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CBSE Questions for Class 12 Commerce Maths Matrices Quiz 4 - MCQExams.com
CBSE
Class 12 Commerce Maths
Matrices
Quiz 4
If
A
=
[
0
1
1
0
]
, then
A
5
=
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0%
I
0%
O
0%
A
0%
A
2
Explanation
Given, A
=
[
0
1
1
0
]
A
2
=
[
1
0
0
1
]
A
5
=
A
2
×
A
2
×
A
A
5
=
[
0
1
1
0
]
=
A
If
A
=
[
1
2
0
3
]
and
B
=
[
3
−
1
]
, then
B
A
=
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[
3
0
0
3
]
0%
[
3
0
]
0%
[
3
3
]
0%
[
0
−
3
]
Explanation
The value of
B
A
=
[
3
−
1
]
[
1
2
0
3
]
=
[
3
×
1
−
1
×
0
3
×
2
−
1
×
3
]
=
[
3
3
]
I
f
A
=
[
a
h
g
h
b
f
g
f
c
]
, then A is
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a nilpotent matrix
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an involutory matrix
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a symmetric matrix
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an idempotent matrix
Explanation
Here The transpose of the matrix A is
=
[
a
h
g
h
b
f
g
f
c
]
∴
And a symmetric matrix is a square matrix that is equal to its transpose.
Hence the answer is option C
If A =
\begin{bmatrix} x & 1\\ 1 & 0 \end{bmatrix}
and
A^{2}
is identity matrix, then
x=
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1
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-1
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\pm 1
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0
Explanation
Since,
A=\begin{bmatrix} x &1 \\ 1 &0 \end{bmatrix}, A^2 =I
A^{2}=\begin{bmatrix} x &1 \\ 1 &0 \end{bmatrix}\begin{bmatrix} x &1 \\ 1 &0 \end{bmatrix}
\Rightarrow \begin{bmatrix} x^{2}+1 &x \\ x &1 \end{bmatrix}= \begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}
\Rightarrow
So
x=0
Option D is the correct answer.
L=\left[\begin{array}{lll} 2 & 3 & 5\\ 4 & 1 & 2\\ 1 & 2 & 1 \end{array}\right] =P+Q
,
P
is a symmetric matrix,
{Q}
is a skew-symmetric matrix then
{P}
is equal to
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\left[\begin{array}{lll} 3 & 5 & 6\\ 5 & 6 & 4\\ 9 & 4 & 3 \end{array}\right]
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\left[\begin{array}{lll} 2 & 3.5 & 3\\ 3.5 & 1 & 2\\ 3 & 2 & 1 \end{array}\right]
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\left[\begin{array}{lll} 6 & 5 & 4\\ 3 & 6 & 3\\ 5 & 2 & 5 \end{array}\right]
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\left[\begin{array}{lll} 6 & 5 & 4\\ 4 & 5 & 3\\ 3 & 4 & 3 \end{array}\right]
Explanation
Given ,
P+Q=\begin{bmatrix} 2 & 3 & 5\\ 4 & 1 & 2\\ 1 & 2 & 1 \end{bmatrix}
-----(i)
ans
P
is symmetric and
Q
is skew symmetric matrix. So, by property of transpose (7),
P=P^{T}
and
Q^{T}=-Q
\therefore
equation (i),
P^{T}-Q=\begin{bmatrix} 2 & 4 & 1\\ 3 & 1 & 2\\ 5 & 2 & 1 \end{bmatrix}
---------(ii)
Add (i) and (ii), we get
P+P^{T}=\begin{bmatrix} 4 & 7 & 6\\ 7 & 2 & 4\\ 6 & 4 & 2 \end{bmatrix}, 2P=\begin{bmatrix} 4 & 7 & 6\\ 7 & 2 & 4\\ 6 & 4 & 2 \end{bmatrix}
\Rightarrow P=\begin{bmatrix} 2 & 3.5 & 3\\ 3.5 & 1 & 2\\ 3 & 2 & 1 \end{bmatrix}
lf
\mathrm{A}=\left[\begin{array}{ll} 2 & -1\\ 3 & -2 \end{array}\right],
then
\mathrm{A}^{5}=
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I
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A
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-A
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A^{2}
Explanation
A^2=\begin{bmatrix} 2 & 3\\ 3 & -2 \end{bmatrix}\begin{bmatrix} 2 & 3\\ 3 & -2 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}
A^3=AI=A
A^5=A^3(A^2)=A^3I=A(A^2I)=A
If
I=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}
and E =
\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}
, then
\left ( 2I+3E \right )^{3}=
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8I+ 18E
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4I+ 36E
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8I +36E
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2I+ 3E
Explanation
Consider,
2I+3E
2\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}+3\begin{bmatrix} 0 &1 \\ 0 &0 \end{bmatrix}
\Rightarrow \begin{bmatrix} 2 &0 \\ 0 &2 \end{bmatrix}+\begin{bmatrix} 0 &3 \\ 0 &0 \end{bmatrix}
\Rightarrow \begin{bmatrix} 2 &3 \\ 0 &2 \end{bmatrix}
Now
(2I+3E)^3
\Rightarrow \begin{bmatrix} 2 &3 \\ 0 &2 \end{bmatrix}\begin{bmatrix} 2 &3 \\ 0 &2 \end{bmatrix}\begin{bmatrix} 2 &3 \\ 0 &2 \end{bmatrix}
\Rightarrow \begin{bmatrix} 4 &6+6 \\ 0 &4 \end{bmatrix}\begin{bmatrix} 2 &3 \\ 0 &2 \end{bmatrix}
=\begin{bmatrix} 8 &12+24 \\ 0 &8 \end{bmatrix}=\begin{bmatrix} 8 &36 \\ 0 &8 \end{bmatrix}=8I+36E
\therefore(C)
Let
A
and
B
be
3\times3
matrices such that
A^{T}=-A, \, B^{T}=B
, then matrix
(\lambda AB+3BA)
is a skew symmertric matrix for
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\lambda =3
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\lambda =-3
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\lambda =3
or
\lambda =-3
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\lambda =3
and
\lambda =-3
Explanation
If the matrix
\lambda AB +3BA
to be skew-symmetry,
Then, it must satisfy
\left ( \lambda AB+3BA\right )^{T}=-\left ( \lambda AB+3BA \right )
Now, considering
\left ( \lambda AB+3BA\right )^{T}
=\lambda \left ( AB \right )^{T}+3\left ( BA \right )^{T}
=\lambda\ B^{T} A^{T}+3\ A^{T}B^{T}
Substituting values of
A^{T}=-A
and
B^{T}=B
, we get
=-\lambda BA-3AB = -(3AB+\lambda BA)
From this, we get
\lambda = 3
clearly.
Hence, option A.
If in a square matrix
A=\left[ { a }_{ ij } \right]
, we find that
{ a }_{ ij }={ a }_{ ji }\quad \forall \quad i,j
, then
A
is
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Symmetric
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Skew Symmetric
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Idempotent
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none of these
Explanation
Given square matrix
A=\left\{ { a }_{ ij } \right\}
and
{ a }_{ ij }={ a }_{ ji }
We know that in a square matrix if
{ a }_{ ij }={ a }_{ ji }
for all i and j, then the matrix is a symmetric matrix
So A is a symmetric matrix
\mathrm{A}
: If
\mathrm{A}=\left\{\begin{array}{ll} 1 & -1\\ -1 & 1 \end{array}\right\}
and
\mathrm{B}=\left\{\begin{array}{ll} 2 & 2\\ 2 & 2 \end{array}\right\},
then
\mathrm{A}\mathrm{B}=0
\mathrm{R}
: If
\mathrm{A}\mathrm{B}=0\Rightarrow \mathrm{A}
or
\mathrm{B}
need not be null matrices.
The correct answer is
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Both
A
and
R
are true,
R
is correct explanation to
A
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Both
A
and
R
are true but
R
is not correct explanation to
A
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A
is true,
R
is false
0%
A
is false,
R
is true
Explanation
If
\mathrm{A}=\left\{\begin{array}{ll} 1 & -1\\ -1 & 1 \end{array}\right\} ;\mathrm{B}=\left\{\begin{array}{ll} 2 & 2\\ 2 & 2 \end{array}\right\}
Then,
\mathrm{AB}=\left\{\begin{array}{ll} 1\times 2-1\times 2 & 1\times 2-1\times 2\\ -1\times 2+1\times 2 & -1\times 2+1\times 2 \end{array}\right\} =O
Hence both statement are correct and Reason is correct explanation of Assertion.
If P =
\begin{bmatrix} 1\\ 3\\ 4\end{bmatrix}
, Q =
\begin{bmatrix} 2 & -1&5 \end{bmatrix}
then PQ =
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\begin{bmatrix}2 & -1 & 5\\ 6& -3& 15\\ 8& -4 & 20\end{bmatrix}
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\begin{bmatrix}2 & -3 & 20\end{bmatrix}
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\begin{bmatrix}2\\ -3\\ 20\end{bmatrix}
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[19]
Explanation
\text{PQ}=\begin{bmatrix} 1\\3\\4\end{bmatrix}\begin{bmatrix} 2&-1&5\end{bmatrix}
\quad = \begin{bmatrix} 1\times 2&1\times(-1)&1\times 5\\3\times2&3\times(-1)&3\times 5 \\4\times 2&4\times(-1)&4\times 5\end{bmatrix}
\quad = \begin{bmatrix} 2 & -1 & 5\\ 6& -3& 15\\ 8& -4 & 20 \end{bmatrix}
If
\mathrm{A}=\left(\begin{array}{lll} x & 1 & 4\\ -1 & 0 & 7\\ -4 & -7 & 0 \end{array}\right)
such that
\mathrm{A}^{\mathrm{T}}=-\mathrm{A}
, then
\mathrm{x}=
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-1
0%
0
0%
1
0%
4
Explanation
Given
A=\begin{pmatrix} x & 1 & 4\\ -1 & 0 & 7\\ -4 & -7 & 0 \end{pmatrix}
and
A^{T}=-A
So By property of traspose (I), and opertaion of matrixes (4),
A^{T}=\begin{pmatrix} x & -1 & -4\\ 1 & 0 & -7\\ 4 & 7 & 0 \end{pmatrix}=-A=\begin{pmatrix} -x & -1 & -4\\ 1 & 0 & -7\\ 4 & 7 & 0 \end{pmatrix}
So, By equality of matrixes,
x=-x
\Rightarrow x=0
\mathrm{If}\mathrm{A}=\left[\begin{array}{lll} 2 & x-3 & x-2\\ 3 & -2 & -1\\ 4 & -1 & -5 \end{array}\right]
is a symmetric matrix then
\mathrm{x}
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0
0%
3
0%
6
0%
8
Explanation
Given
A=\begin{bmatrix} 2 & x-3 & x-2\\ 3 & -2 & -1\\ 4 & -1 & -5 \end{bmatrix}
is a symmetric matrix.
So, by property of symmetric matrices
A=A^{T}
\Rightarrow \begin{bmatrix} 2 & x-3 & x-2\\ 3 & -2 & -1\\ 4 & -1 & -5 \end{bmatrix}=\begin{bmatrix} 2 & 3 & 4\\ x-3 & -2 & -1\\ x-2 & -1 & -5 \end{bmatrix}
So, By operation of matrices - equality,
\Rightarrow x-3=3
\Rightarrow x=6
\begin{bmatrix} 10 & 20 & 30 \\ 20 & 45 & 80 \\ 30 & 80 & 171 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 4 & 1 \end{bmatrix}\begin{bmatrix} x & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}
then
{x}=
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10
0%
20
0%
30
0%
40
Explanation
Given,
\begin{bmatrix} 10 & 20 & 30 \\ 20 & 45 & 80 \\ 30 & 80 & 171 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 4 & 1 \end{bmatrix}\begin{bmatrix} x & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}
\Rightarrow \begin{bmatrix} 10 & 20 & 30 \\ 20 & 45 & 80 \\ 30 & 80 & 171 \end{bmatrix}=\begin{bmatrix} x & 0 & 0 \\ 2x & 5 & 0 \\ 3x & 20 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}
\Rightarrow \begin{bmatrix} 10 & 20 & 30 \\ 20 & 45 & 80 \\ 30 & 80 & 171 \end{bmatrix}=\begin{bmatrix} x & 2x & 3x \\ 2x & 4x+5 & 6x+20 \\ 3x & 6x+20 & 9x+81 \end{bmatrix}
On comparing, we get
x=10
A=\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{bmatrix}
then
A^{2}-A=
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\begin{bmatrix} 3 & 0 & 0\\ 0 & 1 & -1\\ 0 & 5 & 4 \end{bmatrix}
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\begin{bmatrix} 1 & 2 & -1\\ -3 & 1 & 1\\ 3 & 5 & -4 \end{bmatrix}
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\begin{bmatrix} 3 & 1 & -1\\ -3 & 1 & -1\\ 3 & 5 & 4 \end{bmatrix}
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\begin{bmatrix} 3 & -1 & 1\\ 3 & -1 & 1\\ 3 & 5 & 4 \end{bmatrix}
Explanation
A=\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{bmatrix}
{ A }^{ 2 }=\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{bmatrix}=\begin{bmatrix} 4 & 3 & 0 \\ -3 & 2 & -2 \\ 6 & 4 & 5 \end{bmatrix}
Now,
{ A }^{ 2 }-A=\begin{bmatrix} 4 & 3 & 0 \\ -3 & 2 & -2 \\ 6 & 4 & 5 \end{bmatrix}-\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{bmatrix}
=\begin{bmatrix} 3 & 1 & -1 \\ -3 & 1 & -1 \\ 3 & 5 & 4 \end{bmatrix}
1\mathrm{f}\mathrm{A}=\left[\begin{array}{lll} 4 & 1 & 0\\ 1 & -2 & 2 \end{array}\right],\ \mathrm{B}=\left[\begin{array}{lll} 2 & 0 & -1\\ 3 & 1 & 4 \end{array}\right]
,
\mathrm{C}=\left[\begin{array}{l} 1\\ 2\\ -1 \end{array}\right]
and
(3\mathrm{B}-2\mathrm{A})\mathrm{C}+2\mathrm{X}=0
then
\mathrm{X}
is equal to
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{\displaystyle \frac{1}{2}} \begin{bmatrix} 3\\ 13 \end{bmatrix}
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{\displaystyle \frac{1}{2}}\begin{bmatrix} 3\\ -13 \end{bmatrix}
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{\displaystyle \frac{1}{2}} \begin{bmatrix} -3\\ 13 \end{bmatrix}
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\begin{bmatrix} 3\\ -13 \end{bmatrix}
Explanation
Given,
A =\begin{bmatrix} 4 & 1 & 0 \\ 1 & -2 & 2 \end{bmatrix},B=\begin{bmatrix} 2 & 0 & -1 \\ 3 & 1 & 4 \end{bmatrix},{ C }=\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}
Now,
(3{ B }-2{ A }){ C }+2{ X}=0
\Rightarrow \left( \begin{bmatrix} 6 & 0 & -3 \\ 9 & 3 & 12 \end{bmatrix}-\begin{bmatrix} 8 & 2 & 0 \\ 2 & -4 & 4 \end{bmatrix} \right) \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}+2X=0
\begin{bmatrix} -2 & -2 & -3 \\ 7 & 7 & 8 \end{bmatrix}\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}+2X=0
\begin{bmatrix} -3 \\ 13 \end{bmatrix}+2X=0
\Rightarrow 2X=\begin{bmatrix} 3 \\ -13 \end{bmatrix}
\Rightarrow X=\displaystyle \frac{1}{2}\begin{bmatrix} 3 \\ -13 \end{bmatrix}
A=\begin{bmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{bmatrix}
and
\mathrm{A}^{2}=\lambda I
then
\lambda=
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0
0%
1
0%
\dfrac{1}{2}
0%
-2
Explanation
A=\begin{bmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{bmatrix}
{ A }^{ 2 }=\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}
But given
{ A }^{ 2 }=\lambda I
\Rightarrow \lambda=0
A :
\begin{vmatrix} 0 &p-e & e-r\\ e-p& 0 &r-p \\ r-e& p-r & 0 \end{vmatrix}
=0
R : The determinant of a skew symmetric matrix is zero
The correct answer is
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Both A and R are true R is correct explanation to A
0%
Both A and R are true but R is not correct explanation to A
0%
A is true R is false
0%
A is false R is true
Explanation
Let
A=\begin{bmatrix}0 &p-e & e-r\\ e-p& 0 &r-p \\ r-e& p-r & 0\end{bmatrix}
Here,
A^{T}=\begin{bmatrix} 0 & e-p & r-e \\ p-e & 0 & p-r \\ e-r & r-p & 0 \end{bmatrix}
\Rightarrow A^{T}=-A
Hence, A is skew-symmetric matrix of order 3.
Determinant of a skew symmetric matrix of odd order is 0.
A=\left[\begin{array}{lll} 2 & 2 & 1\\ 1 & 2 & 1\\ 3 & 4 & 2 \end{array}\right]
then (
\mathrm{A}-\mathrm{I}
)
(\mathrm{A}-2I)=
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\left[\begin{array}{lll} 5 & 6 & -2\\ 1 & 0 & 7\\ 0 & 1 & 1 \end{array}\right]
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\left[\begin{array}{lll} 5 & 6 & 3\\ 4 & 6 & 2\\ 7 & 10 & 7 \end{array}\right]
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\left[\begin{array}{lll} 1 & 0 & 7\\ 7 & 10 & 7\\ 4 & 6 & 2 \end{array}\right]
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\left[\begin{array}{lll} -1 & 1 & 2\\ 3 & 4 & 1\\ 1 & 1 & 2 \end{array}\right]
Explanation
A-I=\left[\begin{array}{} 1 & 2 & 1\\ 1 & 1 & 1\\ 3 & 4 & 1 \end{array}\right]
A-2I=\left[\begin{array}{} 0 & 2 & 1\\ 1 & 0 & 1\\ 3 & 4 & 0 \end{array}\right]
\therefore (A-I)(A-2I)=\left[\begin{array}{} 1 & 2 & 1\\ 1 & 1 & 1\\ 3 & 4 & 1 \end{array}\right]\left[\begin{array}{} 0 & 2 & 1\\ 1 & 0 & 1\\ 3 & 4 & 0 \end{array}\right]=\left[\begin{array}{} 5 & 6 & 3\\ 4 & 6 & 2\\ 7 & 10 & 7 \end{array}\right]
lf
A=\begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{bmatrix}
and
B=\begin{bmatrix} 10 & -4 & -1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix}
then
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\mathrm{A}\mathrm{B}=\mathrm{B}\mathrm{A}
0%
\mathrm{A}\mathrm{B}=-\mathrm{A}\mathrm{B}
0%
\mathrm{A}\mathrm{B}=2\mathrm{B}\mathrm{A}
0%
\mathrm{A}\mathrm{B}=3\mathrm{B}\mathrm{A}
Explanation
Given,
A=\begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{bmatrix}, B=\begin{bmatrix} 10 & -4 & -1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix}
AB=\begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{bmatrix}\begin{bmatrix} 10 & -4 & -1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix}
=\begin{bmatrix} -3 & 1 & 0 \\ 4 & -2 & -1 \\ -5 & 1 & 1 \end{bmatrix}
BA=\begin{bmatrix} 10 & -4 & -1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{bmatrix}
=\begin{bmatrix} -3 & 1 & 0 \\ 4 & -2 & -1 \\ -5 & 1 & 1 \end{bmatrix}
Hence,
AB=BA
A=\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}
then
A^{3}-35A=
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\mathrm{A}
0%
2\mathrm{A}
0%
3\mathrm{A}
0%
4\mathrm{A}
Explanation
A=\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}
{ A }^{ 2 }=\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}=\begin{bmatrix} 12 & 12 & 12 \\ 12 & 12 & 12 \\ 12 & 12 & 12 \end{bmatrix}
{ A }^{ 3 }=\begin{bmatrix} 12 & 12 & 12 \\ 12 & 12 & 12 \\ 12 & 12 & 12 \end{bmatrix}\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}=\begin{bmatrix} 72 & 72 & 72 \\ 72 & 72 & 72 \\ 72 & 72 & 72 \end{bmatrix}
{ A }^{ 3 }-35A=\begin{bmatrix} 72 & 72 & 72 \\ 72 & 72 & 72 \\ 72 & 72 & 72 \end{bmatrix}-35\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}
=\begin{bmatrix} 72 & 72 & 72 \\ 72 & 72 & 72 \\ 72 & 72 & 72 \end{bmatrix}-\begin{bmatrix} 70 & 70 & 70 \\ 70 & 70 & 70 \\ 70 & 70 & 70 \end{bmatrix}
=\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}
A=\left[\begin{array}{ll} 2 & 1\\ 3 & 0 \end{array}\right]
then
\mathrm{A}^{2}+2\mathrm{A}+I=
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\left[\begin{array}{ll} 12 & 4\\ 12 & 4 \end{array}\right]
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\left[\begin{array}{ll} 12 & -4\\ 4 & 12 \end{array}\right]
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\left[\begin{array}{ll} 4 & 12\\ 12 & 4 \end{array}\right]
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\left[\begin{array}{ll} 4 & 12\\ -12 & -4 \end{array}\right]
Explanation
A^2+2A+I=(A+I)^2
Now,
A+I=\begin{bmatrix} 3 & 1\\ 3 & 1 \end{bmatrix}
(A+I)^2=A^2+2A+I=\begin{bmatrix} 3 & 1\\ 3 & 1 \end{bmatrix}\begin{bmatrix} 3 & 1\\ 3 & 1 \end{bmatrix}=\begin{bmatrix} 12 & 4\\ 12 & 4 \end{bmatrix}
Let
\left[\begin{array}{ll} 2 & -2\\-2 & \ 5 \end{array}\right]=\left[\begin{array}{ll} 1 & 0\\ -1 & 1 \end{array}\right]\left[\begin{array}{ll} 2 & 0\\ 0 & x \end{array}\right]\left[\begin{array}{ll} 1 & -1\\ 0 & 1 \end{array}\right]
, then the value of
x
is
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-3
0%
-2
0%
0
0%
3
Explanation
Consider,
\begin{bmatrix} 1 & 0\\ -1 & 1 \end{bmatrix}\begin{bmatrix} 2 & 0\\ 0 & x \end{bmatrix}\begin{bmatrix} 1 & -1\\ 0 & 1 \end{bmatrix}
=\begin{bmatrix} 2 & 0\\ -2 & x \end{bmatrix}\begin{bmatrix} 1 & -1\\ 0 & 1 \end{bmatrix}
By given, we have
\begin{bmatrix} 2 & -2\\ -2 & 2+x \end{bmatrix}=\begin{bmatrix} 2 & -2\\ -2 & 5 \end{bmatrix}
By equality of above matrices, we have
2+x=5
\therefore x =3
If
A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}
, then additive inverse of A is
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A^{T}
0%
A^{-1}
0%
-A
0%
-A^{-1}
If A and B are two matrices such that
AB = B
and
BA = A
, then
A^2 + B^2
is equal to
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2AB
0%
2BA
0%
A + B
0%
AB
Explanation
A^{2}+B^{2}=AA+BB
=A(BA)+B(AB)
(Given
AB=B \, and \, BA =A
)
=(AB)A+(BA)B
=BA+AB
=A+B
If
A \times \begin{bmatrix} 1 & 1\\ 0 & 2 \end{bmatrix} = [1 \ \ 2],
then A =
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\left[\dfrac {1}{2} \ \ \ 1\right]
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\begin{bmatrix} 1 & 2\\ 1 & 0 \end{bmatrix}
0%
\begin{bmatrix} 2 & 1\\ 1 & 5 \end{bmatrix}
0%
\left[1 \ \ \ \dfrac {1}{2} \right]
If
A
is a non-singular matrix, then
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A^{-1}
is symmetric if
A
is symmeteric
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A^{-1}
is skew-symmetric if
A
is symmeteric
0%
\left| { A }^{ -1 } \right| =\left| A \right|
0%
\left| { A }^{ -1 } \right| ={ \left| A \right| }^{ -1 }
Explanation
Since A is a non singular matrix
\left| A \right| \neq 0
, thus
{ A }^{ -1 }
exists.
Now
A{ A }^{ -1 }=I={ A }^{ -1 }A
\Rightarrow{ \left( A{ A }^{ -1 } \right) }^{ -1}={ I }^{ -1}={ \left( { A }^{ -1 }A \right) }^{ -1 }
\Rightarrow { \left( { A }^{ -1 } \right) }^{ -1 }{ A }^{ -1 }={ I }={ \left( A \right) }^{ -1 }{ \left( { A }^{ -1 } \right) }^{ ' }\left( \therefore { A }^{ ' }=A \right)
\Rightarrow { \left( { A }^{ -1 } \right) }={ \left( { A }^{ -1 } \right) }^{ ' }
. So
{ A }^{ -1 }
is symmetric
Also since
\left| A \right| \neq 0
{ A }^{ -1 }
exists such that
{ \left( { AA }^{ -1 } \right) }=I={ \left( { A }^{ -1 }A \right) }\Rightarrow \left| A{ A }^{ -1 } \right| =\left| I \right|
\left| A \right| \left| { A }^{ -1 } \right| =1\quad \therefore \left| AB \right| =\left| A \right| \left| B \right|
\left| { A }^{ -1 } \right| =\dfrac { 1 }{ \left| A \right| }
\begin{bmatrix}a\ \ b\end{bmatrix}
x
\begin{bmatrix}x\\y \end{bmatrix} =
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\begin{bmatrix}ax+ay +bx+by \end{bmatrix}
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\begin{bmatrix}ax \\ by \end{bmatrix}
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\begin{bmatrix}ax +by \end{bmatrix}
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\begin{bmatrix}ax\ \ by \end{bmatrix}
lf
\left[3x^{2}+10xy+5y^{2} \right]=\begin{bmatrix}x & y \end{bmatrix}A\begin{bmatrix} x\\ y\end{bmatrix}
, and
{A}
is a symmetric matrix then
\mathrm{A}=
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\left[\begin{array}{ll} 3 & 10\\ 10 & 5 \end{array}\right]
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\left[\begin{array}{ll} 10 & 3\\ 5 & 10 \end{array}\right]
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\left[\begin{array}{ll} +3 & -5\\ -5 & +5 \end{array}\right]
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\left[\begin{array}{ll} 3 & 5\\ 5 & 5 \end{array}\right]
Explanation
Given
\begin{bmatrix} 3{ x }^{ 2 }+10xy & +5{ y }^{ 2 } \end{bmatrix}=\begin{bmatrix} x & y \end{bmatrix}A\begin{bmatrix} x \\ y \end{bmatrix}
We will check using options
If option A is the matrix A, then consider RHS
\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 3 & 10 \\ 10 & 5 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}
=\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 3x+10y \\ 10x+5y \end{bmatrix}
=\begin{bmatrix} 3{ x }^{ 2 }+20xy & +5{ y }^{ 2 } \end{bmatrix}
which is not equal to given LHS
Now, option B, so let
A=\begin{bmatrix} 10 & 3 \\ 5 & 10 \end{bmatrix}
\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 10 & 3 \\ 5 & 10 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}
=\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 10x+3y \\ 5x+10y \end{bmatrix}
=\begin{bmatrix} 3{ x }^{ 2 }+8xy & +5{ y }^{ 2 } \end{bmatrix}
which is not equal to given LHS
Now, we will try option C
Let
A=\begin{bmatrix} 3 & -5 \\ -5 & 5 \end{bmatrix}
\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 3 & -5 \\ -5 & 5 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}
=\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 3x-5y \\ -5x+5y \end{bmatrix}
=\begin{bmatrix} 3{ x }^{ 2 }-10xy & +5{ y }^{ 2 } \end{bmatrix}
which is not equal to given LHS
Now, lastly we will try option D.
Let
A=\begin{bmatrix} 3 & 5 \\ 5 & 5 \end{bmatrix}
\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 3 & 5 \\ 5 & 5 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}
=\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} 3x+5y \\ 5x+5y \end{bmatrix}
=\begin{bmatrix} 3{ x }^{ 2 }+10xy & +5{ y }^{ 2 } \end{bmatrix}
which is equal to given LHS
If
A = \begin{bmatrix} 1 & -1\\ 2 & -1 \end{bmatrix}; B = \begin{bmatrix} 1 & 1\\ 4 & -1 \end{bmatrix},
then
A^2 + B^2 =
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2 I
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4 I
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\begin{bmatrix} 7 & 0\\ 0 & 7 \end{bmatrix}
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\begin{bmatrix} 1 & -1\\ 0 & 5 \end{bmatrix}
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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