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CBSE Questions for Class 12 Commerce Maths Matrices Quiz 9 - MCQExams.com
CBSE
Class 12 Commerce Maths
Matrices
Quiz 9
If $$A = \begin{bmatrix} 1& 4 & 4\\ 4 & 1 & 4\\ 4 & 4 & 1\end{bmatrix}$$, then $$A^{2} - 6A =$$ _____.
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$$27 I_{3}$$
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$$5 I_{3}$$
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$$20 I_{3}$$
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$$30 I_{3}$$
Explanation
Given, $$A=\begin{bmatrix} 1& 4 & 4\\ 4 & 1 & 4\\ 4 & 4 & 1\end{bmatrix}$$
Therefore,
$$A^2=\begin{bmatrix} 1& 4 & 4\\ 4 & 1 & 4\\ 4 & 4 & 1\end{bmatrix}$$
$$\begin{bmatrix} 1& 4 & 4\\ 4 & 1 & 4\\ 4 & 4 & 1\end{bmatrix}$$
$$=\begin {bmatrix} 1+16+16 & 4+4+16 & 4+16+4 \\ 4+4+16 & 16+1+16 & 16+4+4 \\ 4+16+4 & 16+4+4 & 16+16+1\end{bmatrix}$$
$$=\begin {bmatrix} 33 &24 &24\\24&33&24\\24&24&33 \end {bmatrix}$$
and
$$6A=\begin{bmatrix} 1& 4 & 4\\ 4 & 1 & 4\\ 4 & 4 & 1\end{bmatrix}$$
$$=\begin{bmatrix} 6& 24 & 24\\ 24 & 6 & 24\\ 24 & 24 & 6\end{bmatrix}$$
Thus $$A^2-6A$$
$$=\begin {bmatrix} 33 &24 &24\\24&33&24\\24&24&33 \end {bmatrix}$$
$$-\begin{bmatrix} 6& 24 & 24\\ 24 & 6 & 24\\ 24 & 24 & 6\end{bmatrix}$$
$$=\begin {bmatrix} 33-6&24-24&24-24 \\24-24&33-6&24-24\\24-24 &24-24&33-6 \end {bmatrix}$$
$$=\begin {bmatrix} 27&0&0\\ 0&27&0\\ 0&0&27\end {bmatrix}$$
$$= 27I_{3}$$
If $$A =\begin{pmatrix} -2& 2\\ 2 & -2\end{pmatrix}$$, then which one of the following is correct?
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$$A^{2} = -2A$$
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$$A^{2} = -4A$$
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$$A^{2} = -3A$$
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$$A^{2} = 4A$$
Explanation
Given $$A=\begin{bmatrix} -2 & 2 \\ 2 & -2 \end{bmatrix}$$
$$A^2=\begin{bmatrix} -2 & 2 \\ 2 & -2 \end{bmatrix}\begin{bmatrix} -2 & 2 \\ 2 & -2 \end{bmatrix}$$
$$=\begin{bmatrix} (-2)(-2)+2\cdot 2 & (-2)(2)+(2)(-2) \\ (2)(-2)+(-2)(2)& (2)(2)+(-2)(-2)\end{bmatrix}$$
$$=\begin{bmatrix} 8 & -8 \\ -8 & 8 \end{bmatrix}=-4\begin{bmatrix} -2 & 2 \\ 2 & -2 \end{bmatrix}=-4A$$
If $$A=\begin{bmatrix} 0 & 0 \\ 0 & 5 \end{bmatrix}$$, then $${ A }^{ 12 }$$ is
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$$\begin{bmatrix} 0 & 0 \\ 0 & 60 \end{bmatrix}$$
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$$\begin{bmatrix} 0 & 0 \\ 0 & { 5 }^{ 12 } \end{bmatrix}$$
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$$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$
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$$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
Explanation
We have $$A=\begin{bmatrix}0&0\\0&5\end{bmatrix}$$
$$A^2=A\times A=\begin{bmatrix}0&0\\0&5\end{bmatrix} \times \begin{bmatrix}0&0\\0&5\end{bmatrix}$$
$$=\begin{bmatrix}0&0\\0&5^2\end{bmatrix}$$
Similarly,
$$A^3=A^2\times A=\begin{bmatrix}0&0\\0&5^2\end{bmatrix} \times \begin{bmatrix}0&0\\0&5\end{bmatrix}$$
$$=\begin{bmatrix}0&0\\0&5^3\end{bmatrix}$$
Thus we can conclude
$$A^{12}=\begin{bmatrix}0&0\\0&5^{12}\end{bmatrix}$$
If $$A=\begin{bmatrix} 3 & 3 & 3\\ 3 & 3 & 3\\ 3 & 3 & 3\end{bmatrix}$$, then $$A^3=$$ ___________.
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$$243$$A
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$$81$$A
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$$27$$A
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$$729$$A
Explanation
$$\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}\times\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}=\begin{bmatrix}3 & 3 & 3 \\ 3 & 3 & 3 \\3 & 3 & 3 \\\end{bmatrix}=A$$
$$\implies A =\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}\times\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}=\begin{bmatrix}3 & 3 & 3 \\ 3 & 3 & 3 \\3 & 3 & 3 \\\end{bmatrix}=3\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}$$
$$A^2=\begin{bmatrix}3 & 3 & 3 \\ 3 & 3 & 3 \\3 & 3 & 3 \\\end{bmatrix}\times \begin{bmatrix}3 & 3 & 3 \\ 3 & 3 & 3 \\3 & 3 & 3 \\\end{bmatrix}$$
$$=3\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}\times3\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}$$
$$=9\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}\times\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}$$
$$=27\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}$$
$$A^3=A^2.A$$
$$=27\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix} \times 3\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}$$
$$=81\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix} \times \begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}$$
$$=81A$$
The answer is option (B).
If $$A=\begin{bmatrix}x&y&z\end{bmatrix},$$ $$ B=\begin{bmatrix}a&h&g\\h&b&f\\g&f&c\end{bmatrix}, C=\begin{bmatrix}\alpha & \beta & \gamma \end{bmatrix}^{T}$$ then $$ABC$$ is
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Not defined
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a $$1\times1$$ matrix
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a $$3\times3$$ matrix
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a $$3\times2$$ matrix
Explanation
$$A=\begin{bmatrix} x & y & z \end{bmatrix}\quad \quad B=\begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix},\quad \quad C=\begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix}\\ \quad \quad \quad 1\times 3\qquad \qquad \quad \quad \quad \ 3\times 3\qquad \qquad \qquad \ 3\times 1$$
So $$ABC$$ is a $$1 \times1$$ matrix
Is it possible to define the matrix $$A + B $$ when both $$A $$ and $$B $$ are square matrices of the same order?
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True
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False
Explanation
It is always possible to define the sum of two matrix $$X $$ and $$Y$$ if they are of same order.
So, If A and B are square matrices of same order, A + B is always definable.
Let A = $$\begin{bmatrix}
0 & 0 & -1 \\[0.3em]
0 & -1 & 0 \\[0.3em]
-1 & 0 & 0
\end{bmatrix}.$$ Then the only correct statement $$A$$ is
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$$A = 0$$
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$$A = (-1) I$$
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$$A^{-1} $$ does not exist
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$$A^2 = I$$
Explanation
$$A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \, \, \therefore A^2 = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \, \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$$
$$A^2 = \begin{bmatrix} 0 + 0 + 1 & 0 & 0 \\ 0 & 0 + 1 + 0 & 0 \\ 0 & 0 & 0 + 0 + 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
$$\therefore A^2 = I $$
Hence it is a identity matrix
If A = $$ \begin{bmatrix} \alpha & 0 \\ 1 & 1\end{bmatrix}$$ , B = $$ \begin{bmatrix} 1 & 0 \\ 5 & 1\end{bmatrix}$$ whenever $$A^2 \, = \, B$$
then values of $$\alpha$$ is
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$$1$$
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$$-1$$
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$$4$$
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no real value of $$\alpha$$
Explanation
$$A^2$$ = B$$\Rightarrow \begin{bmatrix} \alpha & 0 \\ 1 & 1\end{bmatrix} \begin{bmatrix} \alpha & 0 \\ 1 & 1\end{bmatrix} \, = \, \begin{bmatrix} 1 & 0 \\ 5 & 1\end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} \alpha^2 & 0 \\ \alpha+1 & 1\end{bmatrix} \, = \, \begin{bmatrix} 1 & 0 \\ 5 & 1\end{bmatrix}$$
$$\Rightarrow \alpha^2$$ = 1, $$\alpha$$ + 1 = 5
$$\therefore \alpha \, = \, \pm 1\, , \, \alpha \, = \, 4$$
There is no real value of $$\alpha$$ which satisfies both.
Using elementary row transformation find the inverse of the matrix A = $$\left[\begin {array}{ll} 3 & -1 & -2\\ 2 & 0 & -1\\ 3 & -5 & 0 \end {array}\right]$$
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$$\left[\begin {array} {ll} \dfrac{-5}{8} & \dfrac{5}{4} & \dfrac{1}{8} \\
\dfrac{-3}{8} & \dfrac{3}{4} & \dfrac{-1}{8} \\
\dfrac{-5}{4} & \dfrac{3}{2} &\dfrac{1}{4} \end {array}\right]$$
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$$\dfrac {1}{8}\left [\begin {array} {ll} 5 & -5 & 1 \\ 3 & -3 & 1 \\ 0 & 3 & 1\end {array}\right]$$
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$$\dfrac{1}{8} \left[\begin {array} {ll} 5 & 5 & 1 \\
3 & 6 & -1 \\
10 & -12 & 2 \end {array}\right]$$
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None of these
Obtain the inverse of the following matrix using elementary operation:
$$A = \begin{bmatrix} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{bmatrix}$$
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$$\begin{bmatrix} -3 & -4 & -3 \\ 2 & 3 & -2 \\ 8 & -12 & 9 \end{bmatrix}$$
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$$\begin{bmatrix} 3 & 4 & 3 \\ 2 & 3 & -2 \\ 8 & -12 & 9 \end{bmatrix}$$
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$$\begin{bmatrix} -3 & -4 & 3 \\ 2 & 3 & -2 \\ -8 & -12 & 9 \end{bmatrix}$$
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$$\begin{bmatrix}3& -2 &8 \\ -4& 3&-12\\3 &-2&9\end{bmatrix}$$
Explanation
GIven $$A=\begin{bmatrix}3& 0 &-1 \\ 2& 3&0\\0 &4&1\end{bmatrix}$$
Therefore, $$A^{-1}=\dfrac{1}{|A|}adj(A)$$
$$\implies A^{-1}=\dfrac{1}{3(3-0)-1(8-0)}\begin{bmatrix}3& -2 &8 \\ -4& 3&-12\\3 &-2&9\end{bmatrix}$$
$$\implies A^{-1}=\begin{bmatrix}3& -2 &8 \\ -4& 3&-12\\3 &-2&9\end{bmatrix}$$
If $$I =\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & \quad \quad 0 & \quad 1 \end{pmatrix},$$ $$P=\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & \quad \quad 0 & \quad -2 \end{pmatrix},$$ then the matrix $$P^3+2P^2$$ is equal to
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$$P$$
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$$1-P$$
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$$2I+P$$
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$$2I-P$$
Explanation
Given the matrix $$P=\begin{pmatrix}1&0&0\\0&-1&0\\ 0&0&-2\end {pmatrix}$$.
Now,
$$P^2=\begin{pmatrix}1^2&0&0\\0&(-1)^2&0\\ 0&0&(-2)^2\end {pmatrix}$$ [ Since $$P$$ is a diagonal matrix]
$$\Rightarrow P^2=\begin{pmatrix}1&0&0\\0&1&0\\ 0&0&4\end {pmatrix}$$......(1).
Again,
$$P^3=\begin{pmatrix}1^3&0&0\\0&(-1)^3&0\\ 0&0&(-2)^3\end {pmatrix}$$
$$\Rightarrow P^3=\begin{pmatrix}1&0&0\\0&-1&0\\ 0&0&-8\end {pmatrix}$$.....(2).
So,
$$P^3+2P^2$$
$$=\begin{pmatrix}1+2&0&0\\0&-1+2&0\\ 0&0&-8+8\end {pmatrix}$$
$$=\begin{pmatrix}3&0&0\\0&1&0\\ 0&0&0\end {pmatrix}$$
$$=2I+P$$.
Given $$A= \begin{bmatrix} 3&6 \\ -2&-8 \end{bmatrix}$$ and $$B = \begin{bmatrix} 2 & 16\end{bmatrix}$$, find the matrix $$X$$ such that $$XA=B$$.
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$$\begin{bmatrix} -\dfrac{4}{3} & -3 \end{bmatrix}$$
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$$\begin{bmatrix} \dfrac{4}{3} & 3 \end{bmatrix}$$
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$$\begin{bmatrix} \dfrac{4}{3} & -3 \end{bmatrix}$$
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$$\begin{bmatrix} -\dfrac{4}{3} & 3 \end{bmatrix}$$
Explanation
$$XA=B$$
$${ X }_{ 1\times 2 }{ \begin{bmatrix} 3 & 6 \\ -2 & -8 \end{bmatrix} }_{ 2\times 2 }={ \begin{bmatrix} 2 & 16 \end{bmatrix} }_{ 1\times 2 }$$
$$X$$ should be of order $$1 \times 2$$
Let $$X$$ be $$\begin{bmatrix} a & b \end{bmatrix}$$
$$\therefore \begin{bmatrix} a & b \end{bmatrix}\begin{bmatrix} 3 & 6 \\ -2 & -8 \end{bmatrix}=\begin{bmatrix} 2 & 16 \end{bmatrix}\\ \Rightarrow \begin{bmatrix} 3a-2b & 6a-8b \end{bmatrix}=\begin{bmatrix} 2 & 16 \end{bmatrix}\\ \therefore 3a-2b=2\longrightarrow (1)\times 2\\ \quad 6a-8b=16\longrightarrow (2)\times 1\\ \therefore 6a-4b=4\\ \quad 6a-8b=16\\ \quad \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \ 4b=-12\\ \quad \ \Rightarrow b=-3\\ \therefore 3a+6=2\\ \quad a=\dfrac{-4}{3}\\ \therefore X=\begin{bmatrix} \dfrac{-4}{3} & -3 \end{bmatrix}$$
The matrix equation satisfied by $$A$$ is ______$$, $$ if $$A=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}.$$
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$$A^2-4A-5I=0$$
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$$A^2-4A-5=0$$
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$$A^2+4A-5I=0$$
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$$A^2+4A-5=0$$
Explanation
Consider the given matrix,
$$A=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$$
Consider, $$A^2-4A-5I$$
$$=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}-4\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}-5\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
$$=\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}-\begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix}-\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$$
$$=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$
$$=0$$
State true/false:
If $$A=\begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}$$ and $$B=[-2-1-4],$$ then $$(AB)^t\ne B^tA^t$$.
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True
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False
Explanation
Given,
$$A={ \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} }_{ 3\times 1 }\quad and \quad B=\begin{bmatrix} -2 & -1 & -4 \end{bmatrix}$$
Now,
$$AB={ \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} }\times \begin{bmatrix} -2 & -1 & -4 \end{bmatrix} =\begin{bmatrix} 2 & 1 & 4 \\ -4 & -2 & -8 \\ -6 & -3 & -12 \end{bmatrix}\\ \therefore { \left( AB \right) }^{ t }=\begin{bmatrix} 2 & -4 & -6 \\ 1 & -2 & -3 \\ 4 & -8 & -12 \end{bmatrix}\\ Consider, \quad { B }^{ t }\times { A }^{ t }={ \begin{bmatrix} -2 \\ -1 \\ -4 \end{bmatrix} }\times \begin{bmatrix} -1 & 2 & 3 \end{bmatrix} =\begin{bmatrix} 2 & -4 & -6 \\ 1 & -2 & -3 \\ 4 & -8 & -12 \end{bmatrix}\\ \therefore { \left( AB \right) }^{ t }={ B }^{ t }\times { A }^{ t }$$
Hence, given statement is false.
If $$A = \begin{bmatrix}3&-3&4\\2&-3&4\\0&-1&1\end{bmatrix}$$ and $$B = \begin{bmatrix}3&1&2\\2&0&5\\1&2&0\end{bmatrix}$$, find $$AB$$.
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$$\begin{bmatrix}5&6&-9\\4&3&-9\\-1&3&-2\end{bmatrix}$$
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$$\begin{bmatrix}2&4&7\\1&5&3\\-1&0&-5\end{bmatrix}$$
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$$\begin{bmatrix}7&11&-9\\4&10&-11\\-1&2&-5\end{bmatrix}$$
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$$\begin{bmatrix}7&8&-19\\4&1&-5\\-1&12&-5\end{bmatrix}$$
Explanation
$$A=\begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}\quad B=\begin{bmatrix} 3 & 1 & 2 \\ 2 & 0 & 5 \\ 1 & 2 & 0 \end{bmatrix}\\ AB=\begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}\times \begin{bmatrix} 3 & 1 & 2 \\ 2 & 0 & 5 \\ 1 & 2 & 0 \end{bmatrix}\\ \quad =\begin{bmatrix} 9-6+4 & 3+0+8 & 6-15+0 \\ 6-6+4 & 2+0+8 & 4-15+0 \\ 0-2+1 & 0+0+2 & 0-5+0 \end{bmatrix}\\ \quad =\begin{bmatrix} 7 & 11 & -9 \\ 4 & 10 & -11 \\ -1 & 2 & -5 \end{bmatrix}$$
$$\left[ \begin{matrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{matrix} \right] +\left[ \begin{matrix} 5 & 1& -2 \\ 1 & 1 & 0 \\ -2 & -2 & 1 \end{matrix} \right] \ $$
What will be the sum of the diagonal elements of the resultant matrix?
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10
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6
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9
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7
Explanation
$$\begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix}+ \begin{bmatrix} 5 & 1 & -2 \\ 1 & 1 & 0 \\ -2 & -2 & 1 \end{bmatrix}= \begin{bmatrix} 6 & 1 & 0 \\ 0 & 2 & -2 \\ -2 & 0 & 2 \end{bmatrix}$$
Sum of diagonal $$=6+2+2 = 10$$
If $$A=\begin{bmatrix} 1 & 0 \\ \frac { 1 }{ 2 } & 1 \end{bmatrix}$$, then $$A^{100}$$ is equal to
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$$\begin{bmatrix} 1 & 0 \\ \left( \frac { 1 }{ 2 } \right) \times { 100 } & 1 \end{bmatrix}$$
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$$\begin{bmatrix} 1 & 0 \\ 25 & 1 \end{bmatrix}$$
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$$\begin{bmatrix} 1 & 0 \\ 50 & 1 \end{bmatrix}$$
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$$\begin{bmatrix} 1 & 0 \\ 100 & 1 \end{bmatrix}$$
Explanation
Given $$A=\begin{bmatrix} 1& 0\\ \dfrac{1}{2}& 1\end{bmatrix}$$
$$A^2=\begin{bmatrix} 1& 0\\ \dfrac{1}{2}& 1\end{bmatrix}\times \begin{bmatrix} 1& 0\\ \dfrac{1}{2}& 1\end{bmatrix}=\begin{bmatrix} 1& 0\\ \dfrac{2}{2}& 1\end{bmatrix}$$
$$A^3=\begin{bmatrix} 1& 0\\ \dfrac{2}{2}& 1\end{bmatrix}\times \begin{bmatrix} 1& 0\\ \dfrac{1}{2}& 1\end{bmatrix}=\begin{bmatrix} 1& 0\\ \dfrac{3}{2}& 1\end{bmatrix}$$
$$A^4=\begin{bmatrix} 1& 0\\ \dfrac{3}{2}& 1\end{bmatrix}\times \begin{bmatrix} 1& 0\\ \dfrac{1}{2}& 1\end{bmatrix}=\begin{bmatrix} 1& 0\\ \dfrac{4}{2}& 1\end{bmatrix}$$
$$A^n=\begin{bmatrix} 1& 0\\ \dfrac{n}{2}& 1\end{bmatrix}$$
So
$$A^{100}=\begin{bmatrix} 1& 0 \\ \dfrac{100}{2}&1 \end{bmatrix}=\begin{bmatrix} 1& 0\\ 50& 1\end{bmatrix}$$
If$$A=\begin{bmatrix}1&3\\2&4\end{bmatrix}$$ and $$B=\begin{bmatrix}4&7\\5&6\end{bmatrix},$$ then $$AB=$$
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$$\begin{bmatrix}19&25\\28&38\end {bmatrix}$$
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$$\begin{bmatrix}25&8\\26&14\end {bmatrix}$$
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$$\begin{bmatrix}34&8\\58&23\end {bmatrix}$$
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$$\begin{bmatrix}5&10\\7&10\end {bmatrix}$$
Explanation
Given,
$$A=\begin{bmatrix}1&3\\2&4\end{bmatrix}$$ and $$B=\begin{bmatrix}4&7\\5&6\end{bmatrix}$$
$$AB=\begin{bmatrix}1&3\\2&4\end{bmatrix}\begin{bmatrix}4&7\\5&6\end{bmatrix}$$
$$=\begin{bmatrix}1(4)+3(5)&1(7)+3(6)\\2(4)+4(5)& 2(7)+4(6)\end{bmatrix}$$
$$=\begin{bmatrix}19&25\\28&38\end{bmatrix}$$
$$B=A+A^{2}+A^{3}+A^{4}$$
If order of $$A$$ is $$3$$ then order of $$B$$ is
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$$3$$
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$$6$$
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$$2$$
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$$9$$
Explanation
The order of matrix does not change when the operation are done on it.
So, the order of $$B$$ remains same as the order of $$A$$.
If $$A=\begin{bmatrix} 2 & 5 & -3 \\ -1 & 3 & 1 \\ 4 & 1 & 2 \end{bmatrix}$$ then $${A}^{2}$$ is
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$$\begin{bmatrix} -13 & 22 & -7 \\ 15 & 25 & -7 \\ -1 & 5 & 8 \end{bmatrix}$$
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$$\begin{bmatrix} -13 & 22 & -7 \\ -1 & 5 & 8 \\ 15 & 25 & -7 \end{bmatrix}$$
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$$\begin{bmatrix} -13 & 22 & -7 \\ 15 & 25 & 7 \\ 1 & 5 & 8 \end{bmatrix}$$
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$$\begin{bmatrix} 13 & 22 & 7 \\ 15 & 25 & -7 \\ -1 & 5 & 8 \end{bmatrix}$$
Explanation
$$A^2=A.A=\begin{bmatrix} 2 & 5 & -3 \\ -1 & 3 & 1 \\ 4 & 1 & 2 \end{bmatrix}.\begin{bmatrix} 2 & 5 & -3 \\ -1 & 3 & 1 \\ 4 & 1 & 2 \end{bmatrix}$$
$$=\begin{bmatrix} -13\quad \quad & 22\quad \quad & 2\times -3+5\times 1+(-3)\times 2 \\ -1 & 5 & 8 \\ 15 & 25 & -7 \end{bmatrix}$$
$$\begin{bmatrix} -13 & 22 & -7 \\ -1 & 5 & 8 \\ -15 & 25 & -7 \end{bmatrix}$$
Hence, B will be correct answer.
If the graph of $$y = f(x)$$ is symmetrical about the lines $$x = 1$$ and $$x = 2$$, then which of the following is true?
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$$f(x + 1) = f(x)$$
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$$f(x + 3) = f(x)$$
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$$f(x + 2) = f(x)$$
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None of these
Explanation
$$f(x)$$ is symmetrical about the line $$x=1$$
So,
$$f(1-x)=f(1+x)$$ .................... (1)
$$f(x)$$ is symmetrical about the line $$x=2$$
So,
$$f(2-x)=f(2+x)$$ .................... (2)
Replace $$x$$ with $$1-x$$ in equation (1) ,
$$f(1-1+x)=f(1+1-x)$$
$$f(x)=f(2-x)$$ ..........................(3)
From (2) and (3) ,
$$f(x)=f(2+x)$$
Find the inverse of the following matrix.
$$\left[ {\begin{array}{ccccccccccccccc}0&1&2\\1&2&3\\3&1&1\end{array}} \right]$$
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$$\displaystyle \left[ {\begin{array}{ccccccccccccccc}{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{ 4}&3&{ 1}\\{ - \frac{5}{2}}&{ - \frac{3}{2}}&{\frac{1}{2}}\end{array}} \right]$$
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$$\displaystyle \left[ {\begin{array}{ccccccccccccccc}{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{ - 4}&-3&{ - 1}\\{ - \frac{5}{2}}&{ - \frac{3}{2}}&{\frac{1}{2}}\end{array}} \right]$$
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$$\displaystyle \left[ {\begin{array}{ccccccccccccccc}{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{ - 4}&3&{ - 1}\\{ - \frac{5}{2}}&{ - \frac{3}{2}}&{\frac{1}{2}}\end{array}} \right]$$
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$$\left[ {\begin{array}{ccccccccccccccc}0&1&2\\1&-2&3\\3&1&-1\end{array}} \right]$$
State true or false:
The product matrix of $$\left[ \begin{array}{l}1\,\,\,\,\,\,0\\0\,\,\,\,\,\,1\end{array} \right] $$ and $$\left[ \begin{array}{l}0\,\,\,\,\,\,\,\,1\\1\,\,\,\,\,\,\,\,0\end{array} \right]$$ is an identity matrix.
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True
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False
Explanation
Consider,
$$\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \times \begin{bmatrix} 0& 1\\1&0 \end{bmatrix}$$
$$=\begin{bmatrix}1(0)+0(1) & 1(1)+0(0)\\0(0)+1(1) & 0(1)+1(0)\end{bmatrix}$$
$$=\begin {bmatrix} 0&1\\1&0\end{bmatrix},$$ which is not an identify matrix of order $$2 \times 2$$.
Hence given statement is false.
If $$A$$ and $$B$$ are two non-singular matrices and both are symmetric and commute each other, then
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Both $$A^{-1}B$$ and $$A^{-1}B^{-1}$$ are symmetric
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$$A^{-1}B$$ is symmetric but $$A^{-1}B^{-1}$$ is not symmetric
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$$A^{-1}B^{-1}$$ is symmetric but $$A^{-1}B$$ is not symmetric
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Neither $$A^{-1}B$$ nor $$A^{-1}B^{-1}$$ are symmetric
Explanation
$$\begin{array}{l} \left| A \right| \ne 0\, \, \& \, \, \left| B \right| \ne 0 \\ { \left( { { A^{ -1 } }{ B^{ -1 } } } \right) ^{ T } }={ \left( { { B^{ -1 } } } \right) ^{ T } }{ \left( { { A^{ -1 } } } \right) ^{ T } } \\ { \left( { { A^{ -1 } }{ B^{ -1 } } } \right) ^{ T } }={ \left( { { B^{ T } } } \right) ^{ -1 } }{ \left( { { A^{ T } } } \right) ^{ -1 } }\, \, \, \, \left[ \begin{array}{l} { A^{ T } }=A \\ { B^{ T } }=B \end{array} \right] \\ { \left( { { A^{ -1 } }{ B^{ -1 } } } \right) ^{ T } }={ B^{ -1 } }{ A^{ -1 } } \\ { \left( { { A^{ -1 } }{ B^{ -1 } } } \right) ^{ T } }={ \left( { AB } \right) ^{ -1 } }\, \, \, \, \, \, \left[ { AB=BA } \right] \\ { \left( { { A^{ -1 } }{ B^{ -1 } } } \right) ^{ T } }={ \left( { BA } \right) ^{ -1 } } \\ { \left( { { A^{ -1 } }{ B^{ -1 } } } \right) ^{ T } }={ A^{ -1 } }{ B^{ -1 } } \\ { A^{ -1 } }{ B^{ -1 } }\, is\, \, a\, symmetric\, \, matrix \end{array}$$
similarly $$A^{-1}B$$ is also symmetric matrix.
If $$A = \begin{bmatrix} 0 & 2 \\ 3 & -4 \end{bmatrix}$$ and $$kA = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$$, then the value of $$k, a, b $$, are respectively.
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$$-6, -12, -18$$
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$$-6, 4, 9$$
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$$-6, -4, -9$$
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$$-6, 12, 18$$
Explanation
If $$A=\begin{bmatrix} 0 & 2 \\ 3 & -4 \end{bmatrix}$$
then,
$$kA=\begin{bmatrix} 0 & 2k \\ 3k & -4k \end{bmatrix}$$
But it is given that
$$kA=\begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$$
So,
$$\begin{bmatrix} 0 & 2k \\ 3k & -4k \end{bmatrix}$$
$$=\begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$$
$$\Rightarrow -4k=24$$
$$k=-6$$
$$3a=2k$$
$$3a=-12$$
$$a=-4$$
$$2b=3k$$
$$2b=-18$$
$$b=-9$$
If $$A = \left[ {\begin{array}{*{20}{c}}2 & 0 & 0\\0 & 2 & 0\\0 & 0 & 2\end{array}} \right]$$ then $$A^6$$ = _____________.
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$$6A$$
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$$12A$$
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$$16A$$
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$$32A$$
Explanation
$$A=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}\\ A^{ 2 }=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}=\begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}\\ A^{ 4 }=A^{ 2 }A^{ 2 }=\begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}\begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}=\begin{bmatrix} 16 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 16 \end{bmatrix}\\ A^{ 6 }=A^{ 4 }A^{ 2 }=\begin{bmatrix} 16 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 16 \end{bmatrix}\begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}=\begin{bmatrix} 64 & 0 & 0 \\ 0 & 64 & 0 \\ 0 & 0 & 64 \end{bmatrix}\\ taking\quad 32\quad out\quad we\quad get\\ A^{ 6 }=32\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}=32A$$
If $$\left[ {\begin{array}{*{20}{c}}x & 4 & { - 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2 & 1 & 0\\1 & 0 & 2\\0 & 2 & 4\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\4\\{ - 1}\end{array}} \right] = 0$$ then $$x$$ is equal to
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$$ - 1 + \sqrt 6 $$
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$$ 8 \pm \sqrt 5 $$
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$$ - 2 \pm \sqrt {10} $$
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$$ 3 \pm \sqrt 6 $$
Explanation
$$\left[ \begin{matrix} x & 4 & 1 \end{matrix} \right]_{1\times 3}\left[ \begin{matrix}2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 4 \end{matrix} \right]_{3\times 3}\left[ \begin{matrix} x \\ 4 \\ -1 \end{matrix} \right]_{3\times 1}=0$$
$$\left[ \begin{matrix} x & 4 & 1 \end{matrix} \right]_{1\times 3}\left[ \begin{matrix} 2x+4+0 \\ x+0-2 \\ 0+8-4 \end{matrix} \right]_{3\times 1}=0$$
$$\Rightarrow\,2{x}^{2}+4x+4x-8-4=0$$
$$\Rightarrow\,2{x}^{2}+8x-12=0$$
$$\Rightarrow\,{x}^{2}+4x-6=0$$
$$\Rightarrow\,{x}^{2}+2\times 2x+4-4-6=0$$
$$\Rightarrow\,{\left(x+2\right)}^{2}=10$$
$$\Rightarrow\,x+2=\pm\sqrt{10}$$
$$\therefore\,x=-2\pm\sqrt{10}$$
If $$A=\begin{bmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{bmatrix},B=\begin{bmatrix} { a }^{ 2 } & ab & ac \\ ba & { b }^{ 2 } & bc \\ ca & cb & { c }^{ 2 } \end{bmatrix}$$ then $$AB=$$
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$$O$$
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$$I$$
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$$2I$$
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None of these
Explanation
$$AB=\begin{bmatrix} 0 & c & -b \\ -c & b & a \\ b & -a & 0 \end{bmatrix}\begin{bmatrix} a^{ 2 } & ab & ac \\ ba & b^{ 2 } & bc \\ ca & cb & c^{ 2 } \end{bmatrix}\\ =\begin{bmatrix} abc-abc & b^{ 2 }c^{ 1 }-b^{ 2 }c & bc^{ 2 }-bc^{ 2 } \\ -a^{ 2 }c+a^{ 2 }c & -abc+abc & -ac^{ 2 }+ac^{ 2 } \\ a^{ 2 }b-a^{ 2 }b & ab^{ 2 }-ab^{ 2 } & abc-abc \end{bmatrix}\\ =\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}=0\Rightarrow (A)$$
The value of $$x$$, so that $$\left\lceil 1\quad x\quad 1 \right\rceil \begin{bmatrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix}=0$$ is
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$$\cfrac{-7\pm \sqrt{35}}{2}$$
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$$\cfrac{-9\pm \sqrt{53}}{2}$$
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$$\pm 2$$
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$$\dfrac{-7}{10}$$
Explanation
Consider, $$\begin{bmatrix} 1 & x & 1\end{bmatrix}\begin{bmatrix} 1 & 3 & 2\\ 0 & 5 & 1\\ 0 & 3 & 2\end{bmatrix}=1\begin{bmatrix} 1 & 3+5x+3 & 5\end{bmatrix}$$
$$\begin{bmatrix} 1 & 6+5x & 5\end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ x\end{bmatrix}=7+10x$$
By given, we have
$$7+10x =0$$
$$\Rightarrow x=-\dfrac{7}{10}$$.
If A be square matrix of order n and k is a scalar, then adj (KA) is:
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$$K^{n}(adjA)$$
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K (adj A)
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$$K^{n-1}(adjA)$$
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$$K^{n+1}(adjA)$$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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