Tag: properties of inverses of matrices

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Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

If A and B are invertible matrices, which one of the following statement is/are correct 

  1. $Adj(A) = |A|A^{-1}$
  2. $det(A^{-1}) =|det(A)|^{-1}$
  3. $(A + B)^{-1}= B^{-1 }+ A^{-1}$
  4. $(AB)^{-1} = B^{-1}A^{-1}$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation
Option A
${ A }^{ -1 }=\cfrac { AdjA }{ \left| A \right|  } $
$\Rightarrow AdjA=\left| A \right| { A }^{ -1 }$
Option A is true

Option B
$det\left( AB \right) =\left( detA \right) \left( detB \right) $
$\Rightarrow A{ A }^{ -1 }=I$
$det\left( A{ A }^{ -1 } \right) =detI$
$\Rightarrow detA\left( det{ A }^{ -1 } \right) =1$
$\Rightarrow det{ A }^{ -1 }={ \left( detA \right)  }^{ -1 }$
Option B is true

Option C
${ \left( A+B \right)  }^{ -1 }={ A }^{ -1 }+{ B }^{ -1 }$
Option C is true

Option D
${ \left( AB \right)  }^{ -1 }=?$
$AB\left( { B }^{ -1 }{ A }^{ -1 } \right) =A\left( B{ B }^{ -1 } \right) { A }^{ -1 }$
$=AI{ A }^{ -1 }=\left( A{ A }^{ -1 } \right) =I$
$\Rightarrow { B }^{ -1 }{ A }^{ -1 }={ \left( AB \right)  }^{ -1 }$
Option D is true
Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

If $A=\begin{bmatrix} 1 & -2 \ 3 & 0 \end{bmatrix}$, $B=\begin{bmatrix} -1 & 4 \ 2 & 3 \end{bmatrix}$, and $ABC=\begin{bmatrix} 4 & 8 \ 3 & 7 \end{bmatrix}$, then $C$ equals

  1. $\cfrac { 1 }{ 66 } \begin{bmatrix} 54 & 110 \\ 3 & 11 \end{bmatrix}$
  2. $\cfrac { 1 }{ 66 } \begin{bmatrix} -54 & -110 \\ 3 & 11 \end{bmatrix}$
  3. $\cfrac { 1 }{ 66 } \begin{bmatrix} -54 & 110 \\ 3 & -11 \end{bmatrix}$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given ABC = D, then C = (AB)^-1 * D. Calculate AB first, then find its inverse, and multiply by the matrix D.

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

The value of $(\mathrm{A}$dj $\mathrm{A})^{-1}$ is equal to 

  1. $\mathrm{A}$dj $(\mathrm{A}^{-1})$
  2. $\mathrm{A}$dj $[-\mathrm{A}]$
  3. $(\mathrm{A}$dj$\mathrm{A})^{\mathrm{T}}$
  4. $\mathrm{A}$dj $(\mathrm{A}^{\mathrm{T}})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The property (Adj A)^-1 = Adj(A^-1) is a standard identity in matrix algebra.

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

. $\mathrm{If}$ $\mathrm{A}$ is non-singular matrix such that $A^{2}=A^{-1}$ then $adjA=$ 

  1. $\mathrm{A}$
  2. $\mathrm{A}^{-1}$
  3. $\mathrm{A}^{3}$
  4. $(\mathrm{A}^{-1})^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$A^{2}=A^{-1}$

$A.A^{2}=A.A^{-1}$
$A^{3}=I$
$detA.A^{3}=detA.I$
$detA.A^{3}=A.adjA$
$detA.A^{2}=adjA$
Therefore 
$adjA=A^{-1}$.

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

Let A and B be two non-singular matrices which commute. The $A^{-1}$, $B^{-1}$

  1. do not commute

  2. commute
  3. $AB = A^{-1}B^{-1}$
  4. $(AB)^{-1}=AB$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$A^{-1}B^{-1}=(BA)^{-1}=(AB)^{-1}$ since $A,B$ commute
$\Rightarrow A^{-1}B^{-1}=B^{-1}A^{-1}$
Hence $A^{-1}, B^{-1}$ also commute

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

$\mathrm{A}\mathrm{B}\mathrm{A^{-1}}$ $=\mathrm{X}$ then $\mathrm{B}^{2}=$

  1. $\mathrm{x}^{2}$
  2. $\mathrm{A}\mathrm{x}\mathrm{A}^{-1}$
  3. $\mathrm{A}\mathrm{x}^{2}\mathrm{A}^{-1}$
  4. $\mathrm{A}^{-1}\mathrm{x}^{2}\mathrm{A}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$ABA^{-1}=X$
$\Rightarrow (ABA^{-1})(ABA^{-1})=XX=X^2$
$\Rightarrow ABAA^{-1}BA^{-1} =X^2$
$\Rightarrow ABBA^{-1} =X^2[\because AA^{-1}=I]$
$\Rightarrow AB^2A^{-1} =X^2$
Pre and post multiplying both sides with $A^{-1}$ and $A$ respectively we get,
$\Rightarrow B^2 =A^{-1}X^2A$

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

If $A = \begin{bmatrix} 2 & 3\ 5 & 1 \end{bmatrix},$ then find $A^{-1}$

  1. $\begin{bmatrix} - \frac {1}{13} & \frac {3}{13}\\ - \frac {5}{13} & - \frac {2}{13} \end{bmatrix}$
  2. $\begin{bmatrix} - \frac {1}{13} & \frac {3}{13}\\ \frac {5}{13} & \frac {2}{13} \end{bmatrix}$
  3. $\begin{bmatrix} - \frac {1}{13} & \frac {3}{13}\\ \frac {5}{13} & - \frac {2}{13} \end{bmatrix}$
  4. $\begin{bmatrix} \frac {1}{13} & \frac {3}{13}\\ \frac {5}{13} & - \frac {2}{13} \end{bmatrix}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$A=\begin{bmatrix} 2 & 3\\ 5 & 1\end{bmatrix}$

Cofactor matrix of $A=\begin{bmatrix} 1 & -5\\ -3 & 2\end{bmatrix}$

adj. $A=\begin{bmatrix} 1 & -5\\ -3 & 2\end{bmatrix}'=\begin{bmatrix} 1 & -3\\ -5 & 2\end{bmatrix}$

$|A|=2\times 1-(3\times 5)$

$=2-15=-13$

$A^{-1}=\dfrac{adj. A}{|A|}=\dfrac{-1}{13}\begin{bmatrix} 1 & -3\\ -5 & 2\end{bmatrix}$

$=\begin{bmatrix} \dfrac{-1}{13} & \dfrac{3}{13}\\ \dfrac{5}{13} & \dfrac{-2}{13}\end{bmatrix}$.
Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

If $A$ and $B$ are two non singular matrices of the same order such that ${ B }^{ r }=I$, for some positive integer $r>1$, then ${ A }^{ -1 }{ B }^{ r-1 }{ A }-{ A }^{ -1 }{ B }^{ -1 }A=$

  1. $I$
  2. $2I$
  3. $O$
  4. $-I$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

${ A }^{ -1 }{ B }^{ r-1 }{ A }-{ A }^{ -1 }{ B }^{ -1 }A$

$=A^{-1}B^rB^{-1}A - A^{-1}B^{-1}A $
$={ A }^{ -1 }I{ B }^{ -1 }{ A }-{ A }^{ -1 }{ B }^{ -1 }A       (\because B^{r}=I)$
$={ A }^{ -1 }{ B }^{ -1 }{ A }-{ A }^{ -1 }{ B }^{ -1 }A$
$= O$

Hence, option C.