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Matrices Test - 34

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Matrices Test - 34
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  • Question 1
    1 / -0
    If A=[xyz],B= [ahghbfgfc]A=[x\, y\, z], B = \begin{bmatrix}a&h&g\\h&b&f\\g&f&c\end{bmatrix} and C=[xyz]C=\begin{bmatrix}x\\y\\z\end{bmatrix}
    Then, ABC=0ABC = 0, if
    Solution
    A=[xyz]B=[ahghbfgfc]c=[xyz] A = [x\,y\,z] B = \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix} c = \begin{bmatrix} x \\ y \\ z \end{bmatrix}
    A.B=[xyz][ahghbfgfc] A.B = [x\,y\,z] \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}
    =[ax+hy+gz    hx+by+fz    gx+fy+cz] = [ax+hy+gz\,\,\,\,hx+by+fz\,\,\,\,gx+fy+cz]
    A.B.C=(AB)×[xyz]=[ax2+hyx+gyx+hxy+by2+fzy+gxz+fyz+cz2] A.B.C = (AB)\times \begin{bmatrix} x \\ y \\ z\end{bmatrix} = [ax^{2}+hyx+gyx+hxy+by^{2}+fzy+gxz+fyz+cz^{2}]
    ABC=[ax2+by2+cz2+2hxy+2gzx+2fzy] ABC = [ax^{2}+by^{2}+cz^{2}+2hxy+2gzx+2fzy]  
    ABC=0 ABC = 0  
    when,
    ax2+by2+cz2+2hxy+2gzx+2fzy=0 ax^{2}+by^{2}+cz^{2}+2hxy+2gzx+2fzy = 0  
  • Question 2
    1 / -0
    A=[2412 ]A=\begin{bmatrix} -2&4\\-1&2  \end{bmatrix}, then A2A^2 is equal to 
    Solution
    A=[2412 ]A=\begin{bmatrix} -2&4\\-1&2  \end{bmatrix}
    A.A.=A2=[2412 ][2412 ]=[44  2×4+4×21x2+2x1  1×4+2×2 ]A.A.=A^{2}= \begin{bmatrix} -2&4\\-1&2  \end{bmatrix} \begin{bmatrix} -2&4\\-1&2  \end{bmatrix} = \begin{bmatrix} 4-4\,\,-2\times 4+4\times 2\\-1x-2+2x-1\,\,-1\times 4+2\times 2  \end{bmatrix}
    A2=[0000 ]=O= A^{2}=\begin{bmatrix} 0&0\\0&0  \end{bmatrix} = O = null matrix.
  • Question 3
    1 / -0
    If P=(224134123)P=\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}, then P5{P}^{5} is equal to
    Solution
    Given P=(224134123)P=\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}

    P2=P.P=(224134123)(224134123){P}^{2}=P.P=\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}

    =(4+2446+888+1223+42+984+12122+2326+648+9)=\begin{pmatrix} 4+2-4 & -4-6+8 & -8-8+12 \\ -2-3+4 & 2+9-8 & 4+12-12 \\ 2+2-3 & -2-6+6 & -4-8+9 \end{pmatrix}

    =(224134123)=P=\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}=P

    \therefore P4=P2=P{P}^{4}={P}^{2}=P
     
    \Rightarrow P5=P2=P{P}^{5}={P}^{2}=P
  • Question 4
    1 / -0
    If A=[2112 ]A =\begin{bmatrix} 2 & -1 \\ -1 & 2  \end{bmatrix} and II is the unit matrix of order 22, then A2A^2 equals
    Solution
    A=[2112]A2=A×A=[2112]×[2112]=[4+122221+4]=[5445]A=\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\\ { A }^{ 2 }=A\times A=\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\times \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\\ \quad \quad =\begin{bmatrix} 4+1 & -2-2 \\ -2-2 & 1+4 \end{bmatrix}=\begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}
    From the given option to get first entry i.e 55 of A2A^2, only option (A)(A) is satisfied. Or we can check as below
    4A3I=[8448]3×[1001]=[8448][3003]=[5445]=A24A-3I=\begin{bmatrix} 8 & -4 \\ -4 & 8 \end{bmatrix}-3\times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\\ \quad \quad \quad \quad =\begin{bmatrix} 8 & -4 \\ -4 & 8 \end{bmatrix}-\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\\ \quad \quad \quad \quad =\begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}={ A }^{ 2 }
  • Question 5
    1 / -0
    If AA is a matrix such that (2132)A(1  1)=(1100)\begin{pmatrix}2&1 \\3&2 \end{pmatrix} A(1 \space  \,1)=\begin{pmatrix}1&1 \\0&0 \end{pmatrix} then A=A=
    Solution
    (2132)A(11)=(1100)\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}A \begin{pmatrix} 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}
    To maintain the rules of matrix multiplication, AA has to be of the order 2×12 \times 1
    (2132)(xy)(11)=(1100)\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \begin{pmatrix} 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}

    (2x+y3x+2y)(11)=(1100)\Rightarrow \begin{pmatrix} 2x + y \\ 3x + 2y \end{pmatrix} \begin{pmatrix} 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}

    (2x+y2x+y3x+2y3x+2y)=(1100)\Rightarrow \begin{pmatrix} 2x + y & 2x + y \\ 3x + 2y & 3x + 2y \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}

    2x+y=1\Rightarrow 2x + y = 1 and 3x+2y=03x + 2y = 0
    4x+2y(3x+2y)=20\Rightarrow 4x + 2y - (3x + 2y) = 2 - 0
    x=2\Rightarrow x = 2 and y=3y = -3
  • Question 6
    1 / -0
    If AA and BB are any two matrices, then 
    Solution
                
    Step 1: Finding the order\textbf{Step 1: Finding the order}
                    Matrix multiplication is only defined for two matrices whose order is of the form\text{Matrix multiplication is only defined for two matrices whose order is of the form} 
                    a×b and b×c respectivelya\times b \text{ and }b\times c \text{ respectively}
                    Since we cant say anything about the order of the given matrices it may or may not be defined\text{Since we cant say anything about the order of the given matrices it may or may not be defined}
    Hence, AB may or may not be defined\textbf{Hence, AB may or may not be defined}
  • Question 7
    1 / -0
    For the matrices A=[1301]A=\begin{bmatrix} 1 & 3 \\ 0 & -1 \end{bmatrix} B=[2214]B=\begin{bmatrix} 2 & 2 \\ -1 & 4 \end{bmatrix}
    What is ABAB?
    Solution
    AA and BB are given by A=[1301],B=[2214]A=\left[ \begin{matrix} 1 & 3 \\ 0 & -1 \end{matrix} \right] , B=\left[ \begin{matrix} 2 & 2 \\ -1 & 4 \end{matrix} \right]
    Thus, AB=[(1×2)+(3×1)(1×2)+(3×4)(0×2)+(1×1)(0×2)+(1×4)]AB=\left[ \begin{matrix} (1\times 2)+(3\times -1) & (1\times 2)+(3\times 4) \\ (0\times 2)+(-1\times -1) & (0\times 2)+(-1\times 4) \end{matrix} \right]
    =[232+120+104]=\left[ \begin{matrix} 2-3 & 2+12 \\ 0+1 & 0-4 \end{matrix} \right]
    =[11414]=\left[ \begin{matrix} -1 & 14 \\ 1 & -4 \end{matrix} \right] 
  • Question 8
    1 / -0
    If A=[ 011 0]A = \begin{bmatrix} 0& 1\\ 1 & 0\end{bmatrix}, then A2A^{2} is equal to ______
    Solution
    We have, A=[ 011 0]A = \begin{bmatrix} 0& 1\\ 1 & 0\end{bmatrix}

    Hence A2=AA=[0110][0110]A^2=AA=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix} 0&1\\1&0\end{bmatrix} 
    =[0×0+1×10×1+1×01×0+0×11×1+0×0]\quad = \begin{bmatrix}0\times 0+1\times 1&0\times 1+1\times 0\\1\times 0+0\times 1 &1\times 1+0\times 0\end{bmatrix}
    =[1001]=\begin{bmatrix}1&0\\0&1\end{bmatrix}
  • Question 9
    1 / -0
    [ 321 24 5] [ 0ab c ]= [4 94 710 19]\begin{bmatrix} 3& 2\\ -1 & -2\\ 4 & 5\end{bmatrix} \begin{bmatrix} 0& a\\ b & c \end{bmatrix} = \begin{bmatrix}-4 &9 \\ 4 & -7\\ -10 & 19\end{bmatrix}
    Find a,ba, b and cc.
    Solution
    The product is [321245][0abc]=\begin{bmatrix} 3 & 2 \\ -1 & -2 \\ 4 & 5 \end{bmatrix}\begin{bmatrix} 0 & a \\ b & c \end{bmatrix}= [2b3a+2c2b2ca5b5c+4a]\begin{bmatrix} 2b & 3a+2c \\ -2b & -2c-a \\ 5b & 5c+4a \end{bmatrix}
    By given, we have

     [2b3a+2c2b2ca5b5c+4a]\begin{bmatrix} 2b & 3a+2c \\ -2b & -2c-a \\ 5b & 5c+4a \end{bmatrix} =[49471019]=\begin{bmatrix} -4 & 9 \\ 4 & -7 \\ -10 & 19 \end{bmatrix}

    By comparing, we get 
    b=2,a=1,c=3b=-2 , a = 1 , c = 3
  • Question 10
    1 / -0
    If P=(224134123)P=\begin{pmatrix}2&-2&-4 \\-1&3&4\\1&-2&-3\end{pmatrix} then P5P^5 equals
    Solution
    P=[224134123]P=\begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}
    P2=[224134123][224134123]{ P }^{ 2 }=\begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}\begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}
    =[224134123]=P=\begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}=P
    Now,
    P2=PP5=P2×P3P5=P4{ P }^{ 2 }=P\quad { P }^{ 5 }={ P }^{ 2 }\times { P }^{ 3 }\Rightarrow { P }^{ 5 }={ P }^{ 4 }
    (P2=P)P5=P2×P2\left( \because { P }^{ 2 }=P \right) \Rightarrow { P }^{ 5 }={ P }^{ 2 }\times { P }^{ 2 }
    P2=P×P{ P }^{ 2 }=P\times P
    P5=P2=P\Rightarrow { P }^{ 5 }={ P }^{ 2 }=P
    P5=P\therefore { P }^{ 5 }=P
    \therefore Option A is correct
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