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CBSE Questions for Class 12 Commerce Maths Matrices Quiz 1 - MCQExams.com
CBSE
Class 12 Commerce Maths
Matrices
Quiz 1
Let $$X$$ and $$Y$$ be two arbitrary, $$3\times 3$$, non-zero, skew-symmetric matrices and $$Z$$ be an arbitrary $$3\times 3$$, non-zero, symmetric matrix. Then which of the following matrices is (are) skew symmetric?
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$$Y^3Z^4-Z^4Y^3$$
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$$X^{44}+Y^{44}$$
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$$X^4Z^3-Z^3X^4$$
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$$X^{23}+Y^{23}$$
Explanation
Given: $$X^T = -X, Y^T = -Y$$ and $$Z^T = Z$$
Using Properties of transpose:
$$(A+B)^T = A^T + B^T$$ and $$(AB)^T = B^TA^T$$
Option A:
$$(Y^3Z^4 - Z^4Y^3)^T = (Y^3Z^4)^T - (Z^4Y^3)^T$$
$$\Rightarrow (Z^4)^T(Y^3)^T - (Y^3)^T(Z^4)^T = (Z^T)^4(Y^T)^3 - (Y^T)^3(Z^T)^4 = Y^3Z^4 - Z^4Y^3$$
Its a symmetric.
Option B:
$$(X^{44} + Y^{44})^T = (X^T)^{44} + (Y^T)^{44} = X^{44} + Y^{44}$$
Its a symmetric.
Option C:
$$(X^4Z^3 - Z^3X^4)^T = (X^4Z^3)^T - (Z^3X^4)^T$$
$$\Rightarrow (Z^3)^T(X^4)^T - (X^4)^T(Z^3)^T = (Z^T)^3(X^T)^4 - (X^T)^4(Z^T)^3 = Z^3X^4 - X^4Z^3 = - (X^4Z^3 - Z^3X^4)$$
Its a skew symmetric.
Option D:
$$(X^{23} + Y^{23})^T = (X^T)^{23} + (Y^T)^{23} = -(X^{23} + Y^{23})$$
Its a skew symmetric.
Hence, option C,D.
If $$\displaystyle A=\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \end{matrix} \right] $$ and $$\displaystyle I=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] $$, then the correct statement is:
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$$\displaystyle { A }^{ 2 }+5A-7I=O$$
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$$\displaystyle -{ A }^{ 2 }+5A+7I=O$$
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$$\displaystyle { A }^{ 2 }-5A+7I=O$$
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$$\displaystyle { A }^{ 2 }+5A+7I=O$$
Explanation
Now, $$\displaystyle { A }^{ 2 }=\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \end{matrix} \right] \left[ \begin{matrix} 3 & 1 \\ -1 & 2 \end{matrix} \right] $$
$$\displaystyle =\left[ \begin{matrix} 8 & 5 \\ -5 & 3 \end{matrix} \right] $$
$$\displaystyle \therefore \quad { A }^{ 2 }-5A+7I=\left[ \begin{matrix} 8 & 5 \\ -5 & 3 \end{matrix} \right] -\left[ \begin{matrix} 15 & 5 \\ -5 & 10 \end{matrix} \right] +\left[ \begin{matrix} 7 & 0 \\ 0 & 7 \end{matrix} \right] $$
$$\displaystyle =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] $$
If AB = AC then
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B = C
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$$B\neq C$$
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B need not be equal to C
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B = -C
Explanation
Value of B and C depends on the value of A. if A is non-singular, i.e non zero, then B=C.
But if A=0, then both terms AB and AC becomes zero, so B and C need not be necessarily equal.
If $$\mathrm{A}^{2}=\mathrm{A},\ \mathrm{B}^{2}=\mathrm{B},\ \mathrm{A}\mathrm{B}=\mathrm{B}\mathrm{A}=O$$ (Null Matrix), then $$(\mathrm{A}+\mathrm{B})^{2}=$$
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$$\mathrm{A}-\mathrm{B}$$
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$$\mathrm{A}+\mathrm{B}$$
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$$\mathrm{A}^{2}-\mathrm{B}^{2}$$
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$$0$$
Explanation
$$(A + B)^{2} = A^{2} + B^{2} + AB + BA$$
$$= A + B$$
$$I$$ $$A=\left[\begin{array}{ll}
0 & 1\\ 1 & 0 \end{array}\right]$$, $$A^{4}=$$
($$I$$ is an identity matrix.)
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$$I$$
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$$0$$
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$$\mathrm{A}$$
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$$4{I}$$
Explanation
Given, $$IA=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$
$$\Rightarrow A=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$
Now, $$A^{2}=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$
$$\Rightarrow A^{2}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
$$\Rightarrow A^{2}=I$$
$$\Rightarrow A^{4}=I$$
lf $$\mathrm{A}= \left[\begin{array}{lll}
o & c & -b\\
-c & o & a\\
b & -a & o
\end{array}\right]\mathrm{a}\mathrm{n}\mathrm{d}$$ $$ \mathrm{B}=\left[\begin{array}{lll}
a^{2} & ab & ac\\
ab & b^{2} & bc\\
ac & bc & c^{2}
\end{array}\right],$$ then $$\mathrm{A}\mathrm{B}=$$
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$$\mathrm{A}$$
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$$\mathrm{B}$$
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$$I$$
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$$\mathrm{O}$$
Explanation
Multiplying the two matrices we get,
$$\begin{bmatrix} 0+abc-abc & 0+c{ b }^{ 2 }-c{ b }^{ 2 } & 0+b{ c }^{ 2 }-b{ c }^{ 2 } \\ -{ a }^{ 2 }c+0+{ a }^{ 2 }c & -abc+0+abc & -a{ c }^{ 2 }+0+a{ c }^{ 2 } \\ { a }^{ 2 }b-{ a }^{ 2 }b+0 & a{ b }^{ 2 }-a{ b }^{ 2 }+0 & abc-abc+0 \end{bmatrix}$$
$$=\begin{bmatrix} 0&0&0 \\ 0&0&0 \\ 0&0&0\end{bmatrix} $$
$$= O$$
Hence the answer is option D
lIf $$\mathrm{A} =\left[\begin{array}{ll}
a & 0\\
a & 0
\end{array}\right],\ \mathrm{B}=\left[\begin{array}{ll}
0 & 0\\
b & b
\end{array}\right],$$ then $$\mathrm{A}\mathrm{B}=$$
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$$O$$
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$$\mathrm{B}\mathrm{A}$$
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$$\mathrm{A}\mathrm{B}$$
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$$\mathrm{A}\mathrm{B}\mathrm{A}\mathrm{B}$$
Explanation
$$\displaystyle AB = \begin{bmatrix} a&0\\a&0 \end{bmatrix} \begin{bmatrix} 0&0\\b&b \end{bmatrix}$$
$$=\begin{bmatrix} a\times 0+0\times b&a\times 0+0\times b\\a\times 0+0\times b&a\times 0+0\times b \end{bmatrix} $$
$$ =\begin{bmatrix} 0&0\\0&0 \end{bmatrix} = O$$
If $$A=\left[\begin{array}{lll}
1 & -2 & 3\\
-4 & 2 & 5
\end{array}\right]$$ and $$B=\left[\begin{array}{ll}
2 & 3\\
4 & 5\\
2 & 1
\end{array}\right],$$ then
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$$\mathrm{A}\mathrm{B},\ \mathrm{B}\mathrm{A}$$ exist and equal
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$$\mathrm{A}\mathrm{B},\ \mathrm{B}\mathrm{A}$$ exist and are not equal
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$$\mathrm{A}\mathrm{B}$$ exists and $$\mathrm{B}\mathrm{A}$$ does not exist
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$$\mathrm{A}\mathrm{B}$$ does not exist and $$\mathrm{B}\mathrm{A}$$ exists
Explanation
Since, we have $$A = 2 \times 3 , \: B = 3 \times 2$$
So, both $$AB$$ and $$BA$$ exist.
But both are not equal to each other because bith are if different order
$$\because AB \rightarrow 2 \times 2$$ and $$BA \rightarrow 3 \times 3.$$
$$\left[\begin{array}{ll}
x & 0\\
0 & y
\end{array}\right]\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]=$$
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$$\left[\begin{array}{ll} ax & b_{X}\\ yc & dy \end{array}\right]$$
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$$\left[\begin{array}{ll} ax & 0\\ 0 & dy \end{array}\right]$$
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$$\left[\begin{array}{ll} ay & cy\\ bx & dy \end{array}\right]$$
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$$\left[\begin{array}{ll} 0& ax\\ dy & 0 \end{array}\right]$$
Explanation
The value of$$ \begin{bmatrix} x&0\\0&y \end{bmatrix} \begin{bmatrix} a&b\\c&d \end{bmatrix}$$
$$=\begin{bmatrix} x\times a+0\times c&x\times b+0\times d\\ 0\times a+y\times c&0\times b+y\times d \end{bmatrix} $$
$$ =\begin{bmatrix} ax&bx\\cy&dy \end{bmatrix} $$
If $$\mathrm{A}=\left[\begin{array}{lll}
1 & -3 & -4\\
-1 & 3 & 4\\
1 & -3 & -4
\end{array}\right]$$, then $$\mathrm{A}^{2}=$$
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$$A$$
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$$- A$$
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Null matrix
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$$2A$$
Explanation
$$A^2=A.A$$
$$=\begin{bmatrix}
1 & -3 &-4 \\
-1 & 3 & 4\\
1 & -3 & -4
\end{bmatrix}\begin{bmatrix}
1 & -3 & -4\\
-1 & 3 & 4\\
1 & -3 & -4
\end{bmatrix}$$
$$=\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}$$
If $$A=[x,y], B=\left[\begin{array}{ll}
a & h\\
h & b
\end{array}\right], C=\left[\begin{array}{l}
x\\
y
\end{array}\right]$$,
then $$\mathrm{A}\mathrm{B}\mathrm{C}=$$
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$$(ax+hy+bxy)$$
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$$(ax^{2}+2hxy+by^{2})$$
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$$(ax^{2}-2hxy+by^{2})$$
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$$(bx^{2}-2hxy+ay^{2})$$
Explanation
Since, $$A = \begin{bmatrix}
x &y
\end{bmatrix}$$
$$B = \begin{bmatrix}
a &h \\
h &b
\end{bmatrix}$$
$$C =\begin{bmatrix}
x\\
y
\end{bmatrix}$$
$$ABC = \begin{bmatrix}
x &y
\end{bmatrix}\begin{bmatrix}
a &h \\
h &b
\end{bmatrix}\begin{bmatrix}
x\\
y
\end{bmatrix}$$
$$=\begin{bmatrix}
xa+yh &xh+yb
\end{bmatrix}\begin{bmatrix}
x\\
y
\end{bmatrix}$$
$$=\begin{bmatrix}
x^{2}a+xyh+xyh+y^{2}b
\end{bmatrix}$$
$$=\begin{bmatrix}
x^{2}a+2xyh+y^{2}b
\end{bmatrix}$$
If $$\mathrm{A}$$ is skew-symmetric matrix and $$\mathrm{n}$$ is even positive integer, then $$\mathrm{A}^{\mathrm{n}}$$ is
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a symmetric matrix
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skew-symmetric matrix
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diagonal matrix
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triangular matrix
Explanation
If A is skew symmetric or symmetric matrix then $$A^{2}$$ is a symmetric matrix.
$$A^{n} = (A^{2})^{n/2}$$
n is even.
$$\therefore A^{n}$$ is symmetric matrix.
$$\because A$$ is a skew symmetric matrix.
If $$A=[1\ \ 2\ \ 3\ \ 4]$$ and $$AB = [3 \ \ 4\ \ -1],$$ then the order of
matrix $$B$$ is
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$$2 \times 3$$
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$$3\times 3$$
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$$4 \times 3$$
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$$1 \times 3$$
Explanation
Given matrix A $$=\left [ 1 \ \ \ 2 \ \ 3\ \ 4 \right ]$$ of order 1 $$\times 4$$
and AB = $$\left [ 3 \ \ 4 \ \ -1 \right ]$$ of order 1 $$\times 3$$
by multiplication of matrix
$$\left ( AB \right )_{m\times n}=\left ( A \right )_{m\times k}\left ( B \right )_{k\times n}$$
i.e Number of columns of A has to be equal to number of row of B
$$\therefore$$ Order of B is $$4 \times 3$$
$$A=\begin{bmatrix}x& -7\\ 7& y\end{bmatrix}$$ is a skew-symmetric matrix,
then (x,y) =
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(1,-1)
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(7,-7)
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(0,0)
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(14,-14)
Explanation
$$A=\begin{bmatrix}x& -7\\ 7& y\end{bmatrix}$$
Now $$A^T=\begin{bmatrix}x& 7\\ -7& y\end{bmatrix}$$
Now for $$A$$ to be skew symmetric matrix, $$A+A^T=O$$
$$\Rightarrow \begin{bmatrix}x& -7\\ 7& y\end{bmatrix}+\begin{bmatrix}x& 7\\ -7& y\end{bmatrix}=\begin{bmatrix}0& 0\\ 0& 0\end{bmatrix}$$
$$\Rightarrow \begin{bmatrix}2x& 0\\ 0& 2y\end{bmatrix}=\begin{bmatrix}0& 0\\ 0& 0\end{bmatrix}$$
Now equating the corresponding elements we get, $$(x,y)=(0,0)$$
$$A=\left[\begin{array}{lll} 0 & 1 & -2\\1 & 0 & 3\\2 &-3 & 0 \end{array}\right]$$ then $$\mathrm{A}+\mathrm{A}^{\mathrm{T}}=$$
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$$\left[\begin{array}{lll}
0 & 2 & 0\\
2 & 0 & 0\\
0 & 0 & 0
\end{array}\right]$$
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$$\left[\begin{array}{lll}
1 & 0 & 0\\
0 & 3 & 0\\
0 & 0 & 4
\end{array}\right]$$
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$$\left[\begin{array}{lll}
2 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 2
\end{array}\right]$$
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$$\left[\begin{array}{lll}
2 & 0 & 2\\
0 & 2 & 0\\
0 & 0 & 2
\end{array}\right]$$
Explanation
Given, $$A=\begin{bmatrix}
0 & 1 & -2\\
1 & 0& 3\\
2 & -3 & 0
\end{bmatrix}$$
So, By property of transpose (1),
$$A^{T}=\begin{bmatrix}
0 & 1 & 2\\
1 & 0 & -3\\
-2 & 3 & 0
\end{bmatrix}$$
So, By operation of matrixes (2),
$$A+A^{T}=\begin{bmatrix}
0 & 2 & 0\\
2 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}$$
The order of $$[\mathrm{x} \space \mathrm{y} \space\mathrm{z}] \left[\begin{array}{lll}
\mathrm{a} & \mathrm{h} & \mathrm{g}\\
\mathrm{h} & \mathrm{b} & \mathrm{f}\\
\mathrm{g} & \mathrm{f} & \mathrm{c}
\end{array}\right]\left[\begin{array}{l}
\mathrm{x}\\
\mathrm{y}\\
\mathrm{z}
\end{array}\right]$$ is
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$$3\mathrm{x}1$$
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$$1\mathrm{x}1$$
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$$1\mathrm{x}3$$
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$$3\mathrm{x}3$$
Explanation
$$[x y z]_{1 \times 3}\begin{Bmatrix}
a & h & g\\
h & b & f\\
g & f & c
\end{Bmatrix}_{3 \times 3}\begin{Bmatrix}
x\\
y\\
z
\end{Bmatrix}_{3 \times 1}$$
$$\Rightarrow [p \space q \space r]_{1 \times 3}\begin{Bmatrix}
x\\
y\\
z
\end{Bmatrix}_{3 \times 1}$$
$$\Rightarrow [v]_{1 \times 1}$$
If $$A$$ and $$B$$ are two matrices such that $$A + B$$ and $$AB$$ are both defined, then
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$$A$$ and $$B$$ are two matrices not necessarily of same order
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$$A$$ and $$B $$ are square matrices of same order
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$$A$$ and $$B $$ are matrices of same type
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$$A$$ and $$B$$ are rectangular matrices of same order
Explanation
Addition of matrix is define when both matrixes are of same order.
If A and B are two matrices such that $$A+B$$ is define, then $$A $$ and $$B$$ are matrices of same order.
So, If order of $$A=a \times b $$
then order of $$B=a \times b $$
and $$AB$$ is defined, [by operation of matrixes] then the number of columns of first matrix is equal to the number of rows of second matrix.
$$\Rightarrow$$ $$b=a.$$
So, $$A$$ and $$B$$ both are square matrix of same order.
If the transpose of a matrix is equal to the additive
inverse, then matrix is called _________
matrix.
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symmetric
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skew symmetric
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identity
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inverse
If $$I = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix},$$ then find $$I^3$$
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$$1$$
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$$I$$
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$$0$$
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does not exist
Explanation
$$I^n$$ ( for any natural $$n$$ ) $$= I.$$
$$I$$ is known as the identity matrix.
If $$A= \begin{bmatrix}
1 & 2 & 3\\
4 & 5 & 6
\end{bmatrix}$$ and $$B= \begin{bmatrix}
1\\
0\\
5\end{bmatrix},$$ then $$AB = $$
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$$\begin{bmatrix}
1 & 0 & 15
\end{bmatrix}$$
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$$\begin{bmatrix}
4 & 0 & 30
\end{bmatrix}$$
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$$\begin{bmatrix}
16\\
34\end{bmatrix}$$
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$$\begin{bmatrix}
16 & 34
\end{bmatrix}$$
Explanation
$$AB= \begin{bmatrix}
1 &2 &3 \\
4 &5 &6
\end{bmatrix}\begin{bmatrix}
1\\
0\\
5\end{bmatrix}=\begin{bmatrix}
1+ 0+ 15\\
4+ 0+30
\end{bmatrix}=\begin{bmatrix}
16\\
34
\end{bmatrix}$$
Thus, C will be correct answer.
If $$A = \begin{bmatrix}a & b\end{bmatrix},\space B = \begin{bmatrix}-b & -a \end{bmatrix}$$ and $$C = \begin{bmatrix}a \\ -a\end{bmatrix}$$, then the correct statement is
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$$A = -B$$
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$$A+B = A-B$$
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$$AC = BC$$
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$$CA = CB$$
Explanation
Given $$A = \begin{bmatrix}a & b\end{bmatrix}, B = \begin{bmatrix}-b & -a \end{bmatrix}$$ and $$C = \begin{bmatrix}a \\ -a\end{bmatrix}$$
We will check by options.
Clearly , $$A \ne -B$$ as their corresponding elements are different only.
$$A+B= \begin{bmatrix}a-b & b-a\end{bmatrix}$$
$$A-B=\begin{bmatrix}a+b & b+a\end{bmatrix}$$
So, $$A+B \ne A-B$$
Now,$$AC=\begin{bmatrix}a & b\end{bmatrix}\begin{bmatrix}a \\ -a\end{bmatrix}$$
$$\Rightarrow AC=\begin{bmatrix}a^{2}-ab\end{bmatrix}$$
$$BC=\begin{bmatrix}-b & -a \end{bmatrix}\begin{bmatrix}a \\ -a\end{bmatrix}$$
$$\Rightarrow BC=\begin{bmatrix}a^{2}-ab\end{bmatrix}$$
Hence, $$AC=BC$$
Option $$C$$ is correct
Now, $$CA=\begin{bmatrix}a \\ -a\end{bmatrix}\begin{bmatrix}a & b\end{bmatrix}$$
$$\Rightarrow CA=\begin{bmatrix}a^{2}&ab\\-a^{2}&-ab \end{bmatrix}$$
$$CB=\begin{bmatrix}a \\ -a\end{bmatrix}\begin{bmatrix}-b & -a \end{bmatrix}$$
$$\Rightarrow CB=\begin{bmatrix}-ab&-a^{2}\\ab & a^{2}\end{bmatrix}$$
Hence, $$CA\ne CB$$
If $$\displaystyle A = \begin{bmatrix} 1 & -2 & 4 \\ 2 & 3 & 2 \\ 3 & 1 & 5 \end{bmatrix}$$ and $$\displaystyle B = \begin{bmatrix} 0 & -2 & 4 \\ 1 & 3 & 2 \\ -1 & 1 & 5 \end{bmatrix}$$, then $$A + B$$ is
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$$\displaystyle \begin{bmatrix}
1 & -2 & 4 \\
3 & 3 & 2 \\
2 & 1 & 5
\end{bmatrix}$$
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$$\displaystyle \begin{bmatrix}
1 & -2 & 8 \\
3 & 3 & 4 \\
2 & 1 & 10
\end{bmatrix}$$
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$$\displaystyle \begin{bmatrix}
1 & -4 & 8 \\
3 & 6 & 4 \\
2 & 2 & 10
\end{bmatrix}$$
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none of these
Explanation
Given, $$\displaystyle A = \begin{bmatrix} 1 & -2 & 4 \\ 2 & 3 & 2 \\ 3 & 1 & 5 \end{bmatrix}$$ and $$\displaystyle B = \begin{bmatrix} 0 & -2 & 4 \\ 1 & 3 & 2 \\ -1 & 1 & 5 \end{bmatrix}$$
Therefore, $$A+B=\begin {bmatrix} 1+0 & -2-2 & 4+4 \\2+1 & 3+3 & 2+2 \\3-1 & 1+1 & 5+5\end {bmatrix}$$
$$=\begin{bmatrix} 1 & -4 & 8 \\ 3 & 6 & 4 \\ 2 & 2 & 10 \end{bmatrix}$$
If $$A=\begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} $$, then $$ A^{3}-A^{2}=$$
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$$2A$$
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$$2I$$
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$$A$$
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$$I$$
Explanation
$$A^{2}=A\times A=\begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}$$
$$A^{3}=A^{2}\times A=\begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}\begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}=\begin{bmatrix} -1 & 0 \\ 0 & 8 \end{bmatrix}$$
$$A^{3}-A^{2}=\begin{bmatrix} -1 & 0 \\ 0 & 8 \end{bmatrix}-\begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}=\begin{bmatrix} -2 & 0 \\ 0 & 4 \end{bmatrix}$$
$$=2\begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}=2A$$
If $$\begin{bmatrix}3 & -1 \\ 2 & 5\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}4 \\ -3\end{bmatrix},$$ find $$x$$ and $$y$$
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$$x=3,\space y = -1$$
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$$x=2,\space y = 5$$
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$$x=1,\space y = -1$$
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$$x=-1,\space y = 1$$
Explanation
$$\begin{bmatrix}3 & -1 \\ 2 & 5\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}4 \\ -3\end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} 3x-y \\ 2x+5y \end{bmatrix}=\begin{bmatrix} 4 \\ -3 \end{bmatrix}$$
$$\Rightarrow 3x-y=4, 2x+5y=-3$$
On solving we get $$x=1, y=-1$$
If $$\displaystyle \:A= \left [ \begin{matrix}4 &x+2 \\2x-3 &x+1 \end{matrix} \right ]$$ is symmetric, then x=
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3
0%
5
0%
2
0%
4
Explanation
Given $$\displaystyle \:A= \left [ \begin{matrix}4 &x+2 \\2x-3 &x+1 \end{matrix} \right ]$$ is symmteric.
$$A^{T}=A$$
$$\Rightarrow \left[ \begin{matrix} 4 & 2x-3 \\ x+2 & x+1 \end{matrix} \right] =\left [ \begin{matrix}4 &x+2 \\2x-3 &x+1 \end{matrix} \right ]$$
$$\Rightarrow x+2=2x-3$$
$$\Rightarrow x=5$$
If $$\begin{bmatrix} 1 & 2 & 3 \end{bmatrix} B=\begin{bmatrix} 3 & 4 \end{bmatrix}$$, then the order of the matrix $$B$$ is
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$$3\times 1$$
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$$1\times 3$$
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$$2\times 3$$
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$$3\times 2$$
Explanation
Let order of the matrix $$B$$ be $$m\times n$$
$$[\array{1 & 2&3]}_{1\times 3}[B]_{m\times n}=\array{[3&4]}_{1\times 2}$$
For the above multiplication to be valid,
$$m$$ must be equal to $$3$$ i.e., $$m=3$$
Order of the resultant matrix is gievn by $$ 1\times n$$.
Comparing this with the order of the given matrix we get, $$n=2$$
Therefore, order of the matrix $$B $$ is $$3 \times 2$$
If $$A=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$, then $$A^{4}=$$
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$$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
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$$\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$$
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$$\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}$$
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$$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$
Explanation
Given, $$A=\begin{bmatrix} 0&1 \\1&0\end{bmatrix}$$
$$A^{2}=A\times A=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
$$A^{4}=A^{2}\times A^{2}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
Given that $$\displaystyle M=\begin{bmatrix}3 &-2 \\-4 &0 \end{bmatrix}\:and\:N=\begin{bmatrix}-2 &2 \\5 &0 \end{bmatrix}$$
, then $$M+N$$ is a
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null matrix
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unit matrix
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$$\displaystyle \begin{bmatrix}
1 & 0 \\1
&0
\end{bmatrix}$$
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$$\displaystyle \begin{bmatrix}
0 &1 \\1
&1
\end{bmatrix}$$
Explanation
Given:
$$\displaystyle M=\begin{bmatrix}3 &-2 \\-4 &0 \end{bmatrix}\:and\:N=\begin{bmatrix}-2 &2 \\5 &0 \end{bmatrix}$$
Adding $$M$$ and $$N$$, we get
$$M+N=\begin{bmatrix} 3 & -2 \\ -4 & 0 \end{bmatrix}+\begin{bmatrix} -2 & 2 \\ 5 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$$
If $$A=\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$$ and $$B=\begin{bmatrix} 1 & 3 & 2 \\ 2 & 3 & 4 \end{bmatrix}$$, then $$AB$$ equal to
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$$\begin{bmatrix} 8 & 15 & 16 \\ 5 & 9 & 10 \end{bmatrix}$$
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$$\begin{bmatrix} 8 & 5 \\ 15 & 9 \\ 16 & 10 \end{bmatrix}$$
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$$\begin{bmatrix} 8 & 5 \\ 15 & 9 \end{bmatrix}$$
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None of these
Explanation
$$AB=\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 1 & 3 & 2 \\ 2 & 3 & 4 \end{bmatrix}=\begin{bmatrix} 2.1+2.2 & 2.3+3.3 & 2.2+3.4 \\ 1.1+2.2 & 1.3+2.3 & 1.2+2.4 \end{bmatrix}=\begin{bmatrix} 8 & 15 & 16 \\ 5 & 9 & 10 \end{bmatrix}$$
Ans: A
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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$$A$$
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$$-A$$
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null matrix
0%
$$I$$
Explanation
Given, $$A = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1\end{bmatrix}$$
$$A^{2}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix}$$
$$=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
$$\Rightarrow A^{2}=I$$
0:0:1
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