Tag: multiplicative inverse of a matrix

Questions Related to multiplicative inverse of a matrix

Find the inverse f the following matrices by using transformation method.

business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $\begin{bmatrix}
    1 &2 \
    2 &-1
    \end{bmatrix}$

  2. $\begin{bmatrix}
    2 &-3 \
    -1 &2
    \end{bmatrix}$

  3. $\begin{bmatrix}
    0& 1 &2 \
    1& 2 &3 \
    3& 1 &1
    \end{bmatrix}$

  4. $\begin{bmatrix}
    2& 0 &-1 \
    5& 1 &0 \
    0& 1 &3
    \end{bmatrix}$

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Correct Option: B

If $A=\begin{bmatrix} \cos { x }  & \sin { x }  \ -\sin { x }  & \cos { x }  \end{bmatrix}$ and $A(AdjA)=k\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$ then the value of $k$ is

business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $\sin{x}\cos{x}$

  2. $1$

  3. $-1$

  4. $2$

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Correct Option: B
Explanation:

$A(AdjA)=\begin{bmatrix}c^{2}+s^{2}&cs-cs\-cs+cs&c^{2}+s^{2}\end{bmatrix}=\begin{bmatrix}1&0\0&1\end{bmatrix}\implies k=1$

If A be square matrix of order n and k is a scalar, then adj (KA) is:

business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $K^{n}(adjA)$

  2. K (adj A)

  3. $K^{n-1}(adjA)$

  4. $K^{n+1}(adjA)$

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Correct Option: C

If $A=\left[ \begin{matrix} 2 & -3 \ -4 & 7 \end{matrix} \right] $, then ${2A}^{-1}=$

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $81-2A$

  2. $91-A$

  3. $31-2A$

  4. $A-91$

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Correct Option: B
Explanation:

$A=\left[ { \begin{array} { *{ 20 }{ c } }2 & { -3 } \ { -4 } & 7 \end{array} } \right]  \ \left| { A-\lambda I } \right| =0 \ \left| { \begin{array} { *{ 20 }{ c } }{ 2-\lambda  } & { -3 } \ { -4 } & { 7-\lambda  } \end{array} } \right| =0 \ \left( { 2-\lambda  } \right) \left( { 7-\lambda  } \right) -12=0 \ \left( { \lambda -7 } \right) \left( { \lambda -2 } \right) -12=0 \ { \lambda ^{ 2 } }-9\lambda +2=0 \ { A^{ 2 } }-9A+2I=0 \ A-9I+2{ A^{ -1 } }=0 \ 2{ A^{ -1 } }=9I-0 \ 2{ A^{ -1 } }=9I-A$


$ \ Hence,\, option\, B\, is\, the\, correct\, answer.$

If AB = AC then 

business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. B = C

  2. $B\neq C$

  3. B need not be equal to C

  4. B = -C

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Correct Option: C
Explanation:

Value of B and C depends on the value of A. if A is non-singular, i.e non zero, then B=C. 
But if A=0, then both terms AB and AC becomes zero, so B and C need not be necessarily equal.

$A=\begin{bmatrix} \cos\theta & -\sin\theta \ \sin\theta & \cos\theta\end{bmatrix}$ and $AB=BA=I$, then B is equal to

business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $\begin{bmatrix} -\cos\theta & \sin\theta \ \sin\theta & \cos\theta\end{bmatrix}$

  2. $\begin{bmatrix} \cos\theta & \sin\theta \ -\sin\theta & \cos\theta\end{bmatrix}$

  3. $\begin{bmatrix} -\sin\theta & \cos\theta \ \cos\theta & \sin\theta\end{bmatrix}$

  4. $\begin{bmatrix} \sin\theta & -\cos\theta \ -\cos\theta & \sin\theta\end{bmatrix}$

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Correct Option: B
Explanation:

Given, $A=\begin{bmatrix} \cos\theta & -\sin\theta \ \sin\theta & \cos\theta \end{bmatrix}$ and $AB=BA=I$
$\Rightarrow B=A^{-1}I=A^{-1}$
$=\displaystyle\frac{1}{\cos^2\theta +\sin^2\theta}\begin{bmatrix} \cos\theta & \sin\theta \ -\sin\theta & \cos\theta\end{bmatrix}$
$\Rightarrow B=\begin{bmatrix}\cos\theta & \sin\theta \ -\sin\theta & \cos\theta\end{bmatrix}$

$A=\begin{bmatrix} 2&2&1\0&1&4\0&2&6\end{bmatrix}$, $B=\begin{bmatrix} 2&2&1\0&1&4\0&0&1\end{bmatrix}$

To obtain B from the matrix A, order of operations would be   

business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $R _3 \rightarrow R _3-3R _1$, $R _3\rightarrow R _2-R _1$

  2. $R _3 \rightarrow R _1-2R _2$, $R _3 \rightarrow (R _3 \times {-2})$

  3. $R _2 \rightarrow R _2-2R _2$, $R _3 \rightarrow (R _3 \div {2})$

  4. $R _3 \rightarrow R _3-2R _2$, $R _3 \rightarrow (R _3 \div {-2})$

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Correct Option: D
Explanation:
$A=\begin{bmatrix} 2 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 2 & 6 \end{bmatrix}B=\begin{bmatrix} 2 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}$
From A to B
${ R } _{ 1 }$ & ${ R } _{ 2 }$ are same
changing ${ R } _{ 3 }$ w.r.t. ${ R } _{ 2 }$
Clearly the term $'2'$ in ${ R } _{ 3 }{ C } _{ 2 }$ in A is change to $'0'$ in comparing them we find
${ R } _{ 3 }\rightarrow { R } _{ 3 }-2{ R } _{ 2 }$
We get the matrix
$\begin{bmatrix} 2 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 0 & -2 \end{bmatrix}$
Now if we divide ${ R } _{ 3 }$ w.r.t. $'-2'$
We get the derived term $1$ in ${ R } _{ 3 }{ C } _{ 3 }$ of B
$\begin{bmatrix} 2 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}=B$
$\therefore $Operations are
${ R } _{ 3 }\rightarrow { R } _{ 3 }-2{ R } _{ 2 }$
 and ${ R } _{ 3 }\rightarrow { R } _{ 3 }-(-2)$
Option D

A= $\begin{bmatrix} 1&2&3\4&5&6\7&8&9\end{bmatrix}$. 

B is matrix obtained by subtracting $4 \ times \ 1^{st}\ row\ from \ 2^{nd} \ row$ of A. Find matrix B

business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $\begin{bmatrix} 1&2&3\0&3&6\0&6&12\end{bmatrix}$

  2. $\begin{bmatrix} 1&2&3\7&0&0\4&5&6\end{bmatrix}$

  3. $\begin{bmatrix} 1&2&3\0&1&2\3&4&5\end{bmatrix}$

  4. $\begin{bmatrix} 1&2&3\0&-3&-6\7&8&9\end{bmatrix}$

Show answer
Correct Option: D
Explanation:
Given $A=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$
$\Rightarrow { R } _{ 2 }\rightarrow { R } _{ 2 }-4{ R } _{ 1 }$ (for matrix B)
$\Rightarrow B-\begin{bmatrix} 1 & 2 & 3 \\ 4-4(1) & 5-4(2) & 6-4(3) \\ 7 & 8 & 9 \end{bmatrix}=\begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 7 & 8 & 9 \end{bmatrix}$
Option D is correct

$A=\begin{bmatrix} 2&2&1\4&5&6\6&8&9\end{bmatrix}$, $B=\begin{bmatrix} 2&2&1\0&1&4\0&2&6\end{bmatrix}$

To convert matrix A into matrix B, the order of row operations are 

business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $R _1\rightarrow R _2-2R _1$, $R _3 \rightarrow R _3-R _1$

  2. $R _2\rightarrow R _2-2R _1$, $R _3 \rightarrow R _3-3R _1$

  3. $R _1\rightarrow R _1-2R _2$, $R _3 \rightarrow R _1-R _3$

  4. $R _2\rightarrow R _3-2R _3$, $R _3 \rightarrow R _1-R _1$

Show answer
Correct Option: B
Explanation:
$A=\begin{bmatrix} 2 & 2 & 1 \\ 4 & 5 & 6 \\ 6 & 8 & 9 \end{bmatrix}\quad B=\begin{bmatrix} 2 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 2 & 6 \end{bmatrix}$
Clearly in order to convert matrix A to B. Both ${ R } _{ 2 }$ and ${ R } _{ 3 }$ are changed w.r.t. ${ R } _{ 1 }$
${ R } _{ 2 }^{ 1 }\leftrightarrow X{ R } _{ 2 }-Y{ R } _{ 1 }$
For ${ R } _{ 2 }=4$ and ${ R } _{ 1 }=2$ then ${ R } _{ 2 }^{ 1 }=0$
$\Rightarrow 0=X4-2$Y$
$\Rightarrow 2X=Y$
For ${ R }_{ 2 }=5$ and ${ R }_{ 1 }=1$ then ${ R }_{ 2 }^{ 1 }=1$
$1=X(5)-Y(2)$
$\Rightarrow X=1\quad Y=2$
Relation is ${ R }_{ 2 }\leftrightarrow { R }_{ 2 }-2{ R }_{ 21}$
Parallely ${ R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 }$

Multiply the fourth row by $3$.
$\begin{bmatrix}3&4&2&11\9&1&0&0\0&1&0&2\0&0&6&1\end{bmatrix}$

maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

  1. $0, 0, 18, 3$

  2. $0, 3, 0, 6$

  3. $0, 0, 24, 4$

  4. $9, 12, 6, 33$

  5. $0, 0, 12, 2$

Show answer
Correct Option: A
Explanation:

Fourth row = $0,0,6,1$
multiplying by $3$
$(0\times 3 , 0\times 3 , 6\times 3 , 1\times 3)$
$(0,0,18,3)$