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CBSE Questions for Class 12 Commerce Maths Determinants Quiz 1 - MCQExams.com
CBSE
Class 12 Commerce Maths
Determinants
Quiz 1
The number of distinct real roots of the equation, $$\begin{vmatrix} \cos x& \sin x & \sin x\\ \sin x & \cos x & \sin x\\ \sin x & \sin x & \cos x\end{vmatrix} = 0$$
in the interval $$\left [-\dfrac {\pi}{4}, \dfrac {\pi}{4}\right ]$$ is/are :
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$$3$$
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$$2$$
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$$1$$
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$$4$$
Explanation
The given determinant is : $$\begin{vmatrix} \cos x& \sin x & \sin x\\ \sin x & \cos x & \sin x\\ \sin x & \sin x & \cos x\end{vmatrix} = 0$$
Taking $$\cos^3 x$$ common in the above determinant, we get,
$$\cos^3 x\begin{vmatrix} \ 1& tanx & tanx\\ tanx & \ 1 & tanx\\ tanx & tanx & \ 1\end{vmatrix} = 0$$
$$\cos^3x[1(1-\tan^2 x)-tanx(tanx-tan^2 x) + tanx(tan^2x-tanx)]=0$$
$$\cos^3x[1-3tan^2 x+ 2tan^3x]=0$$
We notice that if $$\cos^3x = 0 \rightarrow x=\dfrac{\pi}{2}$$
This is not satisfied by the given interval
Thus, $$x=\dfrac{\pi}{2}$$ is not a solution.
Now, $$1-3tan^2 x+ 2tan^3x=0$$
It is obvious from the above equation that $$tanx=1$$ is a solution.
Thus, $$x=\dfrac{\pi}{4}$$ is a solution.
We just found out that $$(tanx -1)$$ is a factor of the polynomial
$$1-3tan^2 x+ 2tan^3x=0$$
Thus, dividing the polynomial by $$(tanx-1)$$, we get the quotient as $$2tanx^2 - tanx -1 $$ and remainder $$0$$
Thus, this can be written as $$(2tanx+1)(tanx-1)=0$$
Or, $$tanx=\dfrac{-1}{2}$$ or $$tanx=1$$
We already know that
$$tanx=1$$ is a solution to the above equation.
Thus,
$$tanx=\dfrac{-1}{2}$$
Or, $$x= tan^{-1}\left(\dfrac{-1}{2}\right)$$
This also lies in the interval $$\left[\dfrac{-\pi}{4}, \dfrac{\pi}{4}\right]$$
Thus there are two distinct real roots to the above equation.
The points $$\displaystyle \left( 0, \frac{8}{3} \right), (1, 3)$$ and $$(82, 30)$$ :
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form an obtuse angled triangle.
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form a right angled triangle.
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lie on a straight line.
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form an acute angled triangle.
Explanation
Given $$\displaystyle \left( 0, \frac{8}{3} \right), (1, 3)$$ and $$(82, 30)$$
$$\begin{vmatrix} 0 & \dfrac { 8 }{ 3 } & 1 \\ 1 & 3 & 1 \\ 82 & 30 & 1 \end{vmatrix}$$
$$=216-216=0$$
Hence, the given points are collinear.
If $$A=\begin {bmatrix} 2 & -3\\ -4 & 1\end{bmatrix}$$, then adj $$(3A^2+12A)$$ is equal to.
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$$\begin{bmatrix} 72 & -84\\ -63 & 51\end{bmatrix}$$
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$$\begin{bmatrix} 51 & 63\\ 84 & 72\end{bmatrix}$$
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$$\begin{bmatrix} 51 & 84 \\ 63 & 72\end{bmatrix}$$
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$$\begin {bmatrix} 72 & -63\\ -84 & 51\end{bmatrix}$$
Explanation
It is given that $$A =\begin{bmatrix} 2 & -3\\ -4 & 1 \end{bmatrix}$$
Hence, $$ A^2 = \begin {bmatrix} 2 & -3 \\ -4 &1 \end{bmatrix} \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix} =$$
$$\begin{bmatrix} 16 & -9 \\ -12 & 13 \end{bmatrix}$$
Now, $$ 3A^2 = \begin{bmatrix} 48 & -27 \\ -36 & 29\end{bmatrix}$$
$$12A=\begin{bmatrix}24 & -36 \\ -48 & 12\end{bmatrix}$$
Consider, $$3A^2+12A$$
$$=\begin{bmatrix}48 & -27 \\ -36 & 29\end{bmatrix}+\begin{bmatrix}24 & -36 \\ -48 & 12\end{bmatrix}$$
$$=\begin{bmatrix}72 & -63 \\ -84 & 51\end{bmatrix}$$
$$\therefore 3A^2+12A=\begin{bmatrix}72 & -63 \\ -84 & 51\end{bmatrix}$$
Hence,
$$\text{adj}(3A^2+12A)=\begin{bmatrix} 51 & 84 \\ 63 & 72\end{bmatrix}^T=\begin{bmatrix}51&63\\84&72\end{bmatrix}$$
Let $$P = [a_{ij}]$$ be a 3 $$\times$$ 3 matrix and let $$Q = [b_{ij}]$$, where $$b_{ij} = 2^{i + j} a_{ij}$$ for $$1 \leq i, j \leq 3$$. If the determinant of P is 2, then the determinant of the matrix Q is
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$$2^{10}$$
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$$2^{11}$$
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$$2^{12}$$
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$$2^{13}$$
Explanation
Given, $$[P]=a_{ij}$$ is a order 3 matrix.
Let $$Q=\begin{bmatrix} { 2 }^{ 2 }{ a }_{ 11 } & { 2 }^{ 3 }{ a }_{ 12 } & { 2 }^{ 4 }{ a }_{ 13 } \\ { 2 }^{ 3 }{ a }_{ 21 } & { 2 }^{ 4 }{ a }_{ 22 } & { 2 }^{ 5 }{ a }_{ 23 } \\ { 2 }^{ 4 }{ a }_{ 31 } & { 2 }^{ 5 }{ a }_{ 32 } & { 2 }^{ 6 }{ a }_{ 33 } \end{bmatrix}$$
$$|Q|=\begin{vmatrix} { 2 }^{ 2 }{ a }_{ 11 } & { 2 }^{ 3 }{ a }_{ 12 } & { 2 }^{ 4 }{ a }_{ 13 } \\ { 2 }^{ 3 }{ a }_{ 21 } & { 2 }^{ 4 }{ a }_{ 22 } & { 2 }^{ 5 }{ a }_{ 23 } \\ { 2 }^{ 4 }{ a }_{ 31 } & { 2 }^{ 5 }{ a }_{ 32 } & { 2 }^{ 6 }{ a }_{ 33 } \end{vmatrix}$$
$$\Rightarrow Q={ 2 }^{ 2 }{ 2 }^{ 3 }{ 2 }^{ 4 }\begin{bmatrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { 2 }{ a }_{ 21 } & { 2 }{ a }_{ 22 } & { 2 }{ a }_{ 23 } \\ { 2 }^{ 2 }{ a }_{ 31 } & { 2 }^{ 2 }{ a }_{ 32 } & { 2 }^{ 2 }{ a }_{ 33 } \end{bmatrix}$$
$$\Rightarrow |Q|={ 2 }^{ 9 }{ \times 2 }\times { 2 }^{ 2 }\begin{vmatrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{vmatrix}$$
$$\Rightarrow |Q|=2^{12}|P|$$
$$\Rightarrow |Q|=2^{13}$$
Consider three points $${P}=(-\sin(\beta-\alpha), -\cos\beta) , {Q}=(\cos(\beta-\alpha), \sin\beta)$$ and $${R}=(\cos(\beta-\alpha +\theta), \sin(\beta-\theta))$$ , where $$0< \alpha,\ \beta,\ \theta <\displaystyle \frac{\pi}{4}$$. Then
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$$P$$ lies on the line segment $$RQ$$
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$$Q$$ lies on the line segment $$PR$$
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$$R$$ lies on the line segment $$QP$$
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$$P, Q, R$$ are non-collinear
Explanation
$$\mathrm{P}\equiv(-\sin(\beta-\alpha), -\cos\beta)\equiv(\mathrm{x}_{1}, \mathrm{y}_{1})$$
$$\mathrm{Q}\equiv(\cos(\beta-\alpha), \sin\beta)\equiv(\mathrm{x}_{2}, \mathrm{y}_{2})$$
and $$\mathrm{R}\equiv(\mathrm{x}_{2}\cos\theta+\mathrm{x}_{1}\sin\theta, \mathrm{y}_{2}\cos\theta+\mathrm{y}_{1}\sin\theta)$$
We see that $$\displaystyle \mathrm{T}\equiv(\frac{\mathrm{x}_{2}\cos\theta+\mathrm{x}_{1}\sin\theta}{\cos\theta+\sin\theta}, \frac{\mathrm{y}_{2}\cos\theta+\mathrm{y}_{1}\sin\theta}{\cos\theta+\sin\theta})$$ and $$\mathrm{P},\ \mathrm{Q},\ \mathrm{T}$$ are collinear
$$\Rightarrow \mathrm{P},\ \mathrm{Q},\ \mathrm{R}$$ are non-collinear.
The number of $$A$$ in $$T_p$$ such that $$A$$ is either symmetric or skew-symmetric or both, and det $$(A)$$ divisible by $$p$$, is
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$$(p - 1)^2$$
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$$2(p - 1)$$
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$$(p - 1)^2 + 1$$
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$$2 p - 1$$
Explanation
Given $$A=\begin{bmatrix} a\quad & b \\ c & a \end{bmatrix},a,b,c\in \left\{ 0,1,2,...,p-1 \right\} $$
If A is skew -symmetric matrix, then a =0, b=-c
$$\therefore \left| A \right| =-{ \left| b \right| }^{ 2 }$$
Thus, p divides $$\left| A \right| $$ only when $$b=0$$ ...(1)
Again, if A is symmetric matrix, then $$v=c$$ and $$\left| A \right| ={ a }^{ 2 }-{ b }^{ 2 }$$
Thus p divides $$\left| A \right| $$ is either p divides $$(a-b)$$ or o divides $$(a+b)$$
p divides $$(a-b)$$ only when $$a=b$$
i.e $$a=b\in \left\{ 0,1,2,...,\left( p-1 \right) \right\} $$
i.e p choice
p divides $$(a+b)$$ $$\Rightarrow $$ p choice including $$a=b=0$$ in (1)
Therefore total number of choices are $$\left( p+p+1 \right) =2p-1$$
Let k be a positive real number and let $$A = \begin{bmatrix}2k-1 & 2\sqrt{k} & 2\sqrt{k}\\ 2\sqrt{k} & 1 & -2k\\ -2\sqrt{k} & 2k & -1\end{bmatrix}$$ and $$B=\begin{bmatrix}0 & 2k-1 & \sqrt{k}\\ 1-2k & 0 & 2\sqrt{k}\\ -\sqrt{k} & -2\sqrt{k} & 0\end{bmatrix}$$ .
If det $$(adj A) + det (adj B) = 10^{6}, then [k]$$ is equal to
[Note: adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equal to k].
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$$3$$
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$$4$$
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$$5$$
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$$6$$
Explanation
$$|\mathrm{A}|=(2\mathrm{k}+1)^{3},\ |\mathrm{B}|=0$$ (Since $$\mathrm{B}$$ is a skew-symmetric matrix of order 3)
$$\Rightarrow \det(adj \mathrm{A}) =|\mathrm{A}|^{\mathrm{n}-1}=((2\mathrm{k}+1)^{3})^{2}=10^6\Rightarrow 2\mathrm{k}+1=10\Rightarrow 2\mathrm{k}=9$$
$$[\mathrm{k}]=4$$.
The value of $$|\mathrm{U}|$$ is
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3
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$$-3$$
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3/2
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2
Explanation
$$A=\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$$
$$|A|=1$$
$$Adj\: A=\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}$$
$$\Rightarrow A^{-1}=\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}$$
$$AU_1=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$
$$\Rightarrow U_1=A^{-1}\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}=\begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}$$
$$AU_2=\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}$$
$$\Rightarrow U_2=A^{-1}\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}=\begin{bmatrix} 2 \\ -1 \\ -4 \end{bmatrix}$$
$$AU_3=\begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}$$
$$\Rightarrow U_3=A^{-1}\begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}\begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}=\begin{bmatrix} 2 \\ -1 \\ -3 \end{bmatrix}$$
$$\therefore U=\begin{bmatrix} 1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3 \end{bmatrix}$$
$$U^{-1}=\displaystyle\frac{1}{3}\begin{bmatrix} -1 & -2 & 0 \\ -7 & -5 & -3 \\ 9 & 6 & 3 \end{bmatrix}$$
$$\therefore$$ Sum of elements of $$U^{-1}$$ is $$0$$.
Hence, option B.
Hence $$U=\begin{bmatrix} 1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3 \end{bmatrix}\Rightarrow \left| U \right| =3$$
$$A = \begin{bmatrix}1&2\\3&4\end{bmatrix}, B = \begin{bmatrix}2&1\\3&4\end{bmatrix}$$ then $$\left|(B^TA^T)^{-1}\right|$$ is equal to
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$$10$$
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$$\dfrac{1}{10}$$
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$$1$$
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$$- 1$$
Explanation
$$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$
$$B = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$$
$$B^{T}A^{T} = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 2 + 6 & 6 + 12 \\ 1 + 8 & 3 + 16 \end{bmatrix} = \begin{bmatrix} 8 & 18 \\ 9 & 19 \end{bmatrix}$$
$$det (B^{T}A^{T}) = 8 \times 19 - 9 \times 18 = 152 - 162 = -10$$
$$(B^TA^T)^{-1} = \dfrac1{det(B^TA^T)} adj(B^TA^T) = \dfrac{1}{-10}\begin{bmatrix}19&-18\\-9&8\end{bmatrix}$$
$$det( (B^{T}A^{T})^{-1}) =\dfrac{19\times8 - 18\times9}{-10} = 1$$
If three points $$(k, 2k), (2k, 3k), (3, 1)$$ are collinear, then $$k$$ is equal to:
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$$-2$$
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$$1$$
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$$\displaystyle\frac{1}{2}$$
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$$-\displaystyle\frac{1}{2}$$
Explanation
If given points are collinear, then
$$\begin{vmatrix} k&2k&1\\2k&3k&1\\3&1&1\end{vmatrix}=0$$
$$\Rightarrow \begin{vmatrix} k&2k&1\\k&k&0\\3&1&1\end{vmatrix}=0, R_2\rightarrow R_2-R_1$$
$$\Rightarrow 1(k-3k)+1(k^2-2k^2)=0$$
$$\Rightarrow k^2+2k=0$$
$$\Rightarrow k=0,-2$$
$$\begin{vmatrix} 4\sin^{2}\theta & \cos^{2}\theta \\ 3\sec^{2}\theta & \text{cosec}^{2}\theta \end{vmatrix}=$$
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$$8\sin^{2}\theta \cos^{2}\theta $$
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$$4\sin 2\theta \cos 2\theta $$
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$$1$$
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$$4\cos^{3}\theta -3\cos \theta $$
Explanation
The value of $$\begin{vmatrix} 4\sin^{2}\theta & \cos^{2}\theta \\ 3\sec^{2}\theta & \text{cosec}^{2}\theta \end{vmatrix}$$ is
$$=4\sin^{2}\theta .\text{cosec}^{2}\theta -3\sec^{2}\theta .\cos^{2}\theta$$
$$=4-3=1$$
$$\left|\begin{array}{lllll}
0 & & \mathrm{c}\mathrm{o}\mathrm{s}\alpha & \mathrm{c}\mathrm{o}\mathrm{s} & \beta\\
\mathrm{c}\mathrm{o}\mathrm{s} & \alpha & 0 & \mathrm{c}\mathrm{o}\mathrm{s} & \gamma\\
\mathrm{c}\mathrm{o}\mathrm{s} & \beta & \mathrm{c}\mathrm{o}\mathrm{s}\gamma & 0 &
\end{array}\right|=$$
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$$\cos\alpha+\cos\beta+\cos\gamma$$
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$$\cos\alpha\cos\beta\cos\gamma$$
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$$ 2\cos\alpha\cos\beta\cos\gamma$$
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$$2 \displaystyle \sum \cos\alpha\cos\beta$$
Explanation
Let $$A=\begin{vmatrix}
0 & cos \alpha & \cos \beta\\
\cos \alpha & 0 & cos \gamma\\
cos \beta & \cos \gamma & 0
\end{vmatrix}$$
Expanding the determinant,
$$A=-cos \alpha[-cos \beta cos \gamma]+cos \beta[cos \alpha \cos \gamma]$$
$$=2 \cos \alpha \cos \beta \cos \gamma$$
$$A=\left\{\begin{array}{ll}
8 & 9\\
10 & 11
\end{array}\right\}$$, then cofactor of $$\mathrm{a}_{12}$$ is:
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11
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10
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-11
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-10
Explanation
By property of cofactor (2),
$$A=\begin{bmatrix}
8 & 9\\
10 & 11
\end{bmatrix}$$
minor of $$a_{12}=M_{12}=10$$
So, cofactor of $$a_{12}=M_{12}(-1)^{1+2}=-M_{12}=-10$$
If P =$$\begin{bmatrix}
1 & 4\\
2 & 6
\end{bmatrix}$$ ,then adj (P)
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$$\begin{bmatrix}1 & 4\\ 2 & 6\end{bmatrix}$$
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$$\begin{bmatrix}6& -4\\ -2 & 1\end{bmatrix}$$
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$$\begin{bmatrix}6& -2\\ -4 & 1\end{bmatrix}$$
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$$\begin{bmatrix}2& 1\\ 6 & 4\end{bmatrix}$$
Explanation
Cofactor of the martix is
$$\begin{bmatrix} M11 & -M12 \\ -M21 & M22 \end{bmatrix}\\$$
Where$$ M11 = 6$$
$$M12 = 2$$
$$M21 = 4$$
$$M22 = 1$$
Substituting all value we get
Cofactor of the matrix as
$$\begin{bmatrix} 6 & -2 \\ -4 & 1 \end{bmatrix}\\$$
The adjoint of the matrix is transpose of cofactor of the given matrix
$$\therefore $$ Adjoint of the matrix is
$$\begin{bmatrix} 6 & -4 \\ -2 & 1 \end{bmatrix}$$
Hence the answer is option B.
$$\begin{vmatrix}
x^{2}+3 &x-1 &x+3 \\
x+3 & -2x &x-4 \\
x-3& x+4 & 3x
\end{vmatrix}$$ $$=px^{4}+qx^3+rx^{2}+sx+t,$$ then $$t = $$
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$$72$$
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$$0$$
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$$24$$
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$$-48$$
Explanation
On expanding the determinant we get,
$$(x^2+3)(-6x^2-x^2+16)-(x-1)(3x^2+9x-x^2+7x-12)+(x+3)(x^2+7x+12-2x^2+6x)$$
Let us consider only the constant terms,
since $$t$$ is constant term
$$\therefore (16)(3)-(-1)(-12)+3(12) =72$$
Hence, $$t = 72 $$
Maximum value of a second order determinant whose every element is either 0,1 or 2 only is:
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0
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1
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2
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4
Explanation
So, $$A=\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}
$$ Given a,b,c & D can only be 0,1,2
det A= ad-bc
So for max. value of A,
a=2 and d=2 and $$b,c \epsilon {0,0}$$
So, Max value of det $$A=\begin{bmatrix}
2 & 0\\
0 & 2
\end{bmatrix}=4$$
If $$\begin{vmatrix}
cos(A+B) & -sin(A+B) &cos2B \\
sin A& cos A &sin B \\
-cos A& sin A & cos B
\end{vmatrix}$$ =0 then B=
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(2n+1)$$\frac{\pi }{2}$$
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n$$\pi $$
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(2n+1)$$\pi $$
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2n$$\pi $$
Explanation
$$\Rightarrow cos(A+B)(cosAcos B-sin A sin B)+sin^2(A+B)+cos2B+0$$
$$\Rightarrow cos^2(A+B)+sin^2(A+B)+cos2B=0$$
$$\Rightarrow cos (2B)=-1$$
$$2B=(2n+1)\pi$$
$$B=(2n+1)\pi/2$$
so $$ (2n+1)\pi/2$$
If A $$ =\begin{bmatrix}
0 & c &-b \\
-c& 0& a\\
b & -a & 0
\end{bmatrix}$$ then $$\left ( a^{2}+b^{2}-c^{2} \right )\left | A \right |=$$
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$$abc$$
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$$a + b + c$$
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$$\left ( a^{3}+b^{3}+c^{3} \right )$$
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$$0$$
Explanation
$$|A|=0\times(a^2)-c(-ab)-b(ac)$$
$$=0+abc-abc$$
$$=0$$
Find x if it is given that:
$$\det \left[\begin{array}{lll}
2 & 0 & 0\\
4 & 3 & 0\\
4 & 6 & x
\end{array}\right]=42$$
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$$8$$
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$$7$$
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$$6$$
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$$21/4$$
Explanation
Given $$det\begin{pmatrix}
2 & 0 & 0\\
4 & 3 & 0\\
4 & 6 & x
\end{pmatrix}=42$$
By operation of matrix (5)
$$x(6)=42$$
$$\Rightarrow x=7$$
$$\mathrm{If}$$ $$\left|\begin{array}{lll}
1 & 0 & 0\\
2 & 3 & 4\\
5 & -6 & x
\end{array}\right|$$ $$= 45$$ $$\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}$$ $$\mathrm{x}=$$
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$$4$$
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$$7$$
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$$- 5$$
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$$-7$$
Explanation
Given $$\begin{vmatrix}
1 & 0 & 0\\
2 & 3 & 4\\
5 & -6 & x
\end{vmatrix}=45$$
By operation of matrix (5),
$$1(3x+24)=45$$
$$3x=21$$
$$\Rightarrow x=7$$
If A is a singular matrix, then A (adj A) is a
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scalar matrix
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zero matrix
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identity matrix
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orthogonal matrix
Explanation
Given $$A$$ is a singular matrix.
$$\Rightarrow |A|=0$$
$$A(adj\:A)=|A|I=0I=O$$
$$\therefore A(adj\:A)$$ is a zero matrix.
Hence, option B.
If A is a singular matrix, then adj A is
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$$\displaystyle non-singular$$
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singular
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symmetric
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not defined
Explanation
Given $$|A|=0$$
We know $$|adj A|=|A|^{n-1}$$
$$\therefore |adj A|=0$$
Hence, adj A is singular
If the value of the determinant $$\begin{vmatrix}m & 2\\ -5 & 7\end{vmatrix}$$ is $$31$$, find $$m$$.
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$$3$$
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$$8$$
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$$6$$
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$$2$$
Explanation
Given,
$$\begin{vmatrix}m & 2\\ -5 & 7\end{vmatrix}$$
$$
\Rightarrow 7m- (-10)=31$$
$$\Rightarrow 7m+ 10= 31$$
$$\Rightarrow 7m= 21$$
$$\Rightarrow m= 3
$$
If $$A$$ is a $$3\times 3$$ matrix and $$\text{det} (3A)=k(\text{det} A)$$, then $$k=$$
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$$9$$
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$$6$$
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$$1$$
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$$27$$
Explanation
The non zero determinant of a scalar multiple of a n×n matrix is given by the following property.
$$|kA|=kn|A|$$
⟹$$|3A|=3^3|A|=27|A|$$
$$k=27$$
If A and B are similar matrices such that $$det (AB)=0, $$ then
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$$det (A)=0 $$ and $$ det (B)=0$$
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$$det (A)=0$$ and $$ det (B) \neq 0$$
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$$A=0$$ and $$B=0$$
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$$A=0 $$ or $$B=0$$
Explanation
Given. two n$$\times$$n square matrices A and B
Given, $$B=P^{-1}AP$$
$$det\left( B \right) =det\left( { P }^{ -1 }AP \right) $$
$$det\left( B \right) =det\left( { P }^{ -1 }AP \right) $$
$$det\left( B \right) =\left( det\left( { P } \right) \right) ^{ -1 }\left( det\left( A \right) \right) \left( det\left( P \right) \right) $$
$$det\left( B \right) =\left( det\left( { P } \right) \right) ^{ -1 }\left( det\left( A \right) \right) \left( det\left( P \right) \right) $$
$$det\left( B \right) =det\left( A \right) $$
$$det\left( AB \right) =\left( det\left( A \right) \right) \left( det\left( B \right) \right) $$
If$$\left( det\left( A \right) \right) =0$$ then $$\left( det\left( B \right) \right) $$ also zero.
Let a, b, c be three complex numbers, and let
$$z=\begin{vmatrix}
0 & -b & -c\\
b & 0 & -a\\
c & a & 0
\end{vmatrix}$$
then z equal
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$$0$$
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purely imaginary
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$$abc$$
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none of these
Explanation
$$z=\begin{vmatrix} 0 & -b & -c \\ b & 0 & -a \\ c & a & 0 \end{vmatrix}=0\left( 0+{ a }^{ 2 } \right) +b\left( 0+ac \right) -c\left( ab-0 \right) =0$$
Two $$n\times n$$ square matrices A and B are said to be similar if there exists a non-singular matrix P such that
$$P^{-1}A P=B$$.
If A and B are similar matrices such that $$det (A)=1$$, then
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$$det (B)=1$$
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$$det (A)=-det (B)$$
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$$det (B)=-1$$
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none of these
Explanation
Given, two n$$\times$$n square matrices A and B
Given, $$B=P^{-1}AP$$
$$det\left( B \right) =det\left( { P }^{ -1 }AP \right) $$
$$det\left( B \right) =\left( det\left( { P }^{ -1 } \right) \right) \left( det\left( A \right) \right) \left( det\left( P \right) \right) $$
$$det\left( B \right) =\left( det\left( { P } \right) \right) ^{ -1 }\left( det\left( A \right) \right) \left( det\left( P \right) \right) $$ $$[\because \quad|A||A^{-1}|=|AA^{-1}|=|I|=1]$$
$$det\left( B \right) =det\left( A \right) =1$$
If A is a square matrix of order $$n$$ then adj $$\left ( adj A \right )$$ is equal to
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$$\displaystyle \left | A \right |^{n}A$$
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$$\displaystyle \left | A \right |^{n-1}A$$
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$$\displaystyle \left | A \right |^{n-2}A$$
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$$\displaystyle \left | A \right |^{n-3}A$$
Explanation
$${ A }^{ -1 }=\cfrac { AdjA }{ detA } =\cfrac { I }{ A } =\cfrac { AdjA }{ \left| A \right| } $$
$$A(adjA)=\left| A \right| $$
Replacing $$A=adjA$$
$$adjA(AdjAdjA)=\left| AdjA \right| $$
$$\Rightarrow Adj\left( AdjA \right) =\cfrac { \left| Adj(A) \right| }{ \left| A \right| } \times A$$
$$=\cfrac { { \left| A \right| }^{ n-1 } }{ { \left| A \right| } } \times A$$
$$={ \left| A \right| }^{ n-2 }\times A$$
$$Adj(AdjA)={ \left| A \right| }^{ n-2 }\times 2$$
$$n\rightarrow $$order
Option C is correct
If$$\displaystyle \left | \begin{matrix}-12 &0 &\lambda \\ 0& 2& -1\\ 2& 1 &15 \end{matrix} \right |=-360$$, then the value of $$\lambda$$,is
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$$-1$$
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$$-2$$
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$$-3$$
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$$4$$
Explanation
$$\left| \begin{matrix} -12 & 0 & \lambda \\ 0 & 2 & -1 \\ 2 & 1 & 15 \end{matrix} \right| =-360$$
Expanding along first column
$$\Rightarrow (-12)(31)-4\lambda =-360$$
$$\Rightarrow \lambda =-3$$
If $$A$$ is any skew-symmetric matrix of odd order then $$\left| A \right| $$ equals
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0%
$$-1$$
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$$0$$
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$$1$$
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none of these
Explanation
if $$A$$ is skew symmetric matrix
then $$A=-A^T$$
Therefore, $$|A|=-|A^T|=-|A|$$
$$\Rightarrow 2|A|=0$$
$$\Rightarrow|A|=0$$
Ans: B
If $$\begin{vmatrix} x & y\\ 4 & 2 \end{vmatrix}=7$$ and $$\begin{vmatrix} 2 & 3\\ y & x \end{vmatrix}=4$$ then
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$$x=-3, y=-\dfrac {5}{2}$$
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$$x=-\dfrac {5}{2}, y=-3$$
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$$x=-3, y=\dfrac {5}{2}$$
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$$x=-\dfrac {5}{2}, y=3$$
Explanation
$$\begin{vmatrix} x & y\\ 4 & 2 \end{vmatrix}=7$$
$$\Rightarrow 2x-4y=7$$
$$\begin{vmatrix} 2 & 3\\ y & x \end{vmatrix}=4$$
$$\Rightarrow 2x-3y=4$$
Therefore, $$x=-5/2,y=-3$$
Ans: B
If $$\begin{vmatrix}a & -b & -c\\-a & b & -c \\ -a & -b & -c\end{vmatrix}+\lambda abc=0$$, then $$\lambda$$ is equal to
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$$2$$
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$$4$$
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$$-2$$
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$$-4$$
Explanation
$$\begin{vmatrix} a & -b & -c \\ -a & b & -c \\ -a & -b & -c \end{vmatrix}+\lambda abc=0$$
Substitute $$a=b=c=1$$
$$\begin{vmatrix} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & -1 \end{vmatrix}+\lambda =0$$
$$(-2-2)+\lambda=0$$
$$\Rightarrow \lambda =4$$
The cofactors of elements in second row of the determinant $$\begin{vmatrix} 2 & -1 & 4 \\ 4 & 2 & -3 \\1 & 1 & 2 \end{vmatrix}$$ are
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$$5, 6, 4$$
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$$6, 0, -3$$
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$$5, 1, 8$$
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$$6, 0, 3$$
Explanation
$$\begin{vmatrix} 2 & -1 & 4 \\ 4 & 2 & -3 \\ 1 & 1 & 2 \end{vmatrix}\\ { C }_{ 21 }=-\begin{vmatrix} -1 & 4 \\ 1 & 2 \end{vmatrix}=6\\ { C }_{ 22 }=\begin{vmatrix} 2 & 4 \\ 1 & 2 \end{vmatrix}=0\\ { C }_{ 23 }=-\begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix}=-3$$
A set of points which do not lie on the same line are called as
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collinear
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non-collinear
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concurrent
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square
Explanation
A set of points which do not lie on the same line are called as non collinear points
$$A$$ and $$B$$ are two points and $$C$$ is any point collinear with $$A$$ and $$B$$. IF $$AB=10$$, $$BC=5$$, then $$AC$$ is equal to:
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either $$15$$ or $$5$$
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necessarily $$5$$
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necessarily $$16$$
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none of these
Explanation
Since $$C$$ is collinear with $$A$$ and $$B, C$$ lies either
(i) to the left of point B or
(ii) to the right of point B
$$\therefore$$ In case (i) $$AC=AB-BC=10-5=5$$
In case (ii) $$AC=AB+BC=10+5=15$$
If $$A=\begin{bmatrix} 3 & -5 \\ -1 & 0 \end{bmatrix}$$, then adj. $$A$$ is equal to
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$$\begin{bmatrix} 3 & 5 \\ 1 & 0 \end{bmatrix}$$
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$$\begin{bmatrix} 0 & -5 \\ -1 & 3 \end{bmatrix}$$
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$$\begin{bmatrix} 0 & 5 \\ 1 & 3 \end{bmatrix}$$
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$$\begin{bmatrix} 0 & -1 \\ -5 & 3 \end{bmatrix}$$
Explanation
co-factor of $$A_{11}=0$$; co-factor of $$A_{12}=5$$; co-factor of $$A_{21}=1$$; co-factor of $$A_{22}=3$$
$$adjA=\begin{bmatrix} 0 & 5 \\ 1 & 3 \end{bmatrix}$$
Ans: C
The value of the determinant $$\begin{vmatrix}a & b & 0\\ 0 & a & b\\ b & 0 &a \end{vmatrix}$$ is equal to
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$$a^3-b^3$$
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$$a^3+b^3$$
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0
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none of these
Explanation
$$\begin{vmatrix} a & b & 0 \\ 0 & a & b \\ b & 0 & a \end{vmatrix}=a\left( { a }^{ 2 }-0 \right) -b\left( 0-{ b }^{ 2 } \right) +0={ a }^{ 3 }+{ b }^{ 3 }$$
The value of the determinant $$\begin{vmatrix} 1 & 2 & 3\\ 3 & 5 & 7\\ 8 & 14 & 20\end{vmatrix}$$ is equal to
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$$20$$
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$$10$$
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$$0$$
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$$45$$
Explanation
The value of the determinant is $$\begin{vmatrix} 1 & 2 & 3 \\ 3 & 5 & 7 \\ 8 & 14 & 20 \end{vmatrix}\\ =1\left( 100-98 \right) -2\left( 60-56 \right) +3\left( 42-40 \right)$$
$$ =2-8+6=0$$
If $$\begin{vmatrix} 6i & -3i & 1\\ 4 & 3i & -1 \\ 20 & 3 & i\end{vmatrix}=x+iy$$, then
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$$x = 3, y = 1$$
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$$x = 1, y = 3$$
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$$x = 0, y = 3$$
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$$x = 0, y = 0$$
Explanation
$$\begin{vmatrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{vmatrix}=x+iy$$
$$ \Rightarrow 6i\left( 3{ i }^{ 2 }+3 \right) +3i\left( 4i+20 \right) +1\left( 12-60i \right) =x+iy$$ $$\Rightarrow 0=x+iy$$
Therefore, $$x=y=0$$
Ans: D
The points which are not collinear are:
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$$(0, 1),\ (8, 3)\ and\ (6, 7)$$
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$$(4, 3),\ (5, 1)\ and\ (1, 9)$$
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$$(2, 5),\ (-1, 2)\ and\ (4, 7)$$
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$$(-3, 2)\, (1, -2)\ and\ (9, -10)$$
Explanation
$$ Let\quad us\quad take\quad three\quad points\quad A,\quad B\quad \& \quad C.\\ Three\quad points\quad A,\quad B\quad \& \quad C\quad will\quad be\quad collinear\\ If\quad AB+BC=AC.\\ Three\quad points\quad A,\quad B\quad \& \quad C\quad will\quad not\quad be\quad collinear\\ If\quad AB+BC=AC.\\ Let\quad us\quad investigate\quad thr\quad given\quad options.\\ Option\quad A\longrightarrow The\quad points\quad are\\ A(0,1)\quad B(8,3)\quad \& \quad C(6,7).\\ \therefore \quad AB=\sqrt { { \left( 0-8 \right) }^{ 2 }+{ \left( 1-3 \right) }^{ 2 } } units=\sqrt { 68 } =8.25units,\\ \quad \quad \quad BC=\sqrt { { \left( 8-6 \right) }^{ 2 }+{ \left( 3-7 \right) }^{ 2 } } units=\sqrt { 28 } =5.29units,\\ and\quad AC=\sqrt { { \left( 0-6 \right) }^{ 2 }+{ \left( 1-7 \right) }^{ 2 } } units=\sqrt { 72 } =8.49units.\\ \therefore \quad AB+BC=\left( 8.25+5.29 \right) units.=13.54units.\\ So\quad AB+BC\neq AC.\\ \therefore \quad A,\quad B\quad \& \quad C\quad are\quad not\quad collinear.\\ Option\quad B\longrightarrow The\quad points\quad are\\ A(4,3)\quad B(5,1)\quad \& \quad C(1,9).\\ \therefore \quad AB=\sqrt { { \left( 4-5 \right) }^{ 2 }+{ \left( 3-1 \right) }^{ 2 } } units=\sqrt { 5 } =2.24units,\\ \quad \quad \quad BC=\sqrt { { \left( 5-1 \right) }^{ 2 }+{ \left( 1-9 \right) }^{ 2 } } units=\sqrt { 80 } =8.94units,\\ and\quad AC=\sqrt { { \left( 4-1 \right) }^{ 2 }+{ \left( 3-9 \right) }^{ 2 } } units=\sqrt { 45 } =6.71units.\\ \therefore \quad AB+AC=\left( 2.24+6.71 \right) units=8.95units.\\ So\quad AB+AC=BC.\\ \therefore \quad A,\quad B\quad \& \quad C\quad are\quad \quad collinear.\\ Option\quad C\longrightarrow The\quad points\quad are\\ A(2,5)\quad B(-1,2)\quad \& \quad C(4,7).\\ \therefore \quad AB=\sqrt { { \left( 2+1 \right) }^{ 2 }+{ \left( 5-2 \right) }^{ 2 } } units=\sqrt { 18 } units=4.24units,\\ \quad \quad \quad BC=\sqrt { { \left( -1-4 \right) }^{ 2 }+{ \left( 2-7 \right) }^{ 2 } } units=\sqrt { 50 } units=7.07units,\\ and\quad AC=\sqrt { { \left( 2-4 \right) }^{ 2 }+{ \left( 5-7 \right) }^{ 2 } } units=\sqrt { 8 } units=2.83units.\\ \therefore \quad AB+AC=\left( 4.24+2.83 \right) units=7.07units.\\ So\quad AB+AC=BC.\\ \therefore \quad A,\quad B\quad \& \quad C\quad are\quad \quad collinear.\\ Option\quad C\longrightarrow The\quad points\quad are\\ A(-3,2)\quad B(1,-2)\quad \& \quad C(9,-10).\\ \therefore \quad AB=\sqrt { { \left( -3-1 \right) }^{ 2 }+{ \left( 2+2 \right) }^{ 2 } } units=\sqrt { 32 } units=5.66units,\\ \quad \quad \quad BC=\sqrt { { \left( 1-9 \right) }^{ 2 }+{ \left( -2+10 \right) }^{ 2 } } units=\sqrt { 128 } units=11.31units,\\ and\quad AC=\sqrt { { \left( -3-9 \right) }^{ 2 }+{ \left( 2+10 \right) }^{ 2 } } units=\sqrt { 288 } units=16.97units.\\ \therefore \quad AB+BC=\left( 5.66+11.31 \right) units=16.97units.\\ So\quad AB+BC=AC.\\ \therefore \quad A,\quad B\quad \& \quad C\quad are\quad \quad collinear.\\ Ans-\quad Option\quad A.\\ \\ \\ $$
If $$A=\begin{bmatrix} a & b \\ b & a \end{bmatrix}$$, then $$\left| A+{ A }^{ T } \right| $$ equals
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$$4({a}^{2}-{b}^{2})$$
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$$2({a}^{2}-{b}^{2}$$
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$${a}^{2}-{b}^{2})$$
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$$4ab$$
Explanation
$$A+{A}^{T}=\begin{bmatrix} a & b \\ b & a \end{bmatrix}\begin{bmatrix} a & b \\ b & a \end{bmatrix}=\begin{bmatrix} 2a & 2b \\ 2b & 2a \end{bmatrix}$$
$$\left| A+{ A }^{ T } \right| =4{ a }^{ 2 }-4{ b }^{ 2 }\quad $$
Ans: A
If $$\Delta_1=\begin{vmatrix} 1 & 0\\ a & b\end{vmatrix}$$ and $$\Delta_2=\begin{vmatrix} 1 & 0\\ c & d\end{vmatrix}$$ then $$\Delta_2 \Delta_1$$ is equal to
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$$ac$$
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$$bd$$
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$$(b-a)(d-c)$$
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none of these
Explanation
$$\Delta_1=\begin{vmatrix} 1 & 0\\ a & b\end{vmatrix}=1.b-0.a=b$$
$$\Delta_2=\begin{vmatrix} 1 & 0\\ c & d\end{vmatrix}=1.d-0.c=d$$
then $$\Delta_2 \Delta_1=bd$$
Ans: B
If $$A$$ and $$B$$ are two matrices of same order $$3\times 3$$, where
$$A=\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 5 & 6 & 8 \end{bmatrix}$$ and $$B=\begin{bmatrix} 3 & 2 & 5 \\ 2 & 3 & 8 \\ 7 & 2 & 9 \end{bmatrix}$$
The value of $$\text{Adj} (\text{Adj}\, A)$$ is equal to
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$$-A$$
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$$A$$
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$$8A$$
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$$16A$$
Explanation
$$\left| A \right| =1\left( 3\times8-4\times6 \right) -2\left( 2\times8-5\times4 \right) +3\left( 2\times6-5\times3 \right) =-1$$
Use following results
$$\text{Adj} (\text{Adj}\, A)={ \left| A \right| }^{ n-2 }A \Rightarrow \text{Adj}(\text{Adj}\,A)={ \left| A \right| }^{ n-2 }A={ \left( -1 \right) }^{ 1 }A=-A$$
Ans: $$A$$
If $$\omega$$ is a cube root of unity and $$\Delta=\begin{vmatrix}1 & 2\omega \\ \omega & \omega^2\end{vmatrix}$$, then $$\Delta^2$$ is equal to
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$$-\omega$$
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$$\omega$$
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$$1$$
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$$\omega^2$$
Explanation
$$\Delta=\omega^2-2\omega^2=-\omega^2$$
$$\Delta^2=\omega^4=\omega$$
Ans: B
If $$\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \\ 3x & 6 \end{vmatrix}$$, then $$x$$ is equal to
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$$6$$
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$$\pm 6$$
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$$-6$$
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$$0$$
Explanation
$$\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \\ 3x & 6 \end{vmatrix}$$
$$\Rightarrow x^2-36=36-6x$$
$$\Rightarrow x^2+6x-72=0$$
$$\Rightarrow \displaystyle x=\frac{-6\pm\sqrt{6^2+4\times 72}}{2}$$
$$\quad \displaystyle =\frac{-6\pm \sqrt{36(1+8)}}{2}$$
$$\quad =-3\pm 9=6\ or\ -12$$
Which of the following is correct?
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Determinant is a square matrix
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Determinant is a number associated to a matrix
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Determinant is a number associated to a square matrix
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None of these
Explanation
Determinant is defined only for a square matrix.
and its denotes the value of that square matrix.
If $$|A| \displaystyle \neq 0$$, then $$A$$ is
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zero matrix
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singular matrix
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non - singular matrix
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diagonal matrix
Explanation
If $$|A| \neq 0$$ then by definition, $$A$$ is non-singular matrix
$$A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$ and $$A(\text{adj}\,A)=KI$$, then the value of $$K$$ is:
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$$2$$
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$$-2$$
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$$10$$
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$$-10$$
Explanation
Given, $$A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\quad $$
Adjoint of $$A=\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}\quad $$
$$\therefore$$ $$A(\text{adj}\,A)=KI$$
$$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 4 & 2 \\ -3 & 1 \end{bmatrix}=K\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
$$\begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix}=\begin{bmatrix} K & 0 \\ 0 & K \end{bmatrix}$$
$$\therefore$$ $$K=-2$$
What is the determinant of the matrix $$\left [\begin{matrix} 3& 6\\ -1 & 2\end {matrix} \right]$$?
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$$0$$
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$$12$$
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$$|0|$$
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$$|6|$$
Explanation
Given,
$$\begin{vmatrix} 3 & 6 \\ -1 & 2 \end{vmatrix}$$
Let determinent be $$\left| d \right| $$
Value of $$\left| d \right| $$ will be
$$\left| d \right| $$$$=$$$$3\times 2-\left( 6\times -1 \right) $$
$$=6+6=12$$
If $$A = \begin{bmatrix} 1& \log_{b}a\\ \log_{a}b & 1\end{bmatrix}$$ then $$|A|$$ is equal to
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$$0$$
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$$\log_{a}b$$
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$$-1$$
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$$\log_{b}a$$
Explanation
On solving the given matrix,
$$|A| = 1 - \log_{a}b . \log_{b}a = 1 - 1 = 0$$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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