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Matrices Test - 36

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Matrices Test - 36
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  • Question 1
    1 / -0
    If $$\begin{bmatrix} 5 & x & 1 \end{bmatrix} \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}  = (20)$$, then the value of $$x$$ is
    Solution
    Given expression: $${ \begin{bmatrix} 5 &x&1\end{bmatrix} }_{ 1\times 3 }\times { \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} }_{ 3\times 1 }={ \begin{pmatrix} 20 \end{pmatrix} }_{ 1\times 1 }\\\Rightarrow { (10+(-x)+3) }_{ 1\times 1 }={ \begin{pmatrix} 20 \end{pmatrix} }_{ 1\times 1 }$$
    $$\Rightarrow 13-x=20$$ [by equality of matrix]
    $$\Rightarrow x=-7$$
  • Question 2
    1 / -0
    Let $$A = \begin{bmatrix}x + y & y\\ 2x & x - y\end{bmatrix}, B = \begin{bmatrix} 2& -1\end{bmatrix}$$ and $$C = \begin{bmatrix} 3& 2\end{bmatrix}.$$ If $$AB = C$$, then $$A^{2}$$ is equal to
    Solution
    $$A=\begin{bmatrix} x+y & y \\ 2x & x-y \end{bmatrix},  B=\begin{bmatrix} 2 & -1 \end{bmatrix}$$ and $$C=\begin{bmatrix} 3 & 2 \end{bmatrix}$$
    Since, 
    $$AB=C\\ \Longrightarrow \begin{bmatrix} x+y & y \\ 2x & x-y \end{bmatrix}\begin{bmatrix} 2 & -1 \end{bmatrix}=\begin{bmatrix} 3 & 2 \end{bmatrix}$$
    The multiplication of $$A$$ and $$B$$ can't be done because it doesn't satisfy the conditions of matrix multipication
  • Question 3
    1 / -0
    Choose the correct statement related to the matrices $$A=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$ and $$B=\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$$
    Solution
    Clearly, $$A$$ is an identity matrix 

    Now, $${ A }^{ 3 }=$$ $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\quad $$

                $$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = A$$

    $$\implies A^3=A$$

    Also, $${ B}^{ 3 }= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = B$$

    $$\implies B^{3}=B$$

    Hence, option C is correct.
  • Question 4
    1 / -0
    If $$\begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix}\times \begin{pmatrix} 5 & -2 \\ -3 & 1 \end{pmatrix}=\begin{pmatrix} 1 & -1 \\ 17 & \lambda  \end{pmatrix}$$ then what is $$\lambda $$ equal to?
    Solution
    GIven,
    $$\begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix}\times \begin{pmatrix} 5 & -2 \\ -3 & 1 \end{pmatrix}=\begin{pmatrix} 1 & -1 \\ 17 & \lambda  \end{pmatrix}$$

    Now,
    $$\begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix}\times \begin{pmatrix} 5 & -2 \\ -3 & 1 \end{pmatrix}=\begin{pmatrix} 10-9 & -4+3 \\ 20-3 & -8+1 \end{pmatrix}=\begin{pmatrix} 1 & -1 \\ 17 & -7 \end{pmatrix}$$
    So comparing both, we get $$\lambda =-7$$
  • Question 5
    1 / -0
    If the matrix A is such that $$\begin{bmatrix} 1 & 3\\ 0 & 1\end{bmatrix} A=\begin{bmatrix} 1 & 1 \\ 0 & -1\end{bmatrix}$$, then what is equal to A?
    Solution
    Given
    $$\begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}A=\begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix}$$
    Let $$A=\begin{bmatrix} \alpha  & \beta  \\ \gamma  & \delta  \end{bmatrix}$$
    $$\begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} \alpha  & \beta  \\ \gamma  & \delta  \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix}$$
    $$\begin{bmatrix} \alpha +3\gamma  & \beta +4\delta  \\ \gamma  & \delta  \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix}$$
    Two matrices are equal if corresponding elements are equal
    $$\Rightarrow \alpha +3\gamma =1, \gamma =0,$$
    $$\beta +3\delta =1$$ and $$ \delta =-1$$
    $$\Rightarrow \alpha =1,$$ $$\beta =4$$
    $$\Rightarrow A=\begin{bmatrix} 1 & 4 \\ 0 & -1 \end{bmatrix}$$
  • Question 6
    1 / -0
    If matrix $$A=\begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}$$ such that $$Ax=I$$, then $$x=$$................
    Solution
    Given : $$A=\begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}$$ such that $$Ax=I$$ ... $$(i)$$
    Since, $$A$$ and $$I$$ are of order $$2\times 2$$. So, $$x$$ will be a matrix of order $$2\times 2$$
    Let $$x=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$
    From $$(i)$$, we get
    $$Ax=I$$
    $$\implies \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

    $$\implies \begin{bmatrix} a+2c & b+2d \\ 4a+3c & 4b+3d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
    $$\implies a+2c=1$$ ...... $$(ii),$$  $$4a+3c=0$$ ........ $$(iii)$$
    $$b+2d=0$$ ....... $$(iv)$$ and $$4b+3d=1$$ ....... $$(v)$$
    Solving $$(ii)$$ and $$(iii)$$ simultaneously we get
    $$a=-\dfrac{3}{5}$$ and $$c=\dfrac{4}{5}$$
    Then solving $$(iv)$$ and $$(v)$$ simultaneously we get
    $$b=\dfrac{2}{5}$$ and $$d=-\dfrac{1}{5}$$
    Substituting all these values in $$x$$, we get
    $$x=\begin{bmatrix} -\dfrac{3}{5} & \dfrac{2}{5} \\ \dfrac{4}{5} & -\dfrac{1}{5} \end{bmatrix}$$

    $$\therefore x=\dfrac{1}{5} \begin{bmatrix} -3 & 2 \\ 4 & -1 \end{bmatrix}$$
  • Question 7
    1 / -0
    If $$[1\,x\,1]  \begin{bmatrix} 1&3&2 \\ 0&5&1\\0&2&0 \end{bmatrix}$$ $$\begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix}=0$$, then the values of $$x$$ are:
    Solution
    Since, $$[1\,x\,1]  \begin{bmatrix} 1&3&2 \\ 0&5&1\\0&2&0 \end{bmatrix}$$ $$\begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix}=0$$
    $$\Rightarrow \begin{bmatrix} 1 & 3+5x+2 & 2+x+0 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix}=0$$
    $$\Rightarrow \begin{bmatrix} 1 &5+5x & 2+x \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix}=0$$
    $$\Rightarrow \left[ 1+5+5x+2x+{ x }^{ 2 } \right] =0$$
    $$\Rightarrow { x }^{ 2 }+7x+6=0$$
    $$\Rightarrow x^2+6x+x+6=0$$
    $$\Rightarrow (x+1)(x+6)=0$$
    $$\therefore x=-1,-6$$ 
  • Question 8
    1 / -0
    The matrix $$A = \begin{bmatrix} 0& 1 & -1\\ -1 & 0 & 1\\ 1 & -1 & 0\end{bmatrix}$$ is a :
    Solution
    The matrix $$A=\quad \begin{bmatrix} 0 & \quad \quad 1 & \quad \quad -1 \\ -1 & \quad \quad 0 & \quad \quad 1 \\ 1 & \quad \quad -1 & \quad \quad 0 \end{bmatrix}$$ is a $${ \quad A }^{ T\quad  }=\quad \begin{bmatrix} 0 & \quad -1 & \quad \quad 1 \\ 1 & \quad \quad 0 & \quad \quad -1 \\ -1 & \quad \quad 1 & \quad \quad 0 \end{bmatrix}$$ $$Hence    \ \ \  { -A }^{ T }\quad =\quad \begin{bmatrix} 0 & \quad \quad 1 & \quad \quad -1 \\ -1 & \quad \quad 0 & \quad \quad 1 \\ 1 & \quad \quad -1 & \quad \quad 0 \end{bmatrix}\quad \\ \\ \quad A\quad =\quad { -A }^{ T }$$
    The matrix A is skew symmetric matrix 
  • Question 9
    1 / -0
    If $$\begin{bmatrix} 3 & -1 \\ 0 & 6 \end{bmatrix}\begin{bmatrix} 3x \\ 1 \end{bmatrix}+\begin{bmatrix} -2x \\ 3 \end{bmatrix}=\begin{bmatrix} 8 \\ 9 \end{bmatrix}$$, then the value of $$x$$ is
    Solution
    Given, $$\begin{bmatrix} 3 & -1 \\ 0 & 6 \end{bmatrix}\begin{bmatrix} 3x \\ 1 \end{bmatrix}+\begin{bmatrix} -2x \\ 3 \end{bmatrix}=\begin{bmatrix} 8 \\ 9 \end{bmatrix}$$
    LHS $$=\begin{bmatrix} 3 & -1 \\ 0 & 6 \end{bmatrix}\begin{bmatrix} 3x \\ 1 \end{bmatrix}+\begin{bmatrix} -2x \\ 3 \end{bmatrix}$$
          $$=\begin{bmatrix} 9x+\left( -1 \right)  \\ 0+6 \end{bmatrix}+\begin{bmatrix} -2x \\ 3 \end{bmatrix}=\begin{bmatrix} 9x-1 \\ 6 \end{bmatrix}+\begin{bmatrix} -2x \\ 3 \end{bmatrix}$$
          $$=\begin{bmatrix} 9x-1-2x \\ 6+3 \end{bmatrix}=\begin{bmatrix} 7x-1 \\ 9 \end{bmatrix}$$
    Now, $$\begin{bmatrix} 7x-1 \\ 9 \end{bmatrix}=\begin{bmatrix} 8 \\ 9 \end{bmatrix}$$
    $$\therefore 7x-1=8$$
    $$\Rightarrow 7x=9$$
    $$\Rightarrow x=\dfrac { 9 }{ 7 } $$.
  • Question 10
    1 / -0
    If $$A=\begin{bmatrix}1&1&-1\\2&-3&4\\3&-2&3\end{bmatrix}$$ and $$B=\begin{bmatrix}-1&-2&-1\\6&12&6\\5&10&5\end{bmatrix}$$, then which of the following is/are correct?
    1. A and B commute.
    2. AB is null matrix.
    Select the correct answer using the code given below :
    Solution
    We have, $$A=\begin{bmatrix}1&1&-1\\2&-3&4\\3&-2&3\end{bmatrix}$$ and $$B=\begin{bmatrix}-1&-2&-1\\6&12&6\\5&10&5\end{bmatrix}$$
    $$AB=\begin{bmatrix}-1+6-5&-2+12-10&-1+6-5\\-2-18+20&-4-36+40&-2-18+20\\-3-12+15&-6-24+30&-3-12+15\end{bmatrix}$$
    $$=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$
    Hence, $$AB$$ is a null matrix.
    and $$AB\neq BA$$
    Hence, B is the correct option.
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