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Matrices Test - 37

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Matrices Test - 37
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  • Question 1
    1 / -0
    If $$\begin{bmatrix} \alpha  & \beta  \\ \gamma  & -\alpha  \end{bmatrix}$$ to the square is two rowed unit matrix, then $$\alpha ,\beta ,\gamma $$ should satisfy the relation
    Solution
    since $$\begin{bmatrix} \alpha  & \beta  \\ \gamma  & -\alpha  \end{bmatrix}$$ is a square root of $${I}_{2}$$ i.e., two rowed unit matrix
    $$\therefore { \begin{bmatrix} \alpha  & \beta  \\ \gamma  & -\alpha  \end{bmatrix} }^{ 2 }=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
    $$\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} \alpha  & \beta  \\ \gamma  & -\alpha  \end{bmatrix}\begin{bmatrix} \alpha  & \beta  \\ \gamma  & -\alpha  \end{bmatrix}$$
    $$\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} { \alpha  }^{ 2 }+\beta \gamma  & 0 \\ 0 & { \alpha  }^{ 2 }+\beta \gamma  \end{bmatrix}$$
    $$\therefore { \alpha  }^{ 2 }+\beta \gamma =1$$
    $$\Rightarrow 1-{ \alpha  }^{ 2 }-\beta \gamma =0$$ 
  • Question 2
    1 / -0
    If $$A=\begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix}$$ and $${ A }^{ 2 }-4A+10I=A$$, then $$k$$ is equal to
    Solution
    Given, $$A=\begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix}$$ and $${ A }^{ 2 }-4A+10I=A$$
    $$\Rightarrow \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix} \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix} -4 \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix} + 10 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix}$$
    $$\Rightarrow \begin{bmatrix} -5 & -3-3k \\ 2+2k & -6+{ k }^{ 2 } \end{bmatrix} - \begin{bmatrix} 4 & -12 \\ 8 & 4k \end{bmatrix} + \begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix} = \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix}$$
    $$\Rightarrow \begin{bmatrix} 1 & 9-3k \\ -6+2k & 4+{ k }^{ 2 }-4k \end{bmatrix} = \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix}$$
    $$\Rightarrow 9-3k=-3, -6+2k=2$$             ....(i)
    and $$4+{ k }^{ 2 }-4k=k$$ 
    $${ k }^{ 2 }-5k+4=0$$ 
    $$\Rightarrow \left( k-4 \right) \left( k-1 \right) =0 \Rightarrow k=4,1$$
    But $$k=1$$ is not satisfied the equation (i).
  • Question 3
    1 / -0
    If $$A=\begin{bmatrix} \alpha  & 0 \\ 1 & 1 \end{bmatrix}$$ and $$B=\begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$$, then value of $$\alpha$$ for which $${A}^{2}=B$$, is
    Solution
    Given : $$A=\begin{bmatrix} \alpha  & 0 \\ 1 & 1 \end{bmatrix}$$ and $$B=\begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$$

    $${ A }^{ 2 }=\begin{bmatrix} \alpha  & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} \alpha  & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} { \alpha  }^{ 2 } & 0 \\ \alpha +1 & 1 \end{bmatrix}$$

    Now $${ A }^{ 2 }=B\quad $$
    $$\begin{bmatrix} { \alpha  }^{ 2 } & 0 \\ \alpha +1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$$

    $$\Rightarrow { \alpha  }^{ 2 }=1$$ and $$\alpha +1=5$$

    Clearly, no real value of $$\alpha$$ exist.
  • Question 4
    1 / -0
    If $$A = \begin{bmatrix}1 &3 \\ 3 & 4\end{bmatrix}$$ and $$A^{2} - kA - 5I_{2} = 0$$, then the value of $$k$$ is
    Solution
    Given, $$A = \begin{bmatrix}1 &3 \\ 3 & 4\end{bmatrix}$$
    Now, $$A^{2} - 5I_{2} = \begin{bmatrix}1 &3 \\ 3 & 4\end{bmatrix}\begin{bmatrix}1 &3 \\ 3 & 4\end{bmatrix} - 5\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$
    $$= \begin{bmatrix}1 + 9 & 3 + 12 \\ 3 + 12 & 9 + 16\end{bmatrix} - \begin{bmatrix}5 &0 \\ 0 & 5\end{bmatrix}$$
    $$= \begin{bmatrix}5 &15 \\ 15 & 20\end{bmatrix} = 5\begin{bmatrix}1 &3 \\ 3 & 4\end{bmatrix}$$
    $$\Rightarrow A^{2} - 5I_{2} = 5A$$
    But it is given that $$A^{2} - 5I_{2} = kA$$ 
    $$k = 5$$. 
  • Question 5
    1 / -0
    If $$A$$ is a non zero square matrix of order $$n$$ with $$det\left( I+A \right) \neq 0$$, and $${A}^{3}=0$$, where $$I,O$$ are unit and null matrices of order $$n\times n$$ respectively, then $${ \left( I+A \right)  }^{ -1 }=$$
    Solution
    $$det(I+A)\neq 0$$
    $$A^3=0$$   where $$0$$ is null matrix, $$I$$ is the identity matrix
    $$A^3+I=I$$ [adding $$I$$ on both sides]
    $$(A+I)(A^2-IA+I^2)=I$$ [by the formula of $$a^3+b^3$$]
    $$(A+I)(A^2-A+I)=I$$
    $$(I+A)(I+A)^{-1}=I$$ [by the rule of inverse matrix]
    hence $$(I+A)^{-1}=(A^2-A+I)$$
    Ans: $$I-A+A^2$$
  • Question 6
    1 / -0
    If $$\quad A=\begin{pmatrix} 1 & 3 \\ 4 & 5 \end{pmatrix}$$ then $${ A }^{ -1 }$$ equals
    Solution
    $$A=\begin{pmatrix} 1 & 3 \\ 4 & 5 \end{pmatrix}$$
    for every square matrix $$A$$, the characterisitc of $$A$$ is given $$\left| A-\lambda I \right| =0$$
    $$\Rightarrow \begin{vmatrix} 1-\lambda  & 3 \\ 4 & 5-\lambda  \end{vmatrix}=0$$
    $$\Rightarrow { \lambda  }^{ 2 }-6\lambda -7=0\quad \Rightarrow { A }^{ 2 }-6A-7I=0\Rightarrow 7I={ A }^{ 2 }-6A\quad \quad $$
    $$\Rightarrow { A }^{ -1 }=1/7(A-6I)\quad $$
  • Question 7
    1 / -0
    If $$A = \begin{bmatrix} 1& 4 & 4\\ 4 & 1 & 4\\ 4 & 4 & 1\end{bmatrix}$$, then $$A^{2} - 6A =$$ _____.
    Solution
    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 $$A^2-6A$$ $$=\begin {bmatrix} 33 &24 &24\\24&33&24\\24&24&33 \end {bmatrix}$$ $$-\begin{bmatrix} 6& 24 & 24\\ 24 & 6 & 24\\ 24 & 24 & 6\end{bmatrix}$$
    $$=\begin {bmatrix} 33-6&24-24&24-24 \\24-24&33-6&24-24\\24-24 &24-24&33-6 \end {bmatrix}$$
    $$=\begin {bmatrix} 27&0&0\\ 0&27&0\\ 0&0&27\end {bmatrix}$$
    $$= 27I_{3}$$
  • Question 8
    1 / -0
    If $$A = \begin{bmatrix} 2& 3\\ -1 & 2\end{bmatrix}$$, then $$A^{3} + 3A^{2} - 4A + 1$$ is equal to
    Solution
    $$A = \begin{pmatrix} 2& 3\\ -1 & 2\end{pmatrix}$$
    $$\therefore A^{2} = \begin{pmatrix} 1& 12\\ -4 & 1\end{pmatrix}, A^{3} = \begin{pmatrix} -10& 27\\ -9 & -10\end{pmatrix}$$
    $$\therefore A^{3} + 3A^{2} - 4A + t = \begin{pmatrix} -10& 27\\ -9 & -10\end{pmatrix} + \begin{pmatrix} 3& 36\\ -12 & 3\end{pmatrix} + \begin{pmatrix} -8& -12\\ 4 & -8\end{pmatrix} + \begin{pmatrix} 1& 0\\ 0 & 1\end{pmatrix}$$
    $$= \begin{pmatrix} -14& 51\\ -17 & -14\end{pmatrix}$$
    Hence (b) is correct choice. 
  • Question 9
    1 / -0
    If $$A=\begin{bmatrix} 0 & 0 \\ 0 & 5 \end{bmatrix}$$, then $${ A }^{ 12 }$$ is
    Solution
    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
    $$A^{12}=\begin{bmatrix}0&0\\0&5^{12}\end{bmatrix}$$
  • Question 10
    1 / -0
    If $$A=\begin{pmatrix} 3 & 1 \\ -9 & -3 \end{pmatrix}$$ then $${ \left( 1+2A+3{ A }^{ 2 }+....\infty  \right)  }^{ -1 }$$ equals
    Solution
    $$A=\begin{pmatrix} 3 & 1 \\ -9 & -3 \end{pmatrix}$$
    $$\quad \therefore { A }^{ 2 }=\begin{pmatrix} 3 & 1 \\ -9 & -3 \end{pmatrix}\begin{pmatrix} 3 & 1 \\ -9 & -3 \end{pmatrix}=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$

    $$\therefore 1+2A+3{ A }^{ 2 }+....\infty =1+2A$$
    $$=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}+\begin{pmatrix} 6 & 2 \\ -18 & -6 \end{pmatrix}=\begin{pmatrix} 7 & 2 \\ -18 & -5 \end{pmatrix}$$

    $$\therefore (1+2A+3{ A }^{ 2 }+....\infty)^{-1}=(1+2A)^{-1} $$
    $$\quad ={ \begin{pmatrix} 7 & 2 \\ -18 & -5 \end{pmatrix} }^{ -1 }=\begin{pmatrix} -5 & -2 \\ 18 & 7 \end{pmatrix}$$
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