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CBSE Questions for Class 12 Commerce Maths Matrices Quiz 10 - MCQExams.com
CBSE
Class 12 Commerce Maths
Matrices
Quiz 10
If $$A = \left[ \begin{array} { c c } { a b } & { b ^ { 2 } } \\ { - a ^ { 2 } } & { - a b } \end{array} \right]$$ and $$A ^ { n } = 0$$ then the minimum value of $$n$$ is
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$$2$$
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$$3$$
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$$4$$
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$$5$$
Explanation
$$A=\begin{bmatrix}ab&b^2\\-a^2&-ab\end{bmatrix}$$
$$A^2=\begin{bmatrix}ab&b^2\\-a^2&-ab\end{bmatrix}\begin{bmatrix}ab&b^2\\-a^z&-ab\end{bmatrix}=\begin{bmatrix}a^2b^2-a^2b^2&ab^3-ab^3\\-a^3b+a^3b&-a^2b^2+a^2b^2\end{bmatrix}$$
$$=\begin{bmatrix}0&0\\0&0\end{bmatrix}$$
$$A^2=0$$
$$\therefore n=2$$.
If $$A=\left[ \begin{matrix} a & b \\ b & a \end{matrix} \right] $$ and $$A^{2}=\left[ \begin{matrix} \alpha & \beta \\ \beta & \alpha \end{matrix} \right] $$ then
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$$\alpha=a^{2}+b^{2},\ \beta=2ab$$
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$$\alpha=a^{2}+b^{2},\ \beta=a^{2}b^{2}$$
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$$\alpha=2ab,\ \beta=a^{2}+b^{2}$$
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$$\alpha=a^{2}+b^{2},\ \beta=ab$$
If $$A=\left[ \begin{matrix} 6 & 9 \\ -4 & -6 \end{matrix} \right] $$, then $$A^{2}$$=
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$$\left[ \begin{matrix} 6 & 9 \\ -4 & 6 \end{matrix} \right]$$
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$$\left[ \begin{matrix} 6 & 9 \\ 4 & -6 \end{matrix} \right]$$
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$$\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right]$$
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$$\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right]$$
If the order of $$A$$ is $$4 \times 3$$, the order of $$B$$ is $$4 \times 5$$ and the order of $$C ,\ 7\times 3$$ then the order of $$(A^{T}B^{T})^{T}C^{T}$$ is
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$$4 \times 5$$
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$$3 \times 7$$
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$$4 \times 3$$
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$$5 \times 7$$
Explanation
Order of $$A=4\times 3$$
$$\therefore $$ Order of $$A^{T}=3\times 4$$
Order of $$B=4\times 5$$
$$\therefore$$ Order of $$B^{T}=5\times 4$$
Order of $$C=7\times 3$$
$$\therefore$$Order of $$C^{T}=3\times 7$$
Order of $$(A^{T}B^{T})=4\times 5$$
If $$\left[ \begin{matrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{matrix} \right] =\left[ \begin{matrix} 5 & a & -2 \\ 1 & 1 & 0 \\ -2 & -2 & b \end{matrix} \right] ,$$ then
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$$a =4$$
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$$a =1$$
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$$b =4$$
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$$b=1$$
The inverse of a symmetric matrix is
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symmetric
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skew-symmetric
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diagonal matrix
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singular matrix
Explanation
$$ A^{T}=A$$
$$A^{-1}$$ exists.
$$(A^{-1}A)^{T}=A^{T}(A^{-1})^{T}$$
$$=A(A^{-1})^{T}=I^{T}=I$$
$$A^{-1}A(A^{-1})^{T}=A^{-1}I$$
$$I(A^{-1})^{T}=(A^{-1})^{T}=A^{-1}$$
$$\therefore$$ The inverse of a symmetric matrix is also symmetric.
If $$A = \left[ {\begin{array}{*{20}{c}}1&2\\3&4\end{array}} \right]$$, then $$8A^{-4}$$ is equal to
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$$145A^{-1}+27I$$
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$$145A^{-1}-27I$$
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$$27I - 145A^{-1}$$
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$$29A^{-1} +9I$$
If $$A=\left[ \begin{matrix} 1 & 0 & -1 \\ 3 & 4 & 5 \\ 0 & 6 & 7 \end{matrix} \right]$$ and $$A^{-1}=[\alpha_{ij}]_{3\times 3}$$ then $$\alpha_{23}=$$
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$$-1/5$$
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$$1/5$$
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$$-2/5$$
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$$2/5$$
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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$$3A-4I$$
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$$A-I$$
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$$A+I$$
Explanation
$$\begin{array}{l} A=\left( { \begin{array} { *{ 20 }{ c } }2 & { -1 } \\ { -1 } & 2 \end{array} } \right) \\ I=\left( { \begin{array} { *{ 20 }{ c } }1 & 0 \\ 0 & 1 \end{array} } \right) \\ { A^{ 2 } }=A\cdot A \\ =\left( { \begin{array} { *{ 20 }{ c } }2 & { -1 } \\ { -1 } & 2 \end{array} } \right) \left( { \begin{array} { *{ 20 }{ c } }2 & { -1 } \\ { -1 } & 2 \end{array} } \right) \\ =\left( { \begin{array} { *{ 20 }{ c } }{ 2\times 2+1 } & { -2-2 } \\ { -2-2 } & { 1+4 } \end{array} } \right) \\ =\left( { \begin{array} { *{ 20 }{ c } }5 & { -4 } \\ { -4 } & 5 \end{array} } \right) \\ Now, \quad 4A-3I \\= 4\left( { \begin{array} { *{ 20 }{ c } }8 & { -4 } \\ { -4 } & 8 \end{array} } \right) -\left( { \begin{array} { *{ 20 }{ c } }3 & 0 \\ 0 & 3 \end{array} } \right) \\ = \left( { \begin{array} { *{ 20 }{ c } }{ 8-3 } & { -4-0 } \\ { -4-0 } & { 8-3 } \end{array} } \right) \\ = \left( { \begin{array} { *{ 20 }{ c } }5 & { -4 } \\ { -4 } & 5 \end{array} } \right) ={ A^{ 2 } } \end{array}$$
If $$\begin{bmatrix} 3 & 2 & -1 \\ 4 & 9 & 2 \\ 5 & 0 & -2 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0 \\ 7 \\ 2 \end{bmatrix}$$, then $$(x,\ y,\ z)=$$
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$$(1,\ -1,\ 1)$$
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$$(2,\ -1,\ -4)$$
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$$(3,\ 0,\ 6)$$
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$$(2,\ -1,\ 4)$$
Explanation
$$\begin{bmatrix} 3 & 2 & -1\\ 4 & 9 & 2\\ 5 & 0 & -2\end{bmatrix}\begin{bmatrix} x \\ y\\ z\end{bmatrix}=\begin{bmatrix} 0\\ 7\\ 2\end{bmatrix}$$
Thus, multiplying
$$3x+2y-z=0$$ …………..$$(1)$$
$$4x+9y+2z=7$$ ………..$$(2)$$
$$5x-2z=2$$ …………$$(3)$$
Thus $$z=\dfrac{5x-2}{2}$$
$$\therefore 3x+2y=\dfrac{5x-2}{2}$$
$$\therefore 6x+4y=5x-2$$
$$\therefore x+4y=-2$$ …………$$(4)$$
Also,
$$4x+9y+2\left(\dfrac{5x-2}{2}\right)=7$$
$$9x+9y=9$$
$$\therefore x+y=1$$ ………..$$(5)$$
Solving $$(4)$$ and $$(5)$$
$$3y=-3$$
$$y=-1$$
$$\therefore x=1-y$$
$$=1-(-1)=2$$
$$\therefore x=2$$
$$\therefore z=\dfrac{5x-2}{2}=\dfrac{5(2)-2}{2}$$
$$\therefore z=4$$
$$=4$$
$$(x, y, z)=(2, -1, 4)$$.
If A is a 2 X 2 matrix such that $$A^{2009} + A^{2008}$$= I, then : $$(A^{2008})^{-1}$$=
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$$A^{2008} + I$$
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$$A^{2009} + 1$$
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A + I
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A
If $$\left[ \begin{matrix} 2 & -3 \\ 1 & \lambda \end{matrix} \right] \times \left[ \begin{matrix} 1 & 5 & \mu \\ 0 & 2 & -3 \end{matrix} \right] =\left[ \begin{matrix} 2 & 4 & 1 \\ 1 & -1 & 13 \end{matrix} \right],$$ then
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$$\lambda=3, \mu= -4$$
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$$\lambda=4, \mu=-3$$
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no real values of $$\lambda, \mu$$ are possible
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none of these
Explanation
Given,
$$\pmatrix{2&-3\\1&\lambda}\times \pmatrix{1&5& \mu\\0&2&-3}$$=$$\pmatrix{2&4&1\\1&-1&13}$$
$$\pmatrix{2&4&2\mu-9\\1&5+2\lambda&\mu-3\lambda}=\pmatrix{2&4&1\\1&-1&13}$$
Comparing both sides we get
$$2\mu+9=1\implies 2\mu=-8\implies \mu=-4$$
Also $$5+2\lambda=-1\implies 2\lambda=6\implies \lambda=3$$
Hence, we get $$\mu=-4,\lambda=3.$$
Let p be a non-singular matrix, $$1+p+p^{2}+....+p^{n}=0$$ (0 denotes the null matrix) then $$p^{-1}=$$
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$$p^{n}$$
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-$$p^{n}$$
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-(1+p+...+$$p^{n}$$)
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none
If $$A=\begin{bmatrix} 4 & -1 \\ -1 & k \end{bmatrix}$$ such that $$A^{2}-6A+7I=0$$, then $$k=$$
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$$1$$
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$$3$$
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$$2$$
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$$4$$
Explanation
$$A = \begin{bmatrix} 4 & -1 \\ -1 & k \end{bmatrix}$$
$${A}^{2} = \begin{bmatrix} 4 & -1 \\ -1 & k \end{bmatrix} \times \begin{bmatrix} 4 & -1 \\ -1 & k \end{bmatrix}$$
$$\Rightarrow {A}^{2} = \begin{bmatrix} 16+ 1 & -4 - k \\ -4 - k & 1 + {k}^{2} \end{bmatrix} = \begin{bmatrix} 17 & - \left( 4 + k \right) \\ - \left( 4 + k \right) & 1 + {k}^{2} \end{bmatrix}$$
$$6A = 6 \times \begin{bmatrix} 4 & -1 \\ -1 & k \end{bmatrix} = \begin{bmatrix} 24 & -6 \\ -6 & 6k \end{bmatrix}$$
$$7I = 7 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$$
Therefore,
$${A}^{2} - 6A + 7I = 0$$
$$\Rightarrow {A}^{2} = 6A - 7I$$
$$\begin{bmatrix} 17 & - \left( 4 + k \right) \\ - \left( 4 + k \right) & 1 + {k}^{2} \end{bmatrix} = \begin{bmatrix} 24 & -6 \\ -6 & 6k \end{bmatrix} - \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$$
$$\begin{bmatrix} 17 & - \left( 4 + k \right) \\ - \left( 4 + k \right) & 1 + {k}^{2} \end{bmatrix} = \begin{bmatrix} 17 & -6 \\ -6 & 6k - 7 \end{bmatrix}$$
Comparing both sides, we get
$$- \left( 4 + k \right) = -6$$
$$\Rightarrow k = 6 - 4 = 2$$
If $$A\begin{bmatrix} 1 & 1\\ 2 & 0\end{bmatrix}=\begin{bmatrix} 3 & 2\\ 1 & 1\end{bmatrix}$$, then $$A^{-1}$$ is given by?
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$$\begin{bmatrix} 0 & -1\\ 2 & -4\end{bmatrix}$$
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$$\begin{bmatrix} 0 & -1\\ -2 & -4\end{bmatrix}$$
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$$\begin{bmatrix} 0 & 1\\ 2 & -4\end{bmatrix}$$
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None of these
A is an involuntary matrix given by $$A=\begin{bmatrix} 0 & 1 & -1\\ 4 & -3 & 4\\ 3 & -3 & 4\end{bmatrix}$$ then the inverse of $$\dfrac{A}{2}$$ will be?
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$$2A$$
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$$\dfrac{A^{-1}}{2}$$
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$$\dfrac{A}{2}$$
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$$A^{-2}$$
A is an involutory matrix given by $$A=\begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix}$$ then the inverse of $$\dfrac{A}{2}$$ will be
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$$2A$$
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$$\dfrac{A^{-1}}{2}$$
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$$\dfrac{A}{2}$$
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$$A^{2}$$
Explanation
Given $$A$$ to be involutory matrix, then according to the definition of involutory matrix we have, $$A^2 =I$$. [ $$I$$ being identity matrix of order $$3$$].
So,
$$A^2=I$$
or, $$A=A^{-1}$$ [ Since involutory matrix is always invertible]
or, $$\dfrac{A}{2}=\dfrac{A^{-1}}{2}$$.
Let A=$$\left[ \begin{matrix} 1 \\ 2 \end{matrix}\begin{matrix} 2 \\ 1 \end{matrix} \right] and\quad B=\left[ \begin{matrix} 4 \\ 5 \\ 0 \end{matrix}\begin{matrix} -3 \\ 6 \\ 1 \end{matrix} \right] $$ then
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$$AB$$ exists
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$$AB$$ and $$BA$$ both exists
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Neither $$AB$$ nor $$BA$$ exists
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$$BA$$ exists, but $$AB$ does not exists
Explanation
Given,
Order of $$A=2\times 2$$ matrix
Order of $$B=3\times 2$$ matrix
$$AB$$ does not exist, as the number of columns of matrix $$A$$ is not equal to number of rows of $$B$$
But, $$BA$$ exists, as the number of columns of matrix $$B$$ is equal to number of rows of $$A$$
If $$\left[ \begin{matrix} 1 & x & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{matrix} \right] \left[ \begin{matrix} 1 \\ 1 \\ x \end{matrix} \right] =0,$$ then $$x=$$
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$$\cfrac{-9\pm\sqrt{35}}{2}$$
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$$\cfrac{-7\pm\sqrt{53}}{2}$$
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$$\cfrac{-9\pm\sqrt{53}}{2}$$
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$$\cfrac{-7\pm\sqrt{35}}{2}$$
Explanation
Given,
$$\begin{pmatrix}1&x&1\end{pmatrix}\begin{pmatrix}1&3&2\\ 0&5&1\\ 0&3&2\end{pmatrix}\begin{pmatrix}1\\ 1\\ x\end{pmatrix}=0$$
Upon multiplying the first 2 matrices, we get,
$$\begin{pmatrix}1\cdot \:1+x\cdot \:0+1\cdot \:0&1\cdot \:3+x\cdot \:5+1\cdot \:3&1\cdot \:2+x\cdot \:1+1\cdot \:2\end{pmatrix}\begin{pmatrix}1\\ 1\\ x\end{pmatrix}=0$$
Upon further simplification, we get,
$$\begin{pmatrix}1&6+5x&4+x\end{pmatrix}\begin{pmatrix}1\\ 1\\ x\end{pmatrix}=0$$
Again multiplying the above matrices, we get,
$$\begin{pmatrix}1\cdot \:1+\left(6+5x\right)\cdot \:1+\left(4+x\right)x\end{pmatrix}=0$$
Upon further simplification, we get,
$$\begin{pmatrix}x^2+9x+7\end{pmatrix}=0$$
Solving the above quadratic equation, we get,
$$x=\dfrac{-9\pm \sqrt{53}}{2}$$
If $$A$$ and $$B$$ are two matrices such that $$AB$$ and $$A+B$$ are both defined then $$A$$ and $$B$$ are
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Square matrices of the same order
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Square matrices of different order
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Rectangular matrices of same order
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Rectangular matrices of different order
Explanation
For the addition of two matrices,it is required that
the matrices should be of the same order.
For multiplication of two matrices A of order $$m\times$$ n and
B of order $$p\times q,$$ it requires that
$$ n=p----(1) $$
$$ m = p----(2) $$
$$ n = q----(3) $$
form (1) and (3)
$$ p = q-------(4) $$
from (1)and (2)
$$ n = m-----(5) $$
From (4)and (5),we can conclude that
A and B are square matrices and their orders
are one and same .
$$\therefore $$ The addition and multiplication of two matrices
should be of same order and they should be
square matrices.
$$\therefore $$ so, the answer is A. square matrices of the same
order.
if A=$$\left[ \begin{matrix} 2 & 3 \\ 5 & -7 \end{matrix} \right] then\quad \left( { A }^{ '} \right) ^{ 2 }=$$
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$$\left[ \begin{matrix} 5 & -7 & 12 \\ 1 & 4 & 22 \end{matrix} \right] $$
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$$\left[ \begin{matrix} 1 & 17 \\ 1 & -4 \\ 0 & 2 \end{matrix} \right] $$
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$$\left[ \begin{matrix} -19 & -25 \\ -15 & 64 \end{matrix} \right] $$
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$$\left[ \begin{matrix} 19 & -25 \\ -15 & 64 \end{matrix} \right] $$
Explanation
Given,
$$A=\begin{pmatrix}2&3\\ \:5&-7\end{pmatrix}$$
$$\Rightarrow A'=\begin{pmatrix}2&5\\ 3&-7\end{pmatrix}$$
Now,$$(A')^2$$
$$=\begin{pmatrix}2&5\\ 3&-7\end{pmatrix}^2$$
$$=\begin{pmatrix}2&5\\ 3&-7\end{pmatrix}\begin{pmatrix}2&5\\ 3&-7\end{pmatrix}$$
$$=\begin{pmatrix}2\cdot \:2+5\cdot \:3&2\cdot \:5+5\left(-7\right)\\ 3\cdot \:2+\left(-7\right)\cdot \:3&3\cdot \:5+\left(-7\right)\left(-7\right)\end{pmatrix}$$
$$=\begin{pmatrix}19&-25\\ -15&64\end{pmatrix}$$
If $$A=\left[ \begin{matrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{matrix} \right] $$, then value of $$A^{-1}$$ is equal to
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$$A$$
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$$A^{2}$$
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$$A^{3}$$
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$$A^{4}$$
If $$U = [ 2, -3 , 4 ]$$ , $$V = \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$$ , $$X = [0 , 2 , 3]$$ and $$Y = \begin{bmatrix} 2 \\ 2 \\ 4 \end{bmatrix}$$ , then $$UV + XY$$ =
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$$20$$
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$$[ -20 ]$$
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$$-20$$
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$$[ 20 ]$$
If $$A=\begin{bmatrix} 2 & -1\\ -7 & 4\end{bmatrix}$$ and $$B=\begin{bmatrix} 4 & 1\\ 7 & 2\end{bmatrix}$$ then $$B^TA^T$$ is equal to?
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$$\begin{bmatrix} 1 & 0\\ -12 & 1\end{bmatrix}$$
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$$\begin{bmatrix} 1 & 1\\ 1 & 1\end{bmatrix}$$
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$$\begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix}$$
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$$\begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix}$$
Adjacency matrix of all graphs are symmetric.
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True
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False
If $$A^2-A+1=0$$, then the inverse of A is?
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A
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$$A+I$$
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$$I-A$$
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$$A-I$$
If $$\begin{bmatrix} 1 & 1\\ -1 & 1\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}=\begin{bmatrix} 2 \\ 4\end{bmatrix}$$, then the values of $$x $$ and $$y $$ respectively are?
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$$-3, -1$$
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$$1, 3$$
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$$3, 1$$
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$$-1, 3$$
Explanation
On equating given two matrices, we get
$$x+y=2$$ and
$$-x+y=4$$
Solving for $$x, y : x=-1, y=3$$.
Let $$\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 3\\ 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & n-1\\ 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & 78\\ 0 & 1\end{bmatrix}$$
If $$A=\begin{bmatrix} 1 & n\\ 0 & 1\end{bmatrix}$$ then $$A^{-1}=?$$
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$$\begin{bmatrix} 1 & 12\\ 0 & 1\end{bmatrix}$$
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$$\begin{bmatrix} 1 & -13\\ 0 & 1\end{bmatrix}$$
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$$\begin{bmatrix} 1 & -12\\ 0 & 1\end{bmatrix}$$
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$$\begin{bmatrix} 1 & 0\\ -13 & 1\end{bmatrix}$$
Explanation
$$\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 3\\ 0 & 1\end{bmatrix}..\begin{bmatrix} 1 & n-1\\ 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & 78\\ 0 & 1\end{bmatrix}$$
$$\Rightarrow \dfrac{n(n-1)}{2}=78\Rightarrow n=13$$
$$A=\begin{bmatrix} 1 & 13\\ 0 & 1\end{bmatrix}$$
so $$A^{-1}=\begin{bmatrix} 1 & -13\\ 0 & 1\end{bmatrix}$$.
If $$\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 3\\ 0 & 1\end{bmatrix} ..\begin{bmatrix} 1 & n-1\\ 0 & 1\end{bmatrix} =\begin{bmatrix} 1 & 78\\ 0 & 1\end{bmatrix}$$, then the inverse of $$\begin{bmatrix} 1 & n\\ 0 & 1\end{bmatrix}$$ is?
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$$\begin{bmatrix} 1 & -13\\ 0 & 1\end{bmatrix}$$
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$$\begin{bmatrix} 1 & 0\\ 12 & 1\end{bmatrix}$$
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$$\begin{bmatrix} 1 & -12\\ 0 & 1\end{bmatrix}$$
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$$\begin{bmatrix} 1 & 0\\ 13 & 1\end{bmatrix}$$
Explanation
$$\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 3\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & n-1\\ 0 & 1\end{bmatrix} =\begin{bmatrix} 1 & 78\\ 0 & 1\end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} 1 & 1+2+3+...+n-1\\ 0 & 1\end{bmatrix}=$$ $$\begin{bmatrix} 1 & 78\\ 0 & 1\end{bmatrix}$$
$$\Rightarrow \dfrac{n(n-1)}{2}=78\Rightarrow n=13$$, $$-12$$ (reject)
$$\therefore$$ We have to find inverse of $$\begin{bmatrix} 1 & 13\\ 0 & 1\end{bmatrix}$$
$$\therefore \begin{bmatrix} 1 & -13\\ 0 & 1\end{bmatrix}$$
If $$A = \begin{bmatrix} n& 0 & 0\\ 0 & n & 0\\ 0 & 0 & n\end{bmatrix}$$ and $$B = \begin{bmatrix}a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3}\end{bmatrix}$$, then $$AB$$ is equal to
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$$B$$
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$$nB$$
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$$B^{n}$$
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$$A + B$$
Explanation
Let,
$$A = \begin{bmatrix} n& 0 & 0\\ 0 & n & 0\\ 0 & 0 & n\end{bmatrix}$$ and
$$B = \begin{bmatrix}a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3}\end{bmatrix}$$.
Now,
$$AB = \begin{bmatrix} n& 0 & 0\\ 0 & n & 0\\ 0 & 0 & n\end{bmatrix}$$ $$ \begin{bmatrix}a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3}\end{bmatrix}$$
or,
$$A B= \begin{bmatrix}na_{1} & na_{2} & na_{3}\\ nb_{1} & nb_{2} & nb_{3}\\ nc_{1} & nc_{2} & nc_{3}\end{bmatrix}$$
or, $$AB=nB$$.
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