
3
C
2
D
0

Answer

Correct option: A.

Expanding along $\text{R}_1,$ we get.

$\triangle=\text{-i}(\text{i})-(-1)(-1)=-\text{i}^2-1=-(-1)-1=0.$

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MCQ 1401 Mark

The number of line segments possible with three collinear points is $........$

A
$1$
B
$2$
✓
$3$
D
Infinite

Answer

Correct option: C.

$3$

Let three collinear points be $A, B, C$
They can represent three line segments namely, $AB, BC, AC$.
Thus namely $3$ line segments are possible with three collinear points.

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MCQ 1411 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:

A
$0$
B
$-16$
C
$16$
✓
None of these.

Answer

Correct option: D.

None of these.

$\triangle=\begin{vmatrix}\text{x}&2&\text{x}\\\text{x}^2&\text{x}&6\\\text{x}&\text{x}&6\end{vmatrix}$
$=\text{x}\begin{vmatrix}\text{x}&6\\\text{x}&6\end{vmatrix}-\text{x}^2\begin{vmatrix}2&\text{x}\\\text{x}&6\end{vmatrix}+\text{x}\begin{vmatrix}2&\text{x}\\\text{x}&6\end{vmatrix}$
$= 0 - x^2(12 - x^2) + x(12 - x^2)$
$= x^4 - 12x^2 + 12x - x^3$
$= ax^4 + bx^3 + cx^2 + dx + e$
$\Rightarrow x^4 - 12x^2 + 12x - x^3 $
$= ax^4 + bx^3 + cx^2 + dx + e$
$\Rightarrow a = 1, b = -1, c = -12, d = 12, e = 0$
Thus,
$5a + 4b + 3c + 2d + e $
$= 5 - 4 - 36 + 24 + 0 = -11$

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MCQ 1421 Mark

Find the minor of the element 2 in the determinant $\triangle=\begin{bmatrix}1&9\\2&3\end{bmatrix}$?

A
3
✓
9
C
1
D
2

Answer

Correct option: B.

The minor of the element 2 can be obtained by deleting the first row and the first column

$\therefore\text{M}_{11}=9$

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MCQ 1431 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 $A_{ij}$ is cofactors of $a_{ij},$ 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
$a_{21}A_{11} + a_{22}A_{12} + a_{23}A_{13}$
✓
$a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}$

Answer

Correct 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 $d$.

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MCQ 1441 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

Answer

Correct option: D.

Now-existent

$\begin{bmatrix} 5 & 10 & 3 \\ -2 & -4 & 6 \\ -1 & -2 & \text{b} \end{bmatrix}$ is singular matrix.

$\Rightarrow\begin{vmatrix} 5 & 10 & 3 \\ -2 & -4 & 6 \\ -1 & -2 & \text{b} \end{vmatrix}=0$

⇒ 5(-4b + 12) - 10(-2b + 6) + 3(4 - 4) = 0

⇒ -20b + 60 + 20b - 60 = 0

b does not exist.

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MCQ 1451 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 $C_{ij}$ is cofactor of $a_{ij}$ 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
$a_{21}C_{11} + a_{22}C_{12} + a_{23}C_{13}$
✓
$a_{11}C_{11} + a_{21}C_{21} + a_{13}C_{31}$

Answer

Correct 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.

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MCQ 1461 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
$\frac{1}{\text{xyz}}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$

Answer

Correct option: A.

$\begin{bmatrix}\text{x}^{-1}&0&0\\0&\text{y}^{-1}&0\\0&0&\text{z}^{-1}\end{bmatrix}$

$\text{A}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$
$\therefore |A| = x (yz - 0) = xyz \neq 0$
Now, $A_{11} = yz, A_{12} = 0, A_{13}= 0$
$A_{21} = 0, A_{22} = xz, A_{23} = 0$
$A_{31} = 0, A_{32} = 0, A_{33} = xya$
$\therefore\text{adj. A}=\begin{bmatrix}\text{yz}&0&0\\0&\text{xz}&0\\0&0&\text{xy}\end{bmatrix}$
$\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj. A}$
$=\frac{1}{\text{xyz}}\begin{bmatrix}\text{yz}&0&0\\0&\text{xz}&0\\0&0&\text{xy}\end{bmatrix}$
$=\begin{bmatrix}\frac{\text{yz}}{\text{xyz}}&0&0\\0&\frac{\text{xz}}{\text{xyz}}&0\\0&0&\frac{\text{xy}}{\text{xyz}}\end{bmatrix}$
$=\begin{bmatrix}\frac{1}{\text{x}}&0&0\\0&\frac{1}{\text{y}}&0\\0&0&\frac{1}{\text{z}}\end{bmatrix}=\begin{bmatrix}\text{x}^{-1}&0&0\\0&\text{y}^{-1}&0\\0&0&\text{z}^{-1}\end{bmatrix}$
The correct answer is $a$.

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MCQ 1471 Mark

$ \begin{bmatrix}1 & \text{x} & \text{x}^2 \\1 & \text{y} & \text{y}^2 \\1 & \text{z} & \text{z}^2\end{bmatrix}$

A
$(x - y) (y + z)(z + x)$
B
$(x + y) (y - z)(z - x)$
C
$(x - y) (y - z)(z + x)$
✓
$(x - y) (y - z) (z - x)$

Answer

Correct option: D.

$(x - y) (y - z) (z - x)$

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MCQ 1481 Mark

If $ \displaystyle \begin{vmatrix} 2 &\text{amp;} -4 \\ 9 &\text{amp; d}-3 \end{vmatrix}=4$ then $\text{d}=$

A
10
B
−11
C
12
✓
−13

Answer

Correct option: D.

−13

Given, determinant of the matrix $ \displaystyle \begin{vmatrix} 2 &\text{amp;} -4 \\ 9 &\text{amp; d}-3 \end{vmatrix}=4$
$\Rightarrow2(\text{d}−3)−(9)(−4)=4$

$\Rightarrow2\text{d}-6+36 = 4$

$\Rightarrow 2\text{d}=-26$

$\Rightarrow\text{d} = -13$

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MCQ 1491 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:

✓
0
B
-1
C
2
D
3

Answer

Correct option: A.

We 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} $ $\big[\text{Dividing R}_2\text{ and R}_3\text{ by }'\text{t}'\big]$

$\Rightarrow\ \lim\limits_{\text{t}\rightarrow0}\frac{\text{f(t)}}{\text{t}^2}=\begin{vmatrix} \lim\limits_{\text{t}\rightarrow0}\text{t}\cos\text{t}&\lim\limits_{\text{t}\rightarrow0}\text{t}&\lim\limits_{\text{t}\rightarrow0}1\\\lim\limits_{\text{t}\rightarrow0}\text{t}\frac{2\sin}{\text{t}}&\lim\limits_{\text{t}\rightarrow0}1&\lim\limits_{\text{t}\rightarrow0}2\\\lim\limits_{\text{t}\rightarrow0}\text{t}\frac{\sin\text{t}}{\text{t}}&\lim\limits_{\text{t}\rightarrow0}1&\lim\limits_{\text{t}\rightarrow0}1\end{vmatrix}$

$=\begin{vmatrix}1&0&1\\2&1&2\\1&1&1\end{vmatrix}$ $\bigg(\because\lim\limits_{\text{t}\rightarrow 0}\frac{\sin\text{t}}{\text{t}}=1\bigg)$

$=1(1-2)-0+1(2-1)$

$=0$

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MCQ 1501 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

A
Det $(A) = 0$
B
Det $(A) \in (2, \infty)$
C
Det $(A) \in (2, -4)$
✓
Det $(A) \in [2, 4]$

Answer

Correct option: D.

Det $(A) \in [2, 4]$

$\text{A}=\begin{bmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{bmatrix}$
$\therefore|\text{A}|=1(1+\sin^2\theta)-\sin\theta(-\sin\theta+\sin\theta)+1(\sin^2\theta+1)$
$=1+\sin^2\theta+\sin^2\theta+1$
$= 2 + 2 \sin^2 \theta$
$= 2 (1 + \sin^2\theta$)
Now, $0\leq\theta\leq2\pi$
$\Rightarrow 0\leq\sin\theta\leq1$
$\Rightarrow 0\leq1+\sin^2\theta\leq2$
$\Rightarrow2\leq2(1+\sin^2\theta)\leq4$
$\therefore$ Det $(A) \in [2, 4]$
The correct answer is $d.$

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Maths M.C.Q (1 Marks)