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Matrices Test - 28

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Matrices Test - 28
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  • Question 1
    1 / -0
    If [x−3−9y][4−397]=[1001],\begin{bmatrix} x & -3 \\ -9 & y \end{bmatrix}\begin{bmatrix} 4 & -3 \\ 9 & 7 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, then x=x= .......... ,y, y == ........
    Solution
    [x−3−9y][4−397]=[1001]⇒[4x−27−3x+(−21)−36+9y+27+7y]=[1001]\begin{bmatrix} x & -3 \\ -9 & y \end{bmatrix}\begin{bmatrix} 4 & -3 \\ 9 & 7 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\\ \Rightarrow \begin{bmatrix} 4x-27 & -3x+(-21) \\ -36+9y & +27+7y \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
    Using conditions for equality of matrices we can write
    4x−27=1⇒x=7−(i)−(3x+21)=0⇒x=−7−(ii)−36+9y=0⇒y=4−(iii)27+7y=1⇒y=−4−(iv)4x-27=1\\ \Rightarrow x=7\quad \quad -(i)\\ -(3x+21)=0\\ \Rightarrow x=-7\quad \quad -(ii)\\ -36+9y=0\\ \Rightarrow y=4\quad \quad -(iii)\\ 27+7y=1\\ \Rightarrow y=-4\quad \quad -(iv)
    Possible answer is x=7x=7 & y=4y=4
  • Question 2
    1 / -0
    If $$A= \begin{bmatrix}
    2 &  3   \\
    3  & 2      
    \end{bmatrix}, B= \begin{bmatrix}
    2  & 1   \\
    3  & 5      
    \end{bmatrix}and and C= \begin{bmatrix}
    0  & 1   \\
    1  & 2      
    \end{bmatrix},then then \left ( AB \right )\times C=$$
    Solution

  • Question 3
    1 / -0
    If  A=[2113]A=\begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix} , B=[32010 4 ]B=\begin{bmatrix} 3 & 2 & 0 \\ 1 & 0  & 4  \end{bmatrix}, then AB=AB= is
    Solution
    Given A=[2113]A= \begin{bmatrix} 2 &1 \\1 & 3 \end{bmatrix} and B=[320104]B=\begin{bmatrix} 3&2&0\\1&0&4 \end{bmatrix}
    AB=[2×3+1×12×2+1×02×0+1×4 1×3+3×11×2+3×01×0+3×4]AB=\begin{bmatrix} 2\times 3 +1\times1 & 2\times 2+1\times0 & 2\times0+1\times4  \\ 1\times3+3\times1 & 1\times2+3\times0 & 1\times0+3\times4 \end{bmatrix}
            =[6+14+00+43+32+00+12] =\begin{bmatrix} 6+1 & 4+0 & 0+4 \\ 3+3 & 2+0 & 0+12 \end{bmatrix}
            =[7446212]=\begin{bmatrix} 7 & 4 & 4 \\ 6 & 2 & 12 \end{bmatrix}

    Therefore the correct option is (D)(D)
  • Question 4
    1 / -0
    Let  $$A=\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\   and\  B=\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix},  a,b\in N.$$ Then:
    Solution
    Given
    A=[1234]B=[a00b]A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}B=\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}
    Finding AB
    AB=[1234][a00b]=[a+2×01×0+2×b3×a+4×03×0+4×b]AB=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}=\begin{bmatrix} a+2\times 0 & 1\times 0+2\times b \\ 3\times a+4\times 0 & 3\times 0+4\times b \end{bmatrix}
    AB=[a2b3a4b]AB=\begin{bmatrix} a & 2b \\ 3a & 4b \end{bmatrix}
    Finding BA
    BA=[a00b][1234]=[a+0×32a+0×40×1+3b0×2+4b]BA=\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}=\begin{bmatrix} a+0\times 3 & 2a+0\times 4 \\ 0\times 1+3b & 0\times 2+4b \end{bmatrix}
    BA=[a2a3b4b]BA=\begin{bmatrix} a & 2a \\ 3b & 4b \end{bmatrix}
    If AB=BAAB=BA
    [a2b3a4b]=[a2a3b4b]\begin{bmatrix} a & 2b \\ 3a & 4b \end{bmatrix}=\begin{bmatrix} a & 2a \\ 3b & 4b \end{bmatrix}
    Comparing each element we get
    a=a,2b=2aa=a,2b=2a
    ⇒a=b\Rightarrow a=b
    ∴\therefore There are infinitely many b's
    for which AB=BAAB=BA
    Answer=B
  • Question 5
    1 / -0
    If A = $$\begin{bmatrix}
     2\ \ \ 4 \\
     3\ \ 5
    \end{bmatrix}, B= \begin{bmatrix}
     x\ \ \ y \\
     6\ \ \ 5
    \end{bmatrix}and  and  AB= \begin{bmatrix}8\ \ \ 2 \\
     6\ \ \ -2 
    \end{bmatrix}$$
    then x=x= ______, and y=y = _____.
    Solution
    A= [3435] A = \begin{bmatrix} 3 & 4 \\ 3 & 5 \end{bmatrix} ;
    B= [xy65] B = \begin{bmatrix} x & y \\ 6 & 5 \end{bmatrix}
    and AB= [826−2] AB = \begin{bmatrix} 8 & 2 \\ 6 & -2 \end{bmatrix}
    Now, 
    AB= [2435] [xy65]= [2x+242y+203x+303y+25]   [By the property of multiplication of matrix]  AB = \begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix} \begin{bmatrix} x & y \\ 6 & 5 \end{bmatrix} \\ = \begin{bmatrix} 2x+24 & 2y + 20 \\ 3x + 30 & 3y + 25 \end{bmatrix} \, \, \textrm{ [By the property of multiplication of matrix] }

    Now, comparing the given value we get,

     [826−2]= [2x+242y+203x+303y+25] \begin{bmatrix} 8 & 2 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 2x+24 & 2y + 20 \\ 3x+30 & 3y + 25 \end{bmatrix}

    Equating each element we get,

    2x+24=8⇒x=−82y+20=2⇒y=−9 2x + 24 = 8 \Rightarrow x = -8 \\ 2y + 20 = 2 \\ \Rightarrow y = -9
    ∴ \therefore option C is correct.
  • Question 6
    1 / -0
    If A=[130−121002], B=[234123 −112]A=\begin{bmatrix} 1 & 3 & 0 \\ -1 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix},  B=\begin{bmatrix} 2 & 3 & 4 \\ 1 & 2 & 3 \\  -1 & 1 & 2 \end{bmatrix}  then  AB=AB=
    Solution
    Given,
    A=[130−121002]A=\begin{bmatrix} 1 & 3 & 0 \\ -1 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix} and B=[234123−112]B=\begin{bmatrix} 2 & 3 & 4 \\ 1 & 2 & 3 \\ -1 & 1 & 2\end{bmatrix}

    Then, AB=[130−121002][234123−112]AB=\begin{bmatrix} 1 & 3 & 0 \\-1 & 2 & 1\\0&0&2 \end{bmatrix} \begin{bmatrix} 2&3&4 \\1& 2&3 \\-1&1&2 \end{bmatrix}

                      =[1×2+3×1+0×(−1)1×3+3×2+0×11×4+3×3+0×2(−1)×2+2×1+1×(−1)(−1)×3+2×2+1×1(−1)×4+2×3+1×20×2+0×1+2×(−1)0×3+0×2+2×10×4+0×3+2×2]= \begin{bmatrix} 1\times2+3\times1+0\times(-1) & 1\times3+3\times2+0\times1& 1\times4+3\times3+0\times2 \\ (-1)\times2+2\times1+1\times(-1) & (-1)\times3+2\times2+1\times1 & (-1)\times4+2\times3+1\times2 \\ 0\times2+0\times1+2\times(-1) & 0\times3+0\times2+2\times1 & 0\times4+0\times3+2\times2 \end{bmatrix}

                       =[5913−124−224]= \begin{bmatrix} 5 & 9 & 13 \\ -1 & 2 & 4 \\ -2 & 2 & 4\end{bmatrix}

    Hence, the correct option is (B)(B)
  • Question 7
    1 / -0
    A=[122212221 ],then A3−4A2−6A=A=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1  \end{bmatrix}, \text{then} \,A^{3}-4A^{2}-6A=
    Solution
    A2=[122212 221 ][122212 221 ]=[988898 889 ]A^{2}=\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}=\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8  \\ 8 & 8 & 9  \end{bmatrix}

    A3=A2×A=[988898 889 ][122212 221 ]=[414242424142 424241 ]A^{3}=A^{2}\times A=\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8  \\ 8 & 8 & 9  \end{bmatrix}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2  \\ 2 & 2 & 1  \end{bmatrix}=\begin{bmatrix} 41 & 42 & 42 \\ 42 & 41 & 42  \\ 42 & 42 & 41  \end{bmatrix}

    A3−4A2−6A=[414242424142 424241 ]−[363232323632 323236 ][6121212612 12126 ]A^{3}-4A^{2}-6A=\begin{bmatrix} 41 & 42 & 42 \\ 42 & 41 & 42  \\ 42 & 42 & 41  \end{bmatrix}-\begin{bmatrix} 36 & 32 & 32 \\ 32 & 36 & 32  \\ 32 & 32 & 36  \end{bmatrix}\begin{bmatrix} 6 & 12 & 12 \\ 12 & 6 & 12  \\ 12 & 12 & 6  \end{bmatrix}

    =[−1−2−2−2−1−2 −2−2−1 ]=−A=\begin{bmatrix} -1 & -2 & -2 \\ -2 & -1 & -2  \\ -2 & -2 & -1  \end{bmatrix}=-A
  • Question 8
    1 / -0
    If  A=[12 x0  1 00  0 1 ]and  B[1−2 y0  1 00  0 1 ]\displaystyle \:A= \left [ \begin{matrix}1 &2  &x \\0  &1  &0 \\0  &0  &1 \end{matrix} \right ]and \  B\left [ \begin{matrix}1 &-2  &y \\0  &1  &0 \\0  &0  &1 \end{matrix} \right ] and   AB=I,\displaystyle \:AB= I, then x+yx+y equals 
    Solution
     A=[12 x0  1 00  0 1 ]\displaystyle \:A= \left [ \begin{matrix}1 &2  &x \\0  &1  &0 \\0  &0  &1 \end{matrix} \right ]  and B=[1−2 y0  1 00  0 1 ]B=\left [ \begin{matrix}1 &-2  &y \\0  &1  &0 \\0  &0  &1 \end{matrix} \right ]  
    Also given,  AB=I\displaystyle \:AB= I
    [12x010001][1−2y010001]=[100010001]\left[ \begin{matrix} 1 & 2 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & -2 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] =\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

    ⇒[10x+y010001]=[100010001]\Rightarrow \begin{bmatrix} 1 & 0 & x+y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
    ⇒x+y=0\Rightarrow x+y=0
    Hence, option 'A' is correct.
  • Question 9
    1 / -0
    If A=[1tanx−tanx1]A=\begin{bmatrix} 1 & tanx \\ -tanx & 1 \end{bmatrix}, then ATA−1{ A }^{ T }{ A }^{ -1 } is
    Solution
    A=[1tanx−tanx1]A=\begin{bmatrix} 1 & tanx \\ -tanx & 1 \end{bmatrix}

    ⇒∣A∣=sec⁡2x\Rightarrow |A|=\sec^{2}x

    adjA=CT=[1tan⁡x−tan⁡x1]Tadj A=C^{T}={\begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}}^{T}

    ⇒adjA=[cos⁡2x −sin⁡xcos⁡x sin⁡xcos⁡x cos⁡2x ]\Rightarrow adj A=\begin{bmatrix} \cos ^{ 2 }{ x }  & -\sin { x } \cos { x }  \\ \sin { x } \cos { x }  & \cos ^{ 2 }{ x }  \end{bmatrix}

    A−1=adjA∣A∣A^{-1}=\displaystyle \frac {adj A}{|A|}
                 =[cos⁡2x −sin⁡xcos⁡x sin⁡xcos⁡x cos⁡2x ]=\begin{bmatrix} \cos ^{ 2 }{ x }  & -\sin { x } \cos { x }  \\ \sin { x } \cos { x }  & \cos ^{ 2 }{ x }  \end{bmatrix}

    Now, ATA−1=[1−tan⁡xtan⁡x1] [cos⁡2x −sin⁡xcos⁡x sin⁡xcos⁡x cos⁡2x ]A^{T}A^{-1}=\begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \begin{bmatrix} \cos ^{ 2 }{ x }  & -\sin { x } \cos { x }  \\ \sin { x } \cos { x }  & \cos ^{ 2 }{ x }  \end{bmatrix}

    ⇒ ATA−1=[cos⁡2x −sin⁡2x sin⁡2x cos⁡2x ]\Rightarrow A^{T}A^{-1}=\begin{bmatrix} \cos { 2x }  & -\sin { 2x }  \\ \sin { 2x }  & \cos { 2x }  \end{bmatrix}


  • Question 10
    1 / -0
    If [2−110−34]A=[−1−8−101−2−592215]\begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix}A=\begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}, then sum of all the elements of matrix AA is
    Solution
     [2−110−34]A=[−1−8−101−2−592215]\begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix}A=\begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}
    let A=[abcdef] A=\begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix}
    ⇒[2−110−34][abcdef]=[−1−8−101−2−592215]\Rightarrow \begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix}\begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix}=\begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}
    ⇒[2a−d2b−e2c−fabc−3a+4d−3b+4e−3c+4f]=[−1−8−101−2−592215]\Rightarrow \begin{bmatrix} 2a-d & 2b-e & 2c-f \\ a & b & c \\ -3a+4d & -3b+4e & -3c+4f \end{bmatrix}=\begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}
    ⇒2a−d=−1,2b−e=−8,2c−f=−10\Rightarrow 2a-d=-1,2b-e=-8,2c-f=-10 -----(1)
    and a=1,b=−2,c=−5a=1,b=-2,c=-5 ----(2)
    from (1) and (2)
    a+b+c=−6a+b+c=-6 and d+e+f=7d+e+f=7
    ∴\therefore Sum of all the elements in the matrix a+b+c+d+e+f=1a+b+c+d+e+f=1
    Hence, option B.

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