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CBSE Questions for Class 12 Commerce Maths Matrices Quiz 11 - MCQExams.com
CBSE
Class 12 Commerce Maths
Matrices
Quiz 11
If $$A = \begin{bmatrix}1 & 2 & x\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$$ and $$B = \begin{bmatrix}1 & -2 & y\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$$ and $$AB = I_{3}$$, then $$x + y=$$ _____.
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$$0$$
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$$-1$$
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$$2$$
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None of these
If $$A = \begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ a & b & -1\end{bmatrix}$$, then $$A^{2}$$ is equal to
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null matrix
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unit matrix
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$$-A$$
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$$A$$
If $$ \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}A\begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$ , then A =
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$$ \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} $$
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$$ \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} $$
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$$ \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} $$
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$$ - \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} $$
Explanation
$$ \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}A\begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$
$$ A = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}^{ -1 }\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix}^{-1} $$
$$ \Rightarrow A = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}(-1)\begin{bmatrix} -3 & -2 \\ -5 & -3 \end{bmatrix}$$
$$ = (-1)\begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix}=-\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} $$
If $$ \begin{bmatrix} 1/25 & 0 \\ x & 1/25 \end{bmatrix}\quad =\quad \begin{bmatrix} 5 & 0 \\ -a & 5 \end{bmatrix}^{ -2 } $$, then the value of x is
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$$ a / 125 $$
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$$ 2a / 125 $$
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$$ 2a / 25 $$
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None of these
Explanation
Let $$ A = \begin{bmatrix} 5 & 0 \\ -a & 5 \end{bmatrix} $$
$$ \Rightarrow (A)= \begin{bmatrix} 5 & 0 \\ -a & 5 \end{bmatrix} $$
$$ \Rightarrow A^{-1} = \frac {1}{ |A| } \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix} = \frac {1}{25} \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix} $$
$$ \Rightarrow A^{-2} =(A^{-1})^2 = \frac {1}{25} \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix} \frac {1}{25} \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix} $$
$$ =\frac {1}{625} \begin{bmatrix}25 & 0 \\10a&25 \end{bmatrix} $$
Now ,
$$ \begin{bmatrix} 1/2 & 0 \\ x & 1/25 \end{bmatrix}=\begin{bmatrix} \frac {1}{25} &0 \\ \frac {2a}{125} & \frac {1}{25} \end{bmatrix} $$
$$ \Rightarrow x = 2a /125 $$
If $$ AB = A $$ and $$ BA =B, $$ then
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$$ A^2 B = A^2 $$
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$$ B^2A =B^2 $$
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$$ ABA = A $$
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$$ BAB = B $$
Explanation
We have, $$ AB = A $$ and $$ BA =B $$
Now, checking each option:
Option A: $$ A^2B=A(AB)=AA=A^2$$
Option B: $$B^2A=B(BA)=BB=B^2$$
Option C: $$ ABA=A(BA)=AB=A $$ and
Option D: $$BAB = B(AB)=BA=B. $$
Hence, all options are correct.
If $$A = \begin{bmatrix}3 &-4 \\ -1 & 2\end{bmatrix}$$ and $$B$$ is a square matrix of order $$2$$ such that $$AB = I$$ then $$B = ?$$
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$$\begin{bmatrix}1 &2 \\ 2 & 3\end{bmatrix}$$
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$$\begin{bmatrix}1 &\dfrac {1}{2} \\ 2 & \dfrac {3}{2}\end{bmatrix}$$
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$$\begin{bmatrix}1 &2 \\ \dfrac {1}{2} & \dfrac {3}{2}\end{bmatrix}$$
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None of these
Find the value of $$x,$$ if $$\left [ 1\ x\ 1 \right ]$$ $$\begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix}$$ $$\begin{bmatrix}1 \\ 2 \\ x \end{bmatrix}=O.$$
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$$-2$$
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$$2$$
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$$-14$$
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9
Explanation
Given, $$\left [ 1\ x\ 1 \right ]_{1 \times 3} $$ $$\begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix}_{ _3 \times_ 3}$$ $$\begin{bmatrix}1 \\ 2 \\ x \end{bmatrix}_{_3 \times _1}= O$$
$$\Rightarrow$$ $$[1+ 2x + 15\,\,\, 3 + 5x +3 \,\,\, 2 + x + 2] \begin{bmatrix}1 \\ 2 \\ x \end{bmatrix}= O$$
$$\Rightarrow$$ $$[16 + 2x + 15\,\,\, 3+5x + 3\,\,\, 2 + x +2]$$ $$\begin{bmatrix}1 \\ 2 \\ x \end{bmatrix} = O$$
$$\Rightarrow$$ $$[(16 +2x) \cdot 1 + (6 + 5x ) \cdot 2 + (4 + x ) \cdot x] = 0$$
=$$\Rightarrow$$ $$(16 + 2x ) + (12 + 10x ) + (4x + x2) = 0$$
$$\Rightarrow$$ $$x^2 + 16x + 28 = 0$$ $$\Rightarrow$$ $$(x + 14 ) (x+ 2) = 0$$
$$\Rightarrow$$ $$x + 14 = 0$$ or $$x + 2 = 0$$
Hence, $$x = -14$$ or $$x = -2.$$
The matrix $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix}$$ is a
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Identity matrix
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Symmetric matrix
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Skew symmetric matrix
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None of these
Explanation
Let $$A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix}$$
$$A' =\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix} =A$$
So, the given matrix is a symmetric matrix.
[Since, in a square $$A$$. If $$A'=A$$, then $$A$$ is called symmetric matrix]
If matrix $$A=[a_{ij}]_{2\times 2}$$, where $$a_{ij} *1$$ if $$1*j$$ and $$0$$ if $$i=j$$ then $$A^2$$ is equal to
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$$I$$
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$$A$$
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$$0$$
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$$None\ of\ these$$
Explanation
We have, $$A=[a_{ij}]_{2\times 2}$$. where $$a_{ij}=1$$ if $$1\neq j$$ and if $$i=j$$
$$\therefore A=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$
And $$A^2=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=I$$
$$AA^{'}$$ is always a symmetric matrix for any matrix $$A$$.
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True
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False
Explanation
It is true,
Because,
$$[AA^{'}]^{'}=(A^{'})^{'}A^{'}=[AA^{'}]$$
So, $$AA'$$ is a symmetric matrix for any matrix $$A$$
If $$A$$ and $$B$$ are square matrices of the same order, then $$(A+B)(A-B)$$ is equal to
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$$A^2-B^2$$
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$$A^2-BA-AB-B^2$$
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$$A^2-B^2+BA-AB$$
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$$A^2-BA+B^2+AB$$
Explanation
$$(A+B)(A-B)=A(A-B)+B(A-B)$$.........because matrix product is distributive.
$$=A^2-AB+BA-B^2$$
And as matrix product is not commutative, so $$AB\neq BA$$
Hence option $$C$$ is correct.
If $$A=\begin{bmatrix} 2 & 1 & 3 \\ 4 & 5 & 1 \end{bmatrix}$$ and $$B=\begin{bmatrix} 2 & 3 \\ 4 & 2 \\ 1 & 5 \end{bmatrix}$$, then
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Only $$AB$$ is defined
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only $$BA$$ is defined
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$$AB$$ and $$BA$$ both are defined
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$$AB$$ and $$BA$$ both are not defined
Explanation
Let $$A=[a_{ij}]_{2\times 3}, B=[b_{ij}]_{3\times 2}$$.
For, $$AB$$ columns of $$A=3$$ is equal to rows of $$B=3$$. So, $$AB$$ is defined.
For, $$BA$$ columns of $$B=2$$ is equal to rows of $$A=2$$. So, $$BA$$ is defined.
So, Both $$AB$$ and $$BA$$ are defined.
$$(C)$$ is the correct answer.
State true/false:
If $$A$$ and $$B$$ are two matrices of orders $$ 3 \times 2 $$ and $$ 2 \times 3 $$ respectively, then their sum $$ A + B $$ is possible.
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True
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False
Explanation
The Sum of matrices $$ A+B $$ is possible only when the order of both the matrices $$ A $$ and $$ B $$ are same.
If A=$$\displaystyle \begin{vmatrix} a & b &c \\ x & y & z \\ l & m & n \end{vmatrix}$$ is a skew-symmetric matrix then which of the following is equal to x+y+z?
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$$a+b+c$$
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$$l+m+n$$
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$$-b-m$$
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$$c-l-n$$
Explanation
if $$ A $$ is a skew symmetric matrix, then $$ {A}^{T} = -A $$
Now, $$ {A}^{T} = \begin{vmatrix} a & x & l \\ b & y & m \\ c & z & n \end{vmatrix}\quad $$
So, $$ {A}^{T} = -A $$
$$ => \begin{vmatrix} a & x & l \\ b & y & m \\ c & z & n \end{vmatrix}\quad = \begin{vmatrix} -a & -b & -c \\ -x & -y & -z \\ -l & -m & -n \end{vmatrix} $$
$$ => x = -b ; y = -y $$ or $$ y = 0 ; z = - m $$
So, $$ x + y + z = -b + 0 -m = -b-m $$
Given $$A =\begin{bmatrix} 2& 1\\2 &1 \end{bmatrix} .$$ Find all possible matrix $$X$$ for which
$$AX = A.$$
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$$X =\begin{bmatrix} a&b \\ 2 -2a&1 -2b\end{bmatrix}$$ for a, $$b\in R$$
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$$X =\begin{bmatrix} b&a \\ 2 -2b&1 -2a\end{bmatrix}$$ for a, $$b\in R$$
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$$X =\begin{bmatrix} b&a \\ 1 -2a&2 -2b\end{bmatrix}$$ for a, $$b\in R$$
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none of these
Explanation
Given $$A =\begin{bmatrix} 2& 1\\2 &1 \end{bmatrix}$$
We will check by options
Option A:
Consider, $$AX=\begin{bmatrix} 2& 1\\2 &1 \end{bmatrix} \begin{bmatrix} a&b \\ 2 -2a&1 -2b\end{bmatrix}$$
$$=\begin{bmatrix} 2a+2-2a & 2b+1-2b \\ 2a+2-2a & 2b+1-2b \end{bmatrix}$$
$$\Rightarrow AX=\begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}$$
$$\Rightarrow AX=A$$
Hence, this matrix X satisfies given condition.
Option B:
Consider $$AX=\begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} b & a \\ 2-2b & 1-2a \end{bmatrix}$$
$$=\begin{bmatrix} 2b+2-2b & 2a+1-2a \\ 2b+2-2b & 2a+1-2a \end{bmatrix}$$
$$=\begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}$$
$$\Rightarrow AX=A$$
Hence, this matrix X also satisfies the given equation.
Option C:
Consider, $$AX=\begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} b & a \\ 1-2a & 2-2b \end{bmatrix}$$
$$=\begin{bmatrix} 2b+1-2a & 2a+2-2b \\ 2b+1-2a & 2a+2-2b \end{bmatrix}$$
$$\Rightarrow AX\ne A$$
Hence, option C is not possible matrix X.
If $$\alpha ,\beta ,\gamma $$ are three real numbers and $$A=\begin{bmatrix} 1&\cos (\alpha -\beta ) &\cos (\alpha -\gamma ) \\\cos (\beta -\alpha ) &1 & \cos (\beta -\gamma ) \\ \cos (\gamma -\alpha ) & \cos (\gamma -\beta ) &1 \end{bmatrix}$$ then
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$$A$$ is symmetric
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$$A$$ is orthogonal
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$$A$$ is singular
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$$A$$ is not invertible.
Explanation
As $$\cos(-x)=\cos x$$
we can write $$\cos(\alpha-\beta)=\cos(\beta-\alpha)$$
So clearly $$A=A^T$$
$$A$$ is symmetric
We can write $$A$$ as
$$A=\begin{bmatrix} \cos \alpha &\sin \alpha &0 \\ \cos \beta & \sin \beta &0 \\\cos \gamma &\sin \gamma &0 \end{bmatrix}\begin{bmatrix} \cos \alpha &\cos \beta &\cos \gamma \\ \sin \alpha & \sin \beta &\sin \gamma \\0 &0 &0 \end{bmatrix}$$
so $$|A|=0$$
means $$A$$ is singular and non-invertible.
Let $$\alpha =\pi /5$$ and
$$A=\begin{bmatrix}\cos \alpha & \sin \alpha \\-\sin \alpha &\cos \alpha \end{bmatrix}$$ and $$B = A + A^2 + A^3 + A^4$$ , then
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singular
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non-singular
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skew-symmetric
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$$|B|=1$$
Explanation
Given $$A=\begin{bmatrix}\cos \alpha & \sin \alpha \\-\sin \alpha &\cos \alpha \end{bmatrix}$$ and $$\alpha =\pi /5$$
$$A^2=\begin{bmatrix}\cos\! 2\alpha & \sin\! 2\alpha \\-\sin\! 2\alpha &\cos\! 2\alpha \end{bmatrix}$$,
$$A^3=\begin{bmatrix}\cos\! 3\alpha & \sin\! 3\alpha \\-\sin\! 3\alpha &\cos\! 3\alpha \end{bmatrix}$$
and $$A^4=\begin{bmatrix}\cos\! 4\alpha & \sin\! 4\alpha \\-\sin\! 4\alpha &\cos\! 4\alpha \end{bmatrix}$$
We have $$\cos \alpha +\cos 2\alpha +\cos 3\alpha +\cos 4\alpha $$
$$=\cos \alpha +\cos 2\alpha +\cos(\pi -2\alpha)+\cos(\pi -\alpha) [\:\because 5\alpha=\pi]$$
$$=\cos \alpha +\cos 2\alpha -\cos 2\alpha-\cos \alpha=0$$
and $$\sin \alpha+\sin 2\alpha+\sin 3\alpha+\sin 4\alpha$$
$$=\sin \alpha+\sin 2\alpha+\sin(\pi -2\alpha)+\sin(\pi -\alpha)$$
$$=2 [\sin \alpha+\sin 2\alpha]$$
$$\displaystyle =2\left \{ 2\sin \left [ \frac{3\alpha }{2} \right ]\cos \frac{\alpha }{2} \right \}=4\sin \left [ \frac{3\pi }{10} \right ]\cos \frac{\pi }{10}$$
$$\displaystyle =4\sin \left [ \frac{\pi }{2}-\frac{\pi }{5} \right ]\cos \frac{\pi }{10}$$
$$\displaystyle =4\cos \frac{\pi }{5}\cos \frac{\pi }{10}=a $$ (say)
Thus, $$B=\begin{bmatrix} 0&a \\-a &0 \end{bmatrix}$$
$$\therefore B$$ is skew-symmetric.
Also, $$\displaystyle |B|=a^2 = 16\cos^2 \frac{\pi }{5}\cos^2 \frac{\pi }{10}>0$$
$$\therefore B$$ is non-singular.
Hence, options B and C.
If the matrix $$A = \begin{bmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 0 & 2\end{bmatrix}$$, then $$A^n=\begin{bmatrix}a & 0 & 0 \\ 0 & a & 0 \\ b & 0 & a\end{bmatrix}. n \in N$$ where
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$$a = 2n, b = 2^n$$
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$$a = 2^n, b = 2n$$
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$$a = 2^n, b = n2^{n-1}$$
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$$a = 2^n, b = n2^n$$
Explanation
Given $$A=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 0 & 2 \end{bmatrix}\quad =2\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$$
$$\Rightarrow {A}^{2} =2^2\begin{bmatrix}1&0&0\\0&1&0\\1&0&1 \end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\1&0&1\end{bmatrix}= 2^{2}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{bmatrix}$$
$$\Rightarrow {A}^{3} = 2^{3}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 3 & 0 & 1 \end{bmatrix}$$ and so on
we can observe that $${A}^{n} = 2^{n}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ n& 0 & 1 \end{bmatrix}$$
By comparing we get $$ a= 2^{n} , b= n2^{n} $$
If $$3\begin{bmatrix}2 & 3\\ -4 & 1\end{bmatrix} - 2 \begin{bmatrix} x& y\\ 3 & 4\end{bmatrix} = \begin{bmatrix}10 & 11\\ z & -5\end{bmatrix}$$, then $$x + y - z =$$
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$$-21$$
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$$-15$$
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$$0$$
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$$15$$
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$$21$$
If $$O\left( A \right) =2\times 3,$$ $$O\left( B \right) =3\times 2$$ and $$O\left( C \right) =3\times 3$$, which one of the following is not defined?
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$$CB+{ A }^{ ' }$$
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$$BAC$$
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$$C{ \left( A+{ B }^{ ' } \right) }^{ ' }$$
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$$C\left( A+{ B }^{ ' } \right) $$
Explanation
Given that, $$O\left( A \right) =2\times 3,$$ $$O\left( B \right) =3\times 2$$ and $$O\left( C \right) =3\times 3$$
$$\Rightarrow O\left( { A }^{ ' } \right) =3\times 2, O\left( { B }^{ ' } \right) =2\times 3$$
(a) $$CB+{ A }^{ ' }$$
Now, Order of $$CB=$$ (Order of $$C$$) (Order of $$B$$)
$$=$$ (Order of $$C$$ is $$3\times 3$$) (Order of $$B$$ is $$3\times 2$$)
$$=$$ Order of $$CB$$ is $$3\times 2$$
Since, $$O\left( { A }^{ ' } \right) =3\times 2$$
Therefore, matrix $$CB+{ A }^{ ' }$$ can be determined.
(b) $$O\left( BA \right) =3\times 3$$ and $$O\left( C \right) =3\times 3$$
Therefore, matrix $$BAC$$ can be determined.
(c) $$C{ \left( A+{ B }^{ ' } \right) }^{ ' }$$
$$O\left( A+{ B }^{ ' } \right) =2\times 3$$
$$\Rightarrow O{ \left( A+{ B }^{ ' } \right) }^{ ' }=3\times 2$$
and $$O\left( C \right) =3\times 3$$
Therefore, matrix $$C{ \left( A+{ B }^{ ' } \right) }^{ ' }$$ can be determined.
(d) $$C\left( A+{ B }^{ ' } \right) $$
$$O\left( A+{ B }^{ ' } \right) =2\times 3$$
and $$O\left( C \right) =3\times 3$$
$$\therefore $$ Matrix $$C\left( A+{ B }^{ ' } \right) $$ cannot be determined
Hence, option D is correct.
$$A $$ is of order $$m \times n$$ and $$B$$ is of order $$p \times q,$$ addition of $$A$$ and $$B$$ is possible only if
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$$m = p$$
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$$n = q$$
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$$n = p$$
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$$m = p, n = q$$
Explanation
Since, the order of $$A$$ is $$m\times n $$ and order of
$$ B$$ is $$p\times q$$
Then $$A+B$$ is only possible if $$m=p$$ $$\&$$ $$n=q.$$
Hence, the answer is $$m=p$$ $$\&$$ $$n=q.$$
Let $$A=\begin{bmatrix} 3 & 1\\ -1 & 2\end{bmatrix}$$, then
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$$A^2+7A-5I=0$$
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$$A^2-7A+5I=0$$
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$$A^2+5A-7I=0$$
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$$A^2-5A+7I=0$$
Explanation
$${ A }^{ 2 }=\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}=\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$$
$${ A }= \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$$
$${ I }= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
We find (d) satisfy the condition as
$${ A }^{ 2 }-5A-7I=0$$
$$\quad \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}-\begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}+\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}=0$$
Thus,$${ A }^{ 2 }-5A-7I=0$$ is correct
What is the inverse of the matrix
$$A=\begin{bmatrix} \cos { \theta } & \sin { \theta } & 0 \\ -\sin { \theta } & \cos { \theta } & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ ?
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$$\begin{bmatrix} \cos { \theta } & -\sin { \theta } & 0 \\ \sin { \theta } & \cos { \theta } & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
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$$\begin{bmatrix} \cos { \theta } & 0 & -\sin { \theta } \\ 0 & 1 & 0 \\ \sin { \theta } & 0 & \cos { \theta } \end{bmatrix}$$
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$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos { \theta } & -\sin { \theta } \\ 0 & \sin { \theta } & \cos { \theta } \end{bmatrix}$$
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$$\begin{bmatrix} \cos { \theta } & \sin { \theta } & 0 \\ -\sin { \theta } & \cos { \theta } & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
Explanation
$$A = \begin{bmatrix} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
Calculate first minors.
$$M_{11} = \cos \theta , M_{13} = 0, M_{22} = \cos \theta$$
$$M_{12} = -\sin \theta, M_{21} = \sin \theta, M_{23} = 0$$
$$M_{31} = 0, M_{32} = 0, M_{33} = \cos^{2}\theta + \sin^{2}\theta = 1$$
Cofactor Matrix $$= \begin{bmatrix} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0\\ 0 & 0 & 1 \end{bmatrix} = C$$
$$det|A| = \cos^{2}\theta + \sin^{2}\theta = 1$$
$$adj (A) = C^{T} = \begin{bmatrix} \cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0 & 0 & 1\end{bmatrix}$$
$$A^{-1} = \dfrac {adj(A)}{(A)} = \begin{bmatrix} \cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0& 0 & 1\end{bmatrix}$$.
If $$A=
\begin{bmatrix}
2 & -1 \\
-1 & 2
\end{bmatrix}$$ and $$I$$ is the unit matrix of order $$2$$, then $$A^2$$equals
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$$4A-3I$$
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$$3A-AI$$
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$$A-I$$
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$$A+I$$
Explanation
$$A=\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\\ { A }^{ 2 }=\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\times \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\\ \quad =\begin{bmatrix} 4+1 & -2-2 \\ -2-2 & 1+4 \end{bmatrix}\\ \quad =\begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}$$
Answer will be A option because
$$4A-3I=\begin{bmatrix} 8 & -4 \\ -4 & 8 \end{bmatrix}-\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\\ \quad \quad \quad \ =\begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}$$
Let $$A=\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$$ and $$10B=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$$. If $$B$$ is the inverse of matrix $$A$$, then $$\alpha $$ is
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$$-2$$
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$$1$$
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$$2$$
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$$5$$
Explanation
Since, $$B$$ is the inverse of matrix $$A$$.
So, $$10{ A }^{ -1 }=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$$
$$\Rightarrow 10{ A }^{ -1 }\cdot A=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$$
$$\Rightarrow 10I = \begin{bmatrix} 4+4+2 & -4+2+2 & 4-6+2 \\ -5+0+\alpha & 5+0+\alpha & -5+0+\alpha \\ 1-4+3 & -1-2+3 & 1+6+3 \end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix}=\begin{bmatrix} 10 & 0 & 0 \\ -5+\alpha & 5+\alpha & -5+\alpha \\ 0 & 0 & 10 \end{bmatrix}$$
$$\Rightarrow -5+\alpha =0$$
$$\Rightarrow \alpha =5$$
$$If\quad A=\begin{bmatrix} ab & { b }^{ 2 } \\ -{ a }^{ 2 } & -ab \end{bmatrix},then\quad { A }^{ 2 }\quad is\quad equal\quad to$$
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$$O$$
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$$I$$
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$$-I$$
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$$None\quad of\quad these$$
The value of $$x$$, so that $$\left[ 1 \quad x \quad 1 \right] \begin{bmatrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix}=0$$ is/are
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$$\pm\ 2$$
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$$0$$
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$$\dfrac { -7\pm \sqrt { 35 } }{ 2 }$$
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$$\dfrac { -9\pm \sqrt { 53 } }{ 2 }$$
If the matrix $$\begin{bmatrix} 0 & 2\beta & \Upsilon \\ \alpha & \beta & -\Upsilon \\ \alpha & -\beta & \Upsilon \end{bmatrix}$$is orthogonal, then
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$$\alpha = \pm\dfrac{1}{\sqrt{2}}$$
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$$\beta = \pm\dfrac{1}{\sqrt{6}}$$
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$$\gamma = \pm\dfrac{1}{\sqrt{3}}$$
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all of these
Explanation
$$\begin{bmatrix} 0 & 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{bmatrix}$$
for orthogonal matrix we have
$$A.A^{T}=I$$
$$\begin{bmatrix} 0 & 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{bmatrix}\begin{bmatrix} 0 & \alpha & \alpha \\ 2\beta & \beta & -\beta \\ \gamma & -\gamma & \gamma \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
$$\begin{bmatrix} 0+4{ \beta }^{ 2 }+{ \gamma }^{ 2 } & 0+2{ \beta }^{ 2 }-{ \gamma }^{ 2 } & 0-2{ \beta }^{ 2 }+{ \gamma }^{ 2 } \\ 0+2{ \beta }^{ 2 }-{ \gamma }^{ 2 } & { \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 } & { \alpha }^{ 2 }-{ \beta }^{ 2 }-{ \gamma }^{ 2 } \\ 0-2{ \beta }^{ 2 }+{ \gamma }^{ 2 } & { \alpha }^{ 2 }-{ \beta }^{ 2 }-{ \gamma }^{ 2 } & { \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 } \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
$$4\beta^{2}+\gamma^{2}=1, 2\beta^{2}-\gamma^{2}=0$$,
$$4\left(\dfrac{\gamma^{2}}{2}\right)+\gamma^{2}=1$$ $$\beta^{2}=\dfrac{r^{2}}{2}$$
$$r^{2}[3]=1$$
$$r=\pm \dfrac{1}{\sqrt{3}}$$
$$2\beta^{2}-\gamma^{2}=0, \alpha^{2}+\beta^{2}+\gamma^{2}+\gamma^{2}=1, \alpha^{2}-\beta^{2}-\gamma^{2}=0$$
$$\beta^{2}=\dfrac{\gamma^{2}}{2}, \alpha^{2}+\dfrac{\gamma^{2}}{2}+\dfrac{\gamma^{2}}{1}=1$$
$$\alpha^{2}+\dfrac{3\gamma^{2}}{2}=1$$
$$\beta^{2}=\dfrac{1}{6}\alpha^{2}+\dfrac{3}{2}\times \dfrac{1}{3}=1$$
$$\beta=\pm \dfrac{1}{\sqrt{6}}$$ $$\alpha=\pm \dfrac{1}{\sqrt{2}}$$
If $$\left[ \begin{matrix} x & 4 & -1 \end{matrix} \right] \left[ \begin{matrix} 2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 4 \end{matrix} \right] \left[ \begin{matrix} x \\ 4 \\ -1 \end{matrix} \right] =0,$$ then $$x=$$
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$$-1+\sqrt { 6 } $$
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$$8\pm \sqrt { 5 } $$
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$$-2\pm \sqrt { 10 } $$
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$$3\pm \sqrt { 6 } $$
If $$A=\begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$$, then : $$A^{-1}$$=
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$$A$$
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$$A^{2}$$
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$$A^{3}$$
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$$A^{4}$$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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