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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 $$\mathrm{A}=\left[\begin{array}{ll}
0 & 1\\
1 & 0
\end{array}\right]$$, then $$\mathrm{A}^{5}=$$
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$$I$$
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$$\mathrm{O}$$
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$$\mathrm{A}$$
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$$\mathrm{A}^{2}$$
Explanation
Given, A $$=$$$$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$
$$A^2$$$$=$$
$$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
$$A^5$$$$=$$$${A^2}\times{A^2}\times{A}$$
$$A^5$$$$=$$ $$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$ $$=$$$$A$$
If $$\mathrm{A}=\left[\begin{array}{ll}
1 & 2\\
0 & 3
\end{array}\right]$$ and $$\mathrm{B}=[3 \space-1]$$, then $$\mathrm{B}\mathrm{A}=$$
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$$\left[\begin{array}{ll}
3 & 0\\
0 & 3
\end{array}\right]$$
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$$[3 \ \ \ 0]$$
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$$[3 \ \ \ 3 ]$$
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$$[0 \ \ -3]$$
Explanation
The value of $$\mathrm{BA}$$
$$=[3 \space -1]\begin{bmatrix}1 & 2\\0 & 3\end{bmatrix}$$
$$=[ 3\times 1-1\times 0 \ \ \space 3\times 2-1\times 3 ]$$
$$=[3 \ \ \ \space 3]$$
$$ If \space A= \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}$$
, 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
$$ = \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}$$
$$ \therefore A = A^{T}$$
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
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$$A$$ 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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$$-1$$
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$$0$$
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$$1$$
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$$4$$
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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$$0$$
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$$3$$
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$$6$$
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$$8$$
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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$$10$$
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$$20$$
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$$30$$
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$$40$$
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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$$0$$
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$$1$$
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$$\dfrac{1}{2}$$
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$$-2$$
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
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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
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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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$$\mathrm{A}\mathrm{B}=\mathrm{B}\mathrm{A}$$
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$$\mathrm{A}\mathrm{B}=-\mathrm{A}\mathrm{B}$$
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$$\mathrm{A}\mathrm{B}=2\mathrm{B}\mathrm{A}$$
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$$\mathrm{A}\mathrm{B}=3\mathrm{B}\mathrm{A}$$
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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$$\mathrm{A}$$
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$$2\mathrm{A}$$
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$$3\mathrm{A}$$
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$$4\mathrm{A}$$
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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$$-3$$
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$$-2$$
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$$0$$
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$$3$$
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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$$A^{T}$$
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$$A^{-1}$$
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$$-A$$
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$$-A^{-1}$$
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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$$2AB$$
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$$2BA$$
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$$A + B$$
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$$AB$$
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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$$\begin{bmatrix} 1 & 2\\ 1 & 0 \end{bmatrix}$$
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$$\begin{bmatrix} 2 & 1\\ 1 & 5 \end{bmatrix}$$
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$$\left[1 \ \ \ \dfrac {1}{2} \right]$$
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
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$$\left| { A }^{ -1 } \right| =\left| A \right| $$
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$$\left| { A }^{ -1 } \right| ={ \left| A \right| }^{ -1 }$$
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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