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Matrices Test - 35

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Matrices Test - 35
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  • Question 1
    1 / -0
    The symmetric part of the matrix A= $$\begin{bmatrix}
    1 &2  &4 \\
    6 & 8 & 2\\
    2 & -2 &7
    \end{bmatrix}$$.
    Solution
    Symmetric part of any matrix A is given by $$\dfrac{A+A'}{2}$$, where $$A'$$  transpose of A
    So symmetric part of given matrix $$\begin{bmatrix}1&2&4\\6&8&2\\2&-2&7 \end{bmatrix}$$
    is $$ = \dfrac{1}{2}\bigg ( $$ $$\begin{bmatrix}1&2&4\\6&8&2\\2&-2&7 \end{bmatrix}$$+$$\begin{bmatrix}1&6&2\\2&8&-2\\4&2&7 \end{bmatrix}$$ $$\bigg)$$
    $$=\dfrac{1}{2}$$$$\begin{bmatrix}2&8&6\\8&16&0\\6&0&14 \end{bmatrix}$$
    $$ = \begin{bmatrix}1&4&3\\4&8&0\\3&0&7 \end{bmatrix}$$ 
  • Question 2
    1 / -0
    For a matrix $$A \begin{pmatrix} 1& 0 & 0\\ 2 & 1 & 0\\ 3 & 2 & 1\end{pmatrix}$$, if $$U_{1}, U_{2}$$ and $$U_{3}$$ are $$3\times 1$$ column matrices satisfying $$AU_{1} = \begin{pmatrix}1\\ 0 \\ 0
    \end{pmatrix}, AU_{2} \begin{pmatrix}2\\3 \\ 0
    \end{pmatrix}, AU_{3} = \begin{pmatrix}2\\ 3\\ 1
    \end{pmatrix}$$ and $$U$$ is $$3\times 3$$ matrix whose columns are $$U_{1}, U_{2}$$ and $$U_{3}$$
    Then sum of the elements of $$U^{-1}$$ is
    Solution
    Let $$U_{i} = \begin{pmatrix}a_{i}\\b_{i} \\c_{i} \end{pmatrix} i = 1, 2, 3$$
    $$AU_{1} = \begin{pmatrix}a_{1}\\2a_{1} + b_{1} \\ 3a_{1} + 2b_{1} + c_{1}
    \end{pmatrix} = \begin{pmatrix}1\\0 \\ 0
    \end{pmatrix}$$, So $$a_{1} = 1, b_{1} = -2, c_{1} = 1$$
    $$AU_{2} = \begin{pmatrix}a_{2}\\2a_{2} + b^{2} \\ 3a_{2} + 2b_{2} + c_{2}
    \end{pmatrix} = \begin{pmatrix}2\\ 3\\ 0
    \end{pmatrix}$$,
    So, $$a_{2} = 2, b_{2} = -1, c_{2} = -4$$. Similarly, $$a_{3} = 2, b_{3} = -1, c_{3} = -3$$
    So, $$U = \begin{pmatrix} 1& 2 & 2\\ -2 & -1 & -1\\ 1 & -4 & -3\end{pmatrix}$$. So, sum of elements of $$U^{-1}$$ is zero.
  • Question 3
    1 / -0
    If $$A$$ is a scalar matrix $$kI$$ with scalar $$k\ne 0$$ of order $$3$$, the $${A}^{-1}$$ is:
    Solution
    If $$A$$ is a scalar matrix with scalar $$k$$, then $$A=kI$$
    Thus  $$A^{-1}=(kI)^{-1}=\dfrac{1}{k}I$$

    Note: Inverse of identity matrix is Identity matrix itself 
    And $$(kA)^{-1}=\dfrac{1}{k}I^{-1}=\dfrac1k I$$
  • Question 4
    1 / -0
    If $$A = \begin{bmatrix} 4& -2\\ 6 & -3\end{bmatrix}$$, then $$A^2$$ is
    Solution
    $$A=\begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}$$

    $$ { A }^{ 2 }=\begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}\times\begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}$$ 

    using the multiplication properties of matrices, we get

    $${ A }^{ 2 }=\begin{bmatrix} 4\times4+(-2)\times6\quad \quad\quad\quad  & 4\times(-2)+(-2)\times(-3) \\ 6\times4+(-3)\times6\quad \quad \quad \quad  & 6\times(-2)+(-3)\times(-3) \end{bmatrix}$$

    $${ A }^{ 2 }=\begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}$$
  • Question 5
    1 / -0
    If $$\bigl(\begin{smallmatrix} 1& 2\\ 2 & 1\end{smallmatrix}\bigr) \bigl(\begin{smallmatrix} x \\ y \end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} 2 \\  4 \end{smallmatrix}\bigr)$$, then the values of $$x$$ and $$y$$ respectively, are
    Solution
    $$\left[ \begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 4 \end{matrix} \right] $$
    $$\Rightarrow \left[ \begin{matrix} x & 2y \\ 2x & y \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 4 \end{matrix} \right] $$
    $$\Rightarrow x+2y=2\longrightarrow (1)\;\; \& \;\;2x+y=4\longrightarrow (2)$$
    Multiply $$(1)$$ by 2,$$
    $$\Rightarrow 2x+4y=4\longrightarrow(3)$$
    Solving $$(2)\;\&\;(3),[(2)-(3)],$$ we get
    $$\Rightarrow -3y=0$$
    $$\Rightarrow y=0$$
    Put $$y=0$$ in $$(1),$$ we get
    $$\Rightarrow x+0=2$$
    $$\Rightarrow x=2$$
    $$\Rightarrow y=0$$
    Hence, the answer is $$2,0.$$
  • Question 6
    1 / -0
    If $$A = \bigl(\begin{smallmatrix} 4& -2\\ 6 & -3\end{smallmatrix}\bigr)$$, then $$A^2$$ is
    Solution
    $$A=\left[ \begin{matrix} 4 & -2 \\ 6 & -3 \end{matrix} \right] $$
    $$\Rightarrow { A }^{ 2 }=A.A=\left[ \begin{matrix} 4 & -2 \\ 6 & -3 \end{matrix} \right] \left[ \begin{matrix} 4 & -2 \\ 6 & -3 \end{matrix} \right] $$
                              $$=\left[ \begin{matrix} 16-12 & -8+6 \\ 24-18 & -12+9 \end{matrix} \right] $$
                              $$=\left[ \begin{matrix} 4 & -2 \\ 6 & -3 \end{matrix} \right] $$
    Hence, the answer is $$\left[ \begin{matrix} 4 & -2 \\ 6 & -3 \end{matrix} \right] .$$
  • Question 7
    1 / -0
    If $$A \times \bigl(\begin{smallmatrix} 1& 1\\0  & 2\end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix}1 & 2 \end{smallmatrix}\bigr)$$, then the order of A is
    Solution
    $$A \times \left[ \begin{matrix} 1 & 1 \\ 0 & 2 \end{matrix} \right] =\left[ \begin{matrix} 1 & 2 \end{matrix} \right] $$
    The order of matrix $$A$$ will be $$2\times1.$$
    Hence, the answer is $$2\times1.$$
  • Question 8
    1 / -0
    $$\bigl(\begin{smallmatrix} -1& 0\\ 0 & 1\end{smallmatrix}\bigr) \bigl(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} 1& 0\\ 0 & -1\end{smallmatrix}\bigr)$$, then the values of $$a, b, c $$ and $$d $$ respectively are
    Solution
    $$\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \end{matrix} \right] \left[ \begin{matrix} a & b \\ c & d \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right] $$
    $$\Rightarrow \left[ \begin{matrix} -a & -b \\ c & d \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right] $$
    On comparing both sides, corresponding columns,
    $$\Rightarrow -a=1,$$ $$-b=0$$ and $$c=0$$
            $$a=-1,$$ $$b=0$$ and $$d=-1$$
    Hence, the answer is $$-1,0,0,-1.$$
  • Question 9
    1 / -0
    If $$\bigl(\begin{smallmatrix}a & 3\\ 1 & 2\end{smallmatrix}\bigr) \bigl(\begin{smallmatrix} 2 \\ -1 \end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} 5\\ 0 \end{smallmatrix}\bigr)$$, then the value of $$a$$ is
    Solution
    $$\left[ \begin{matrix} a & 3 \\ 1 & 2 \end{matrix} \right] \left[ \begin{matrix} 2 \\ -1 \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 0 \end{matrix} \right] $$
    $$\Rightarrow \left[ \begin{matrix} 2a-3 \\ 0 \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 0 \end{matrix} \right] $$
    $$\Rightarrow 2a-3=5$$
    $$\Rightarrow 2a=8$$
    $$\Rightarrow a=4$$
  • Question 10
    1 / -0
    If $$A=\left[ \begin{matrix} 1 & -2 & 3 \end{matrix} \right] $$ and $$B=\left[ \begin{matrix} -1 \\ 2 \\ -3 \end{matrix} \right] $$, then $$A + B$$ is
    Solution
    Given, $$A=\left[ \begin{matrix} 1 & -2 & 3 \end{matrix} \right] $$ and $$B=\left[ \begin{matrix} 1 \\ -2 \\ 3 \end{matrix} \right] $$
    Here $$A+B$$  is not possible because the order of the matrices is not same.
    Hence, the answer is not defined.
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