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CBSE Questions for Class 12 Commerce Maths Matrices Quiz 7 - MCQExams.com
CBSE
Class 12 Commerce Maths
Matrices
Quiz 7
Say true or false:
If $$A$$, $$B$$ be two matrices such that they commute, then $$A^2 - B^2 = (A - B)(A + B)$$.
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True
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False
Explanation
$$A$$ and $$B$$ commute when $$AB=BA$$
Hence $$(A-B)(A+B)={ A }^{ 2 }+AB-BA-{ B }^{ 2 }=A^{ 2 }-B^{ 2 }$$
If $$A=\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}$$, then $$(A-2I)(A-3I)=$$
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$$0$$
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$$A$$
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$$I$$
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$$5I$$
Explanation
$$A=\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}$$
$$\Rightarrow$$ $$(A-2I)(A-3I)={A}^{2}-5AI+6{I}^{2}$$
$${A}^{2}-5A+6I$$
by Hemilton rule $$\begin{vmatrix} 4-\lambda & 2 \\ -1 & 1-\lambda \end{vmatrix}=0$$
after expanding determinant we got;
$$\Rightarrow 4-5\lambda +{ \lambda }^{ 2 }+2=0$$
$$\Rightarrow { \lambda }^{ 2 }-5\lambda +6=0$$
$$\Rightarrow { A }^{ 2 }-5A+6I=0$$
Ans: 0
Or
It can be done using scalar multiplication and followed by matrix multiplication.
If $$\displaystyle A\times \begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \end{bmatrix}=\begin{bmatrix} 1 & 2 &3 \\ 3 & 2 & 1 \\ 3 & 1 & 2 \end{bmatrix}$$ then the order of A is _______
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$$\displaystyle 2\times 3$$
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$$\displaystyle 3\times 3$$
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$$\displaystyle 3\times 2$$
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$$\displaystyle 2\times 2$$
Explanation
When a matrix of order $$ m \times a $$ is multiplied with a matrix of order $$ a \times n $$, the resulting matrix is of order $$ m \times n $$
As the second matrix is of order $$ 2 \times 3 $$ and the product matrix is $$ 3 \times 3 $$, we get $$ a = 2 ; m = 3 ; n = 3 $$
Thus, the matrix A should have order $$ 3 \times 2 $$
Inverse of a diagonal matrix is
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Symmetric
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A skew-symmetric
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A diagonal matrix
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None of these
Explanation
Consider a diagonal
$$\begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$
Minor of the matrix
$$\begin{bmatrix} 4 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$
Co factor of matrix
$$\begin{bmatrix} 4 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 2 \end{bmatrix}\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$$
$$\begin{bmatrix} 4 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$
Adjoint of matrix$$=$$ Transpose of co factor$$=\begin{bmatrix} 4 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$
Set of matrix considered$$=6$$
$${ A }^{ -1 }=\cfrac { Adjoint\quad (A) }{ det(A) } $$
$$=\cfrac { 1 }{ 6 } \begin{bmatrix} 4 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$
Clearly $${ A }^{ -1 }$$ is also diagonal matrix
$$\therefore $$ Inverse of a diagonal matrix is a diagonal matrix
Option C is correct
If $$A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$, then $$A^2 - 5A$$ is equal to
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$$2I$$
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$$-2I$$
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$$3I$$
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Null matrix
Explanation
$$ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} ,$$
Also, $$A^{2} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 1\times1+2\times 3\,\,\,\,\,1\times 2+2\times 4 \\ 3\times 1+4\times 3\,\,\,\,\,3\times2+4\times 4 \end{bmatrix} $$
$$ \Rightarrow A^{2} = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix} $$
$$ 5A = 5\times \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 10 \\ 15 & 20 \end{bmatrix} $$
$$ A^{2} -5A = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix}-\begin{bmatrix} 5 & 10 \\ 15 & 20 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & +2 \end{bmatrix} = 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$
$$ \therefore A^{2}-5A = 2I $$
If $$A+I=\begin{bmatrix} 3 & -2 \\ 4 & 1 \end{bmatrix}$$, then $$\left( A+I \right) \cdot \left( A-I \right) $$ is equal to
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$$\begin{bmatrix} -5 & -4 \\ 8 & -9 \end{bmatrix}$$
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$$\begin{bmatrix} -5 & 4 \\ -8 & 9 \end{bmatrix}$$
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$$\begin{bmatrix} 5 & 4 \\ 8 & 9 \end{bmatrix}$$
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$$\begin{bmatrix} -5 & -4 \\ -8 & -9 \end{bmatrix}$$
Explanation
Given, $$A+I=\begin{bmatrix} 3 & -2 \\ 4 & 1 \end{bmatrix}$$
$$\therefore A-I=A+I-2I$$
$$=\begin{bmatrix} 3 & -2 \\ 4 & 1 \end{bmatrix}-\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$$
$$=\begin{bmatrix} 1 & -2 \\ 4 & -1 \end{bmatrix}$$
$$\therefore \left( A+I \right) \cdot \left( A-I \right) =\begin{bmatrix} 3 & -2 \\ 4 & 1 \end{bmatrix}\begin{bmatrix} 1 & -2 \\ 4 & -1 \end{bmatrix}$$
$$=\begin{bmatrix} 3-8 & -6+2 \\ 4+4 & -8-1 \end{bmatrix}$$
$$=\begin{bmatrix} -5 & -4 \\ 8 & -9 \end{bmatrix}$$
Two matrices $$A$$ and $$B$$ are multiplied to get $$AB$$ if
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both are rectangular
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both have same order
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number of columns of $$A$$ is equal to rows of $$B$$
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number of rows of $$A$$ is equal to no of columns of $$B$$
Explanation
According to condition of multiplication of two matrices:
The columns in first matrix should be same in number as the rows in the second matrix [For $$AB$$ to be defined, we must have columns in $$A=$$ rows in $$B$$]
Hence option C is true.
If $$A=\begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}$$, then $${ A }^{ 2 }$$ is equal to
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Null matrix
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Itself $$A$$
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Unit matrix
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Scalar matrix
Explanation
Given, $$A=\begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}$$
$$\therefore { A }^{ 2 }=\begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}\begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}$$
$$=\begin{bmatrix} 4+2-4 & -2-3+4 & 2+2-3 \\ -4-6+8 & 2+9-8 & -2-6+6 \\ -8-8+12 & 4+12-12 & -4-8+9 \end{bmatrix}$$
$$=\begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}=A$$
If $$\displaystyle A=\left[ \begin{matrix} 2 \\ 0 \\ 0 \end{matrix}\,\,\,\begin{matrix} 0 \\ 2 \\ 0 \end{matrix}\,\,\,\begin{matrix} 0 \\ 0 \\ 2 \end{matrix} \right] $$ then
$$\displaystyle { A }^{ 5 }=$$
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$$\displaystyle 5A$$
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$$\displaystyle 10A$$
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$$\displaystyle 16A$$
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$$\displaystyle 32A$$
Explanation
Given, $$A= \displaystyle \left[\begin{matrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{matrix} \right]$$
$$\Rightarrow A= \displaystyle 2\left[\begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{matrix} \right]$$
$$\Rightarrow A= \displaystyle 2I$$
$$\Rightarrow A^5= \displaystyle 2^5I^5$$
$$\Rightarrow A^5= \displaystyle 2^5I$$
$$\Rightarrow A^5= \displaystyle 2^4.2I$$
$$\Rightarrow A^5= \displaystyle 2^4. \displaystyle 2\left[\begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{matrix} \right]$$
$$\Rightarrow A^5= \displaystyle 16\displaystyle \left[\begin{matrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{matrix} \right]$$.
$$\Rightarrow A^5= \displaystyle 16A $$
Option $$C$$ is correct.
If $$A$$ and $$B$$ are symmetric matrices, then $$ABA$$ is
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Symmetric
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Skew-symmetric
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Diagonal
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Triangular
Explanation
Given $$A$$ and $$B$$ are symmetric matrices.
$$\Rightarrow A^T = A$$ and $$B^T = B$$
Now, take $$(ABA)^T$$
$$\Rightarrow (ABA)^T = A^T B^T A^T$$
$$\Rightarrow (ABA)^T = ABA$$
Hence, $$ABA$$ is also a symmetric matrix.
If $$A$$ is an $$m \times n$$ matrix such that $$AB$$ and $$BA$$ are both defined, then order of $$B$$ is
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$$m \times n$$
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$$n \times m$$
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$$n \times n$$
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$$m \times m$$
Explanation
Since $$AB$$ exists, so the number of column in $$A=$$ Number of column in $$B$$.
So $$B$$ has $$n$$ row.
Since $$BA$$ exists, so the number of column in $$B=$$ Number of column in $$A$$.
So $$B$$ has $$m$$ column.
$$\therefore$$ B is an $$n\times m $$ matrix.
If $$\displaystyle A=\begin{bmatrix} 0 & 0 & 1\\ 0 & 1&0 \\ 1& 0 & 0\end{bmatrix}$$, then $$A^{-1}$$ is.
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$$-A$$
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$$A$$
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$$1$$
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None of these
Explanation
We have, $$A=\begin{bmatrix} 0 & 0&1\\ 0 &1 &0\\ 1&0 &0\end{bmatrix}$$
$$\Rightarrow |A|=0(0-0)-0(0-0)+1(0-1)$$
$$\Rightarrow |A|=-1$$
and cofactors of A are
$$A_{11}=0, A_{12}=0, A_{13}=-1,$$
$$A_{21}=0, A_{22}=-1, A_{23}=0,$$
$$A_{31}=-1, A_{32}=0, A_{33}=0$$
$$\therefore A^{-1}=\displaystyle\frac{adj(A)}{|A|}$$
$$=-\displaystyle\frac{1}{1}\begin{bmatrix} 0 & 0 & -1\\0 & -1 &0\\ -1 &0 &0\end{bmatrix}$$
If $$A=\begin{bmatrix}1 & 1\\ 1& 1\end{bmatrix}$$, then $$A^{100}$$ is equal to.
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$$2^{100}A$$
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$$2^{99}A$$
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$$100A$$
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$$299A$$
Explanation
Given, $$A=\begin{bmatrix}1&1\\1&1\end{bmatrix}$$
$$\therefore A^2=A\cdot A=\begin{bmatrix}1&1\\1&1\end{bmatrix}\begin{bmatrix}1&1\\1&1\end{bmatrix}$$
$$=2\begin{bmatrix}1&1\\1&1\end{bmatrix}=2A$$
Now, $$A^4=A^2\cdot A^2=2A\cdot 2A$$
$$=4A^2=4\times 2A$$
$$=8A=2^3A$$
Similarly, $$A^8=2^7A$$
$$\therefore A^{100}=2^{99}A$$
If $$\begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}$$, then $${A}^{-1}$$ is equal to
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$$\begin{bmatrix} \cfrac { 5 }{ 11 } & \cfrac { 2 }{ 11 } \\ \cfrac { 3 }{ 11 } & -\cfrac { 1 }{ 11 } \end{bmatrix}$$
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$$\begin{bmatrix} -\cfrac { 5 }{ 11 } & -\cfrac { 2 }{ 11 } \\ -\cfrac { 3 }{ 11 } & -\cfrac { 1 }{ 11 } \end{bmatrix}$$
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$$\begin{bmatrix} \cfrac { 5 }{ 11 } & \cfrac { 2 }{ 11 } \\ \cfrac { 3 }{ 11 } & \cfrac { 1 }{ 11 } \end{bmatrix}$$
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$$\begin{bmatrix} 5 & 2 \\ 3 & -1 \end{bmatrix}$$
Explanation
Since $$A=\begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}$$
$$\therefore \left| A \right| =\begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}=-5-6=-11$$
and $$adj(A)=\begin{bmatrix} -5 & -2 \\ -3 & 1 \end{bmatrix}$$
$$\therefore { A }^{ -1 }=\cfrac { 1 }{ \left| A \right| } adj(A)$$
$$=-\cfrac { 1 }{ 11 } \begin{bmatrix} -5 & -2 \\ -3 & 1 \end{bmatrix}=\cfrac { 1 }{ 11 } \begin{bmatrix} 5 & 2 \\ 3 & -1 \end{bmatrix}$$
$$\quad =\begin{bmatrix} \cfrac { 5 }{ 11 } & \cfrac { 2 }{ 11 } \\ \cfrac { 3 }{ 11 } & -\cfrac { 1 }{ 11 } \end{bmatrix}$$
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 $$A$$.
ie, $$B=10{ A }^{ -1 }$$
$$\therefore \left( 10 \right) { A }^{ -1 }=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$$
$$\therefore \left( 10 \right) { A }^{ -1 }\cdot A=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}A$$
$$\Rightarrow 10I=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix}=\begin{bmatrix} 10 & 0 & 0 \\ -5+\alpha & 5+\alpha & -5+\alpha \\ 0 & 0 & 10 \end{bmatrix}$$
$$\Rightarrow 5+\alpha =10$$
$$\Rightarrow \alpha =5$$
Inverse of the matrix $$\begin{bmatrix} \cos 2\theta & -\sin 2\theta\\ \sin 2\theta & \cos 2\theta\end{bmatrix}$$ is.
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$$\begin{bmatrix} \cos 2\theta & -\sin 2\theta\\ \sin 2\theta & \cos 2\theta\end{bmatrix}$$
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$$\begin{bmatrix} \cos 2\theta & \sin 2\theta\\ \sin 2\theta & -\cos 2\theta\end{bmatrix}$$
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$$\begin{bmatrix} \cos 2\theta & \sin 2\theta\\ \sin 2\theta & \cos 2\theta\end{bmatrix}$$
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$$\begin{bmatrix} \cos 2\theta & \sin 2\theta\\ -\sin 2\theta & \cos 2\theta\end{bmatrix}$$
Explanation
Let $$A=\begin{bmatrix} \cos 2\theta & -\sin 2\theta\\ \sin 2\theta & \cos 2\theta\end{bmatrix}$$
$$\therefore |A|=\cos^2 2\theta+\sin^2 2\theta=1$$
and $$adj(A)=\begin{bmatrix}\cos 2\theta &\sin 2\theta\\ -\sin 2\theta & \cos 2\theta\end{bmatrix}$$
$$\therefore A^{-1}=\displaystyle\frac{1}{1}\begin{bmatrix}\cos 2\theta & \sin 2\theta\\ -\sin 2\theta &\cos 2\theta\end{bmatrix}$$
$$=\begin{bmatrix}\cos 2\theta & \sin 2\theta\\ -\sin 2\theta & \cos 2\theta\end{bmatrix}$$
If $$A=\begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{bmatrix}$$ and $$B=\begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{bmatrix}$$, then $${ \left( AB \right) }^{ T }$$ is equal to
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$$\begin{bmatrix} -3 & -2 \\ 10 & 7 \end{bmatrix}$$
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$$\begin{bmatrix} -3 & 10 \\ -2 & 7 \end{bmatrix}$$
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$$\begin{bmatrix} -3 & 7 \\ 10 & 2 \end{bmatrix}$$
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None of these
Explanation
Given, $$A=\begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{bmatrix},B=\begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{bmatrix}$$
$$\therefore AB=\begin{bmatrix} 2-6+1 & 1-4+1 \\ 4+3+3 & 2+2+3 \end{bmatrix}=\begin{bmatrix} -3 & -2 \\ 10 & 7 \end{bmatrix}$$
Now, $${ \left( AB \right) }^{ T }=\begin{bmatrix} -3 & 10 \\ -2 & 7 \end{bmatrix}$$
If $$\begin{bmatrix}1 & 1 &1\\ 1&-2 &-2\\1 & 3 &1\end{bmatrix}$$ $$\begin{bmatrix} x\\ y\\z\end{bmatrix}=\begin{bmatrix} 0 \\ 3\\4\end{bmatrix}$$, then $$\begin{bmatrix} x\\ y\\z\end{bmatrix}$$ is equal to.
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$$\begin{bmatrix} 0\\1\\1\end{bmatrix}$$
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$$\begin{bmatrix} 1\\2\\-3\end{bmatrix}$$
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$$\begin{bmatrix} 5\\-2\\1\end{bmatrix}$$
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$$\begin{bmatrix} 1\\-2\\3\end{bmatrix}$$
Explanation
Given, $$\begin{bmatrix} 1&1&1\\1&-2&-2\\1&2&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\3\\4\end{bmatrix}$$
$$\therefore x+y+z=0$$,
$$x-2y-2z=3$$,
$$x+3y=z=4$$
On solving these equations, we get
$$x=1, y=2$$ and $$z=-3$$
$$\therefore \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\-3\end{bmatrix}$$
$$A=\begin{bmatrix} -2&4\\-1&2 \end{bmatrix}$$, then $$A^2$$ is equal to
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Null matrix
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Unit matrix
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$$\begin{bmatrix} 1&0\\0&1 \end{bmatrix}$$
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$$\begin{bmatrix} 0&0\\0&1 \end{bmatrix}$$
Explanation
$$A=\begin{bmatrix} -2&4\\-1&2 \end{bmatrix}$$
$$A.A.=A^{2}= \begin{bmatrix} -2&4\\-1&2 \end{bmatrix} \begin{bmatrix} -2&4\\-1&2 \end{bmatrix} = \begin{bmatrix} 4-4\,\,-2\times 4+4\times 2\\-1x-2+2x-1\,\,-1\times 4+2\times 2 \end{bmatrix} $$
$$ A^{2}=\begin{bmatrix} 0&0\\0&0 \end{bmatrix} = O =$$ null matrix
.
If $$A=[x\, y\, z], B = \begin{bmatrix}a&h&g\\h&b&f\\g&f&c\end{bmatrix}$$ and $$C=\begin{bmatrix}x\\y\\z\end{bmatrix}$$
Then, $$ABC = 0$$, if
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$$[ax^2+by^2+cz^2+2gxy+2fyz+2czx]=0$$
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$$[ax^2+cy^2+bz^2+xy+yz+zx]=0$$
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$$[ax^2+by^2+cz^2+2hxy+2by+2cz]=0$$
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$$[ax^2+by^2+cz^2+2gzx+2fyz+2hxy]=0$$
Explanation
$$ A = [x\,y\,z] B = \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix} c = \begin{bmatrix} x \\ y \\ z \end{bmatrix} $$
$$ A.B = [x\,y\,z] \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix} $$
$$ = [ax+hy+gz\,\,\,\,hx+by+fz\,\,\,\,gx+fy+cz] $$
$$ A.B.C = (AB)\times \begin{bmatrix} x \\ y \\ z\end{bmatrix} = [ax^{2}+hyx+gyx+hxy+by^{2}+fzy+gxz+fyz+cz^{2}] $$
$$ ABC = [ax^{2}+by^{2}+cz^{2}+2hxy+2gzx+2fzy] $$
$$ ABC = 0 $$
when,
$$ ax^{2}+by^{2}+cz^{2}+2hxy+2gzx+2fzy = 0 $$
If $$A$$ and $$B$$ are any two matrices, then
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$$AB=BA$$
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$$AB=I$$
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$$AB=0$$
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$$AB$$ may or may not be defined
Explanation
$$\textbf{Step 1: Finding the order}$$
$$\text{Matrix multiplication is only defined for two matrices whose order is of the form}$$
$$a\times b \text{ and }b\times c \text{ respectively}$$
$$\text{Since we cant say anything about the order of the given matrices it may or may not be defined}$$
$$\textbf{Hence, AB may or may not be defined}$$
If $$P=\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}$$, then $${P}^{5}$$ is equal to
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$$P$$
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$$2P$$
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$$-P$$
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$$-2P$$
Explanation
Given $$P=\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}$$
$${P}^{2}=P.P=\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}$$
$$=\begin{pmatrix} 4+2-4 & -4-6+8 & -8-8+12 \\ -2-3+4 & 2+9-8 & 4+12-12 \\ 2+2-3 & -2-6+6 & -4-8+9 \end{pmatrix}$$
$$=\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}=P$$
$$\therefore$$ $${P}^{4}={P}^{2}=P$$
$$\Rightarrow$$ $${P}^{5}={P}^{2}=P$$
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
$$A=\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\\ { A }^{ 2 }=A\times A=\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\times \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\\ \quad \quad =\begin{bmatrix} 4+1 & -2-2 \\ -2-2 & 1+4 \end{bmatrix}=\begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}$$
From the given option to get first entry i.e $$5$$ of $$A^2$$, only option $$(A)$$ is satisfied. Or we can check as below
$$4A-3I=\begin{bmatrix} 8 & -4 \\ -4 & 8 \end{bmatrix}-3\times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\\ \quad \quad \quad \quad =\begin{bmatrix} 8 & -4 \\ -4 & 8 \end{bmatrix}-\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\\ \quad \quad \quad \quad =\begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}={ A }^{ 2 }$$
If $$A$$ is a matrix such that $$\begin{pmatrix}2&1 \\3&2 \end{pmatrix} A(1 \space \,1)=\begin{pmatrix}1&1 \\0&0 \end{pmatrix}$$ then $$A= $$
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$$\begin{pmatrix}1&1 \\0&1 \end{pmatrix}$$
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$$(2 \,1)$$
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$$\begin{pmatrix}1&0 \\-1&1 \end{pmatrix}$$
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$$\begin{pmatrix}2 \\-3 \end{pmatrix}$$
Explanation
$$\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}A \begin{pmatrix} 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$$
To maintain the rules of matrix multiplication, $$A$$ has to be of the order $$2 \times 1$$
$$\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \begin{pmatrix} 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$$
$$\Rightarrow \begin{pmatrix} 2x + y \\ 3x + 2y \end{pmatrix} \begin{pmatrix} 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$$
$$\Rightarrow \begin{pmatrix} 2x + y & 2x + y \\ 3x + 2y & 3x + 2y \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$$
$$\Rightarrow 2x + y = 1$$ and $$3x + 2y = 0$$
$$\Rightarrow 4x + 2y - (3x + 2y) = 2 - 0$$
$$\Rightarrow x = 2$$ and $$y = -3$$
For the matrices $$A=\begin{bmatrix} 1 & 3 \\ 0 & -1 \end{bmatrix}$$ $$B=\begin{bmatrix} 2 & 2 \\ -1 & 4 \end{bmatrix}$$
What is $$AB$$?
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$$\begin{bmatrix} 3 & 5 \\ -1 & 3 \end{bmatrix}$$
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$$\begin{bmatrix} 1 & -1 \\ -1 & 5 \end{bmatrix}$$
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$$\begin{bmatrix} -1 & 14 \\ 1 & -4 \end{bmatrix}$$
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$$\begin{bmatrix} 2 & 4 \\ -1 & -7 \end{bmatrix}$$
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none of the above
Explanation
$$A$$ and $$B$$ are given by $$A=\left[ \begin{matrix} 1 & 3 \\ 0 & -1 \end{matrix} \right] , B=\left[ \begin{matrix} 2 & 2 \\ -1 & 4 \end{matrix} \right]$$
Thus, $$AB=\left[ \begin{matrix} (1\times 2)+(3\times -1) & (1\times 2)+(3\times 4) \\ (0\times 2)+(-1\times -1) & (0\times 2)+(-1\times 4) \end{matrix} \right] $$
$$=\left[ \begin{matrix} 2-3 & 2+12 \\ 0+1 & 0-4 \end{matrix} \right] $$
$$=\left[ \begin{matrix} -1 & 14 \\ 1 & -4 \end{matrix} \right]$$
The symmetric part of the matrix A= $$\begin{bmatrix}
1 &2 &4 \\
6 & 8 & 2\\
2 & -2 &7
\end{bmatrix}$$.
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$$\begin{bmatrix}
1 &4 &3 \\
2 & 8 & 0\\
3 & 0 &7
\end{bmatrix}$$
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$$\begin{bmatrix}
1 &4 &3 \\
4 & 8 & 0\\
3 & 0 &7
\end{bmatrix}$$
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$$\begin{bmatrix}
0 &-2 &-1 \\
-2 & 0 & -2\\
-1 & -2 &0
\end{bmatrix}$$
0%
$$\begin{bmatrix}
0 &-2 &-1 \\
2 & 0 & 2\\
-1 & 2 &0
\end{bmatrix}$$
Explanation
Symmetric part of any matrix A is given by $$\dfrac{A+A'}{2}$$, where $$A'$$ transpose of A
So symmetric part of given matrix $$\begin{bmatrix}1&2&4\\6&8&2\\2&-2&7 \end{bmatrix}$$
is $$ = \dfrac{1}{2}\bigg ( $$
$$\begin{bmatrix}1&2&4\\6&8&2\\2&-2&7 \end{bmatrix}$$+
$$\begin{bmatrix}1&6&2\\2&8&-2\\4&2&7 \end{bmatrix}$$
$$\bigg)$$
$$=\dfrac{1}{2}$$
$$\begin{bmatrix}2&8&6\\8&16&0\\6&0&14 \end{bmatrix}$$
$$ = \begin{bmatrix}1&4&3\\4&8&0\\3&0&7 \end{bmatrix}$$
If $$A = \begin{bmatrix} 0& 1\\ 1 & 0\end{bmatrix}$$, then $$A^{2}$$ is equal to ______
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$$\begin{bmatrix} 0& 1\\ 1 & 0\end{bmatrix}$$
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$$\begin{bmatrix} 1& 0\\ 1 & 0\end{bmatrix}$$
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$$\begin{bmatrix} 1& 0\\ 0 & 1\end{bmatrix}$$
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$$\begin{bmatrix} 0& 1\\ 0 & 1\end{bmatrix}$$
Explanation
We have, $$A = \begin{bmatrix} 0& 1\\ 1 & 0\end{bmatrix}$$
Hence $$A^2=AA=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix} 0&1\\1&0\end{bmatrix}$$
$$\quad = \begin{bmatrix}0\times 0+1\times 1&0\times 1+1\times 0\\1\times 0+0\times 1 &1\times 1+0\times 0\end{bmatrix}$$
$$=\begin{bmatrix}1&0\\0&1\end{bmatrix}$$
$$\begin{bmatrix} 3& 2\\ -1 & -2\\ 4 & 5\end{bmatrix} \begin{bmatrix} 0& a\\ b & c \end{bmatrix} = \begin{bmatrix}-4 &9 \\ 4 & -7\\ -10 & 19\end{bmatrix}$$
Find $$a, b$$ and $$c$$.
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$$2, -1, 3$$
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$$1, -2, 3$$
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$$3, -2, 1$$
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$$1, 2, 3$$
Explanation
The product is $$\begin{bmatrix} 3 & 2 \\ -1 & -2 \\ 4 & 5 \end{bmatrix}\begin{bmatrix} 0 & a \\ b & c \end{bmatrix}=$$ $$\begin{bmatrix} 2b & 3a+2c \\ -2b & -2c-a \\ 5b & 5c+4a \end{bmatrix}$$
By given, we have
$$\begin{bmatrix} 2b & 3a+2c \\ -2b & -2c-a \\ 5b & 5c+4a \end{bmatrix}$$
$$=\begin{bmatrix} -4 & 9 \\ 4 & -7 \\ -10 & 19 \end{bmatrix}$$
By comparing, we get
$$b=-2 , a = 1 , c = 3$$
If $$P=\begin{pmatrix}2&-2&-4 \\-1&3&4\\1&-2&-3\end{pmatrix}$$ then $$P^5$$ equals
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$$P$$
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$$2P$$
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$$-P$$
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$$-2P$$
Explanation
$$P=\begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}$$
$${ P }^{ 2 }=\begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}\begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}$$
$$=\begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}=P$$
Now,
$${ P }^{ 2 }=P\quad { P }^{ 5 }={ P }^{ 2 }\times { P }^{ 3 }\Rightarrow { P }^{ 5 }={ P }^{ 4 }$$
$$\left( \because { P }^{ 2 }=P \right) \Rightarrow { P }^{ 5 }={ P }^{ 2 }\times { P }^{ 2 }$$
$${ P }^{ 2 }=P\times P$$
$$\Rightarrow { P }^{ 5 }={ P }^{ 2 }=P$$
$$\therefore { P }^{ 5 }=P$$
$$\therefore $$Option A is correct
For a matrix $$A \begin{pmatrix} 1& 0 & 0\\ 2 & 1 & 0\\ 3 & 2 & 1\end{pmatrix}$$, if $$U_{1}, U_{2}$$ and $$U_{3}$$ are $$3\times 1$$ column matrices satisfying $$AU_{1} = \begin{pmatrix}1\\ 0 \\ 0
\end{pmatrix}, AU_{2} \begin{pmatrix}2\\3 \\ 0
\end{pmatrix}, AU_{3} = \begin{pmatrix}2\\ 3\\ 1
\end{pmatrix}$$ and $$U$$ is $$3\times 3$$ matrix whose columns are $$U_{1}, U_{2}$$ and $$U_{3}$$
Then sum of the elements of $$U^{-1}$$ is
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$$6$$
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$$0 (zero)$$
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$$1$$
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$$2/3$$
Explanation
Let $$U_{i} = \begin{pmatrix}a_{i}\\b_{i} \\c_{i} \end{pmatrix} i = 1, 2, 3$$
$$AU_{1} = \begin{pmatrix}a_{1}\\2a_{1} + b_{1} \\ 3a_{1} + 2b_{1} + c_{1}
\end{pmatrix} = \begin{pmatrix}1\\0 \\ 0
\end{pmatrix}$$, So $$a_{1} = 1, b_{1} = -2, c_{1} = 1$$
$$AU_{2} = \begin{pmatrix}a_{2}\\2a_{2} + b^{2} \\ 3a_{2} + 2b_{2} + c_{2}
\end{pmatrix} = \begin{pmatrix}2\\ 3\\ 0
\end{pmatrix}$$,
So, $$a_{2} = 2, b_{2} = -1, c_{2} = -4$$. Similarly, $$a_{3} = 2, b_{3} = -1, c_{3} = -3$$
So, $$U = \begin{pmatrix} 1& 2 & 2\\ -2 & -1 & -1\\ 1 & -4 & -3\end{pmatrix}$$. So, sum of elements of $$U^{-1}$$ is zero.
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