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CBSE Questions for Class 12 Commerce Maths Matrices Quiz 2 - MCQExams.com
CBSE
Class 12 Commerce Maths
Matrices
Quiz 2
If $$A = \begin{bmatrix}2 & 3 & 4 \\ -3 & 4 & 8\end{bmatrix},$$ $$B = \begin{bmatrix}-1 & 4 & 7 \\ -3 & -2 & 5\end{bmatrix}$$ and $$ A+B = \begin{bmatrix}1 & a & b \\ c & 2 & 13\end{bmatrix},$$ then find the value of $$a+b+c.$$
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$$12$$
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$$21$$
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$$15$$
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$$5$$
Explanation
Given,
$$A = \begin{bmatrix}2 & 3 & 4 \\ -3 & 4 & 8\end{bmatrix},$$ $$B = \begin{bmatrix}-1 & 4 & 7 \\ -3 & -2 & 5\end{bmatrix}$$ and $$\quad A+B = \begin{bmatrix}1 & a & b \\ c & 2 & 13\end{bmatrix}$$ ....(1)
Therefore,
$$\quad A+B = \begin{bmatrix}2 & 3 & 4 \\ -3 & 4 &
8\end{bmatrix} + \begin{bmatrix}-1 & 4 & 7 \\ -3 & -2 &
5\end{bmatrix}=\begin{bmatrix}2-1 & 3+4 & 4+7 \\ -3-3 & 4-2 & 8+5\end{bmatrix}= \begin{bmatrix}1 & 7 & 11 \\ -6 & 2 & 13\end{bmatrix}$$ .....(2)
By equating (1) and (2), we get
$$ a+b+c=7+11-6=12$$
If $$A = \begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix}$$, Then $$A^2$$ =
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$$\begin{bmatrix}8 & -5 \\ -5 & 3\end{bmatrix}$$
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$$\begin{bmatrix}8 & -5 \\ 5 & 3\end{bmatrix}$$
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$$\begin{bmatrix}8 & -5 \\ -5 & -3\end{bmatrix}$$
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$$\begin{bmatrix}8 & 5 \\ -5 & 3\end{bmatrix}$$
Explanation
Given $$A = \begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix}$$
$$A^{2}= \begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix} \begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix}$$
$$=\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$$
If $$\displaystyle 2\begin{bmatrix}3 &4 \\5 &x \end{bmatrix}+\begin{bmatrix}1 &y \\0 &2 \end{bmatrix}=\begin{bmatrix}7 &0 \\10 &5 \end{bmatrix}$$
, then the values of $$x\ and\ y$$ are :
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$$\displaystyle x=0,y=-2$$
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$$\displaystyle x=\dfrac{3}{2},y=-8$$
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$$\displaystyle x=-2,y=-8$$
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$$\displaystyle x=2,y=8$$
Explanation
$$2\left[ \begin{matrix} 3 & 4 \\ 5 & x \end{matrix} \right] +\left[ \begin{matrix} 1 & y \\ 0 & 2 \end{matrix} \right] =\left[ \begin{matrix} 7 & 0 \\ 10 & 5 \end{matrix} \right] $$
$$\Rightarrow \left[ \begin{matrix} 6 & 8 \\ 10 & 2x \end{matrix} \right] +\left[ \begin{matrix} 1 & y \\ 0 & 2 \end{matrix} \right] =\left[ \begin{matrix} 7 & 0 \\ 10 & 5 \end{matrix} \right] $$
$$\Rightarrow \left[ \begin{matrix} 7 & 8+y \\ 10 & 2x+2 \end{matrix} \right] =\left[ \begin{matrix} 7 & 0 \\ 10 & 5 \end{matrix} \right] $$
$$\Rightarrow 8+y=0\;\&\;2x+2=5$$
$$\Rightarrow y=-8\;\&\;2x=3$$
$$\Rightarrow x=\dfrac { 3 }{ 2 } $$
$$\therefore x=\dfrac { 3 }{ 2 }\; \&\;y=-8$$
Hence, the answer is $$x=\dfrac { 3 }{ 2 }\; \&\;y=-8.$$
If $$\begin{bmatrix} x & y \\ u & v \end{bmatrix}$$ is symmetric matrix, then
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$$x+v=0$$
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$$x-v=0$$
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$$y+u=0$$
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$$y-u=0$$
Explanation
$$A =\begin{bmatrix} x & y \\ u & v \end{bmatrix}$$
$$A^T =\begin{bmatrix} x & u \\ y & v \end{bmatrix}$$
So for $$A$$ to be symmetric matrix, $$A=A^T$$
$$\Rightarrow y = u\Rightarrow y-u=0$$
If $$\displaystyle A=\begin{bmatrix}x &y \\z &w \end{bmatrix},B=\begin{bmatrix}x &-y \\-z &w \end{bmatrix}$$ and $$C=\begin{bmatrix}-2x &0 \\0 &-2w \end{bmatrix},$$ then $$A+B+C$$ is a:
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identity matrix
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null matrix
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row matrix
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column matrix
Explanation
Given:
$$\displaystyle A=\begin{bmatrix}x &y \\z &w \end{bmatrix},B=\begin{bmatrix}x &-y \\-z &w \end{bmatrix}$$ and $$C=\begin{bmatrix}-2x &0 \\0 &-2w \end{bmatrix}$$
Consider,
$$\displaystyle A+B+C=\begin{bmatrix}X &Y \\Z &W \end{bmatrix}+\begin{bmatrix}X &-Y \\-Z &W \end{bmatrix}+\begin{bmatrix}-2X &0 \\0 &-2W \end{bmatrix}$$
$$=\begin{bmatrix}X+X+\left ( -2X \right ) &Y+\left ( -Y \right )+0 \\Z+\left ( -Z \right )+0 &W+W+\left ( -2W \right ) \end{bmatrix}$$
$$=\begin{bmatrix}0 &0 \\0 &0 \end{bmatrix}$$
Hence, $$A+B+C$$ is a null matrix.
If matrix $$A$$ is of order $$p\times q$$ and matrix $$B$$ is of order $$r\times s$$ ,then $$A-B$$ will exist if
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$$p=q$$
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$$p=r, q=s$$
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$$p=q, r=s$$
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$$p=s, q=r$$
Explanation
If matrix $$A$$ is of order $$p\times q$$ and matrix $$B$$ is of order $$r\times s,$$ then $$A-B$$ will exist if order of $$A$$ and $$B$$ is same
Therefore, $$p=r,q=s$$
Ans: B
If $$\displaystyle A=\left [ a_{ij} \right ]_{m\times\:n}, B=\left [ b_{ij} \right ]_{m\times\:n},$$ then the element $$\displaystyle C_{23}$$ of the matrix $$C=A+B$$ is
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$$\displaystyle C_{32}$$
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$$\displaystyle a_{23}+b_{32}$$
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$$\displaystyle a_{23}+b_{23}$$
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$$\displaystyle a_{32}+b_{23}$$
Explanation
$$\displaystyle A=\left [ a_{ij} \right ]_{m\times\:n}$$ and
$$B=\left [ b_{ij} \right ]_{m\times\:n}$$
$$\therefore C=A+B$$
$$\Rightarrow \left [ c_{ij} \right ]_{m\times\:n}=\left [ a_{ij}+b_{ij} \right ]_{m\times\:n}$$
Hence, $$c_{23}=a_{23}+b_{23}$$
If $$A-A'=0$$, then $$A'$$ is
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orthogonal matrix
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symmetric matrix
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skew-symmetric matrix
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triangular matrix
Explanation
It is fundamental concept that,
If $$A = A'$$ then $$A=A'$$ is called symmetric matrix.
For any square matrix $$A, \ A+{A}^{T}$$ is-
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unit matrix
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symmetric matrix
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skew symmetric matrix
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zero matrix
Explanation
Let, $$B = A+A^T$$
$$\Rightarrow B^T = (A+A^T)^T =A^T+(A^T)^T =A^T+A =B$$
Hence $$B = A+A^T$$ is a symmetric matrix.
If $$\displaystyle \begin{bmatrix} x & y \\ 1 & 6 \end{bmatrix} $$ = $$\displaystyle \begin{bmatrix} 1 & 8 \\ 1 & 6 \end{bmatrix},$$ then $$x+2y=$$
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$$13$$
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$$17$$
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$$19$$
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None of these
Explanation
By equating given matrices, we have
$$x=1$$ and $$y=8$$
$$\displaystyle \therefore x+2y=17$$
If $$A= \displaystyle \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} $$ and B=$$\displaystyle \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, $$ then $$A+B=$$
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$$A$$
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$$B$$
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$$\displaystyle \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} $$
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$$\displaystyle \begin{bmatrix} 0 & 2 \\ 2 & 2 \end{bmatrix} $$
Explanation
Given, $$A= \displaystyle \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} $$ and B=$$\displaystyle \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$
Then, $$A+ B = \displaystyle \begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix}$$
If A is any square matrix, then $$(A\, +\, A^T)$$ is a ............ matrix
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symmetric
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skew symmetric
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scalar
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identity
Explanation
Lets matrix $$A=\begin{bmatrix} a & b & c\\ d & f & g\\ e & h & i\end{bmatrix}$$
Then, matrix $$A^T$$, after transforming rows with each other will be,
$$A^T=\begin{bmatrix} a & d & e\\ b & f & h\\ c & g & i\end{bmatrix}$$
on adding $$A+A^T$$, we get
$$\begin{bmatrix} 2a & (b+d) & (c+e)\\ (b+d) & 2f & (h+g)\\ (e+c) & (h+g) & 2i\end{bmatrix}$$
which is clearly symmetry about its diagonal.
If $$A$$ is a square matrix, then $$A-{A}^{T}$$ is-
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unit matrix
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null matrix
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$$A$$
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a skew symmetric matrix
Explanation
$${ \left( A-{ A }^{ T } \right) }^{ T }={ A }^{ T }-{ \left( { A }^{ T } \right) }^{ T }={ A }^{ T }-A=-\left( A-{ A }^{ T } \right) $$
Therefore, it is a skew symmetric matrix
Ans: D
If $$\displaystyle \begin{bmatrix} 2 & 3 \\ 4 & 4 \end{bmatrix} $$+$$\displaystyle \begin{bmatrix} x & 3 \\ y & 1 \end{bmatrix} $$=$$\displaystyle \begin{bmatrix} 10 & 6 \\ 8 & 5 \end{bmatrix},$$ then $$(x,y)=$$
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$$(4,8)$$
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$$(8,4)$$
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$$(1,2)$$
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$$(2,4)$$
Explanation
By equating given matrices, we have
$$2+x=10 \Rightarrow x=8$$
Also, $$4+y=8 \Rightarrow y=4$$.
For square matrix $$A$$, $$A{A}^{T}$$ is-
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unit matrix
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symmetric matrix
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skew symmetric matrix
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diagonal matrix
Explanation
Since,$${ \left( A{ A }^{ T } \right) }^{ T }={ \left( { A }^{ T } \right) }^{ T }{ A }^{ T }=A{ A }^{ T }$$
Therefore, $$A{ A }^{ T }$$ is symmetric matrix
Ans: B
IF $$A=\begin{bmatrix} -1 & 0 & 2 \\ 3 & 1 & 2 \end{bmatrix}$$ and $$B=\begin{bmatrix} -1 & 5 \\ 2 & 7 \\ 3 & 10 \end{bmatrix},$$ then
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$$AB$$ and $$BA$$ both exist
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$$AB$$ exists but bot $$BA$$
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$$BA$$ exists but not $$AB$$
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but $$AB$$ and $$BA$$ do not exist
Explanation
Since, order of $$A$$ is $$2\times3$$ and or
der of $$B$$ is $$3\times 2$$
Thus, $$AB$$ and $$BA$$ exists.
If $$A$$ and $$B$$ are symmetric matrices of order $$\displaystyle n,\left( A\neq B \right) $$, then
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$$A+B$$ is skew-symmetric
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$$A+B$$ is symmetric
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$$A+B$$ is a diagonal matrix
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$$A+B$$ is a zero matrix
Explanation
Since $$A$$ and $$B$$ are symmetric matrices, $$A' = A$$ and $$B'=B$$
$$\therefore (A+B)'=A'+B'=A+B$$
$$\therefore A+B$$ is symmetric matrix
If $$A$$ is a skew symmetric matrix then $$ \displaystyle A^{T} $$
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$$-A$$
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$$A$$
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$$0$$
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diagonal matrix
Explanation
In mathematics, a skew-symmetric (or anti-symmetric or antimetric) matrix is a square matrix whose transpose is its negation; that is, it satisfies the condition $$-A=A^{T}$$
Hence, $$A^{T}=-A$$.
If $$ A= \begin{bmatrix} 1 & 2\end{bmatrix}, B=\begin{bmatrix} 3 & 4\end{bmatrix}$$ then $$A+B=$$
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$$\begin{bmatrix}1 & 4\end{bmatrix}$$
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$$\begin{bmatrix}4 & 4\end{bmatrix}$$
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$$\begin{bmatrix}4 & 6\end{bmatrix}$$
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None of these
Explanation
Given $$A=\begin{bmatrix} 1 & 2 \end{bmatrix}$$
$$B=\begin{bmatrix} 3 & 4 \end{bmatrix}$$
Now, $$A+B=\begin{bmatrix} 1+3 & 2+4 \end{bmatrix}$$
$$\Rightarrow A+B=\begin{bmatrix} 4 & 6 \end{bmatrix}$$
$$\therefore $$Option C is correct
If $$A$$ is matrix of order $$\displaystyle m\times n$$ and $$B$$ is a matrix of order $$\displaystyle n\times p,$$ then the order of $$AB$$ is
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$$\displaystyle p\times m$$
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$$\displaystyle p\times n$$
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$$\displaystyle n\times p$$
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$$\displaystyle m\times p$$
Explanation
If $$A$$ is matrix of order $$m \times n$$ and $$B$$ is a matrix of order $$ n \times p,$$ then the order of $$AB$$ is $$m \times p.$$
Two matrices $$A$$ and $$B$$ are added if
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both are rectangular
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both have same order
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no of columns of A is equal to columns of B
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no of rows of A is equal to no of columns of B
Explanation
While adding two matrices we add the numbers which belong to some row and column of each matrix or
two matrices can be added if there are equal number of rows and columns in both.
Hence, both matrices should have same order.
$$\therefore$$ option B is correct
Find the output order for the following matrix multiplication $$A_{4 \times 2}\times B_{2\times4}$$?
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$$2 \times 4$$
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$$4 \times 4$$
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$$4 \times2$$
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Multiplication not possible
Explanation
Note:
1. $$m×n$$ is the order of matrix $$AB$$ if the order of matrix $$A$$ is $$m×p$$ and the order of $$B$$ is $$p×n$$.
2.
For matrix multiplication to work, the columns of the second matrix have to have the same number of entries as do the rows of the first matrix.
Here $$A,B$$ are conformant matrices, Multiplication is possible.
Therefore, $$A$$ is of size $$4 \times 2$$, and $$B$$ is of size $$2 \times 4$$, in which case the output is
of size $$4 \times 4$$ and is the matrix product of $$A$$ and $$B$$.
If the matrices $$A=\begin{bmatrix}2 & 1 & 3 \\4 & 1 & 0\end{bmatrix}$$ and $$B=\begin{bmatrix}1 & -1\\ 0 & 2 \\5 & 0\end{bmatrix}$$, then AB will be
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$$\begin{bmatrix}17 & 0 \\4 & -2\end{bmatrix}$$
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$$\begin{bmatrix}4 & 0 \\ 0 & 4\end{bmatrix}$$
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$$\begin{bmatrix}17 & 4 \\0 & -2\end{bmatrix}$$
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$$\begin{bmatrix}0 & 0 \\0 & 0\end{bmatrix}$$
Explanation
Given, $$A=\begin{bmatrix}2 & 1 & 3 \\ 4 & 1 & 0\end{bmatrix}$$ and $$B=\begin{bmatrix}1 & -1 \\ 0 & 2 \\ 5 & 0\end{bmatrix}$$
Now, $$AB=\begin{bmatrix}2\times1+1\times0+3\times5 & 2\times(-1)+1\times2+3\times0 \\ 4\times1+1\times0+0\times5 & 4\times(-1)+1\times2+0\times0\end{bmatrix}$$
$$=\begin{bmatrix}17 & 0 \\ 4 & -2\end{bmatrix}$$
Least number of changes for the expression $$ax^{2} + bxy + cy^{2} + dx + ey + f$$ to be symmetric in x and y is
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$$a = b, c = d$$
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$$b = c, e = f$$
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$$a = c, d = e$$
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$$a = f, b = e, c = d$$
Explanation
$$ax^2+bxy+cy^2+dx+ey+f$$.......(1)
Replace by $$x$$ by $$y$$ and $$y$$ by $$x$$
$$\Rightarrow ay^2+byx+cx^2+dy+ex+f$$.......(2)
if (1) and (2) are equal then
$$\boxed{a=c\,and\,d=e}$$
What is the output for the following matrix multiplication $$A_{3 \times 2}\times B_{2\times 3}$$?
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$$(AB)_{3\times 2}$$
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$$(AB)_{3\times 3}$$
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$$(AB)_{2\times 3}$$
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$$(AB)_{2\times 2}$$
Explanation
Here $$A,B$$ are conformant matrices.
Therefore, $$A$$ is of size $$3 \times 2$$, and $$B$$ is of size $$2 \times 3$$, in which case the output is of size $$3 \times 3$$ and is the matrix product of $$A$$ and $$B$$.
The result will be $$(AB)_{3\times 3}.$$
Find the output order for the following matrix multiplication $$X_{5 \times 3}\times Y_{3\times 5}$$?
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$$5 \times 5$$
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$$3 \times 5$$
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$$5 \times 3$$
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$$3 \times 3$$
Explanation
Since, $$X$$ is of size $$5 \times 3$$, and $$Y$$ is of size $$3 \times 5$$
$$\therefore$$ T
he matrix product of $$X$$ and $$Y$$ has order
$$5 \times 5.$$
Find the value in place of question mark in the following:
$$A_{6 \times 2}\times B_{2\times 6} = C_{?\times6}$$?
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$$2$$
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$$6$$
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$$12$$
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None of these
Explanation
$$A_{6 \times 2}$$ denotes a matrix having $$6$$ rows and $$2$$ columns.
Similarly, $$B_{2 \times 6}$$ denotes a matrix having $$2$$ rows and $$6$$ columns.
When multiplied, we get $$6$$ rows and $$6$$ columns based on the matrix multiplication rule.
Thus, if the product of matrices $$A$$ and $$B$$ is $$C$$, we can write the answer as $$C_{6 \times 6}$$
$$A=\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}$$ and $$AB=BA=I$$, then B is equal to
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$$\begin{bmatrix} -\cos\theta & \sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}$$
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$$\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$$
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$$\begin{bmatrix} -\sin\theta & \cos\theta \\ \cos\theta & \sin\theta\end{bmatrix}$$
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$$\begin{bmatrix} \sin\theta & -\cos\theta \\ -\cos\theta & \sin\theta\end{bmatrix}$$
Explanation
Given, $$A=\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$$ and $$AB=BA=I$$
$$\Rightarrow B=A^{-1}I=A^{-1}$$
$$=\displaystyle\frac{1}{\cos^2\theta +\sin^2\theta}\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$$
$$\Rightarrow B=\begin{bmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$$
If $$A=\begin{bmatrix} 3 & x-1 \\ 2x+3 & x+2 \end{bmatrix}$$ is symmetric matrix then the value of $$x$$ is
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$$4$$
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$$3$$
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$$-4$$
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$$-3$$
Explanation
Since $$A$$ is symmetric matrix, then
$$A=A'$$
$$\Rightarrow$$ $$\begin{bmatrix} 3 & x-1 \\ 2x+3 & x+2 \end{bmatrix}=\begin{bmatrix} 3 & 2x+3 \\ x-1 & x+2 \end{bmatrix}$$
$$\Rightarrow $$ $$x-1=2x+3$$ $$\Rightarrow$$ $$x=-4$$
What is the output order for the following matrix multiplication $$A_{2 \times 1}\times B_{1\times 2}$$?
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$$1 \times 1$$
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$$2 \times 3$$
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$$2 \times 2$$
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$$2 \times 1$$
Explanation
We know that $$m\times n $$ is the order of matrix $$AB$$ if the order of matrix $$A$$ is $$m\times p$$ and the order of $$B$$ is $$p\times n.$$
Since, $$A$$ is of size $$2 \times 1$$, and $$B$$ is of size $$1 \times 2$$, in which case the output is of size $$2 \times 2$$
and is the matrix product of $$A$$ and $$B$$.
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