Tag: properties of multiplication of matrix

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Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $A = \begin{bmatrix} -2& 3\ 1 & 1\end{bmatrix}$ then $|A^{-1}| = ?$

  1. $-5$
  2. $\dfrac {-1}{5}$
  3. $\dfrac {1}{25}$
  4. $25$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$AA^{-1} = I\Rightarrow |AA^{-1}| = |I|\Rightarrow |A|\cdot |A^{-1}| = 1\Rightarrow |A^{-1}| = \dfrac {1}{|A|}$.
$|A| = \begin{vmatrix}-2 & 3\ 1 & 1\end{vmatrix} = (-2 - 3) = (-5) \Rightarrow |A^{-1}| = \dfrac {-1}{5}$.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

Let $A$ and $B$ be two $2 \times 2$ matrices. Consider the statements
          $(i)$ $AB =0 \Rightarrow A = 0 :or :B = 0$
         $ (ii)$ $AB =I \Rightarrow A =B^{-1}$
          $(iii)$ $(A + B)^2 = A^2 + 2AB + B^2$

  1. $(i)$ is false, $(ii)$ and $(iii)$ are true
  2. $(i)$ and $(iii)$ are false, $(ii)$ is true
  3. $(i)$ and $(ii)$ are false, $(iii)$ is true
  4. $(ii)$ and $(iii)$ are false, $(i)$ is true
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

(i) is false.


If $A=\begin{bmatrix}0& 1\0 & -1\end{bmatrix}$ and $B =\begin{bmatrix}1& 1\0 & 0\end{bmatrix}$, then

 $AB=\begin{bmatrix}0& 0\0 & 0\end{bmatrix}=0$

Thus,  $AB = 0$ but neither $A = 0$ nor $B = 0$

(iii) is false since matrix multiplication is not commutative.

$(A + B)^2 = A^2 + AB + BA + B^2$

(ii) is true as the product $AB$ is an identity matrix, if and only $B$ is inverse of the matrix $A$.

Hence, option B.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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

  1. $\begin{bmatrix}\dfrac {3}{7}
    &\dfrac {-1}{7} \\
    \dfrac {1}{7} & \dfrac {2}{7}
    \end{bmatrix}$
  2. $\begin{bmatrix}\dfrac {3}{7}
    &\dfrac {1}{7} \\
    \dfrac {-1}{7} & \dfrac {2}{7}
    \end{bmatrix}$
  3. $\begin{bmatrix}\dfrac {3}{7}
    &\dfrac {1}{7} \\
    \dfrac {1}{7} & \dfrac {2}{7}
    \end{bmatrix}$

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

$|A| = \begin{vmatrix}2 & -1\ 1 & 3\end{vmatrix} = (6 + 1) = 7\neq 0$
$M _{11} = 3, M _{12} = 1, M _{21} = -1$ and $M _{22} = 2$


$\therefore C _{11} = 3, C _{12} = -1, C _{21} = 1$ and $C _{22} = 2$

$\Rightarrow Adj\ A = \begin{bmatrix} 3& -1\ 1 & 2\end{bmatrix}^T = \begin{bmatrix} 3& 1\ -1 & 2\end{bmatrix}$

$\Rightarrow A^{-1} = \dfrac {1}{|A|} (adj\ A) = \dfrac {1}{7} \begin{bmatrix}3 & 1\ -1 & 2\end{bmatrix} = \begin{bmatrix}\dfrac {3}{7} & \dfrac {1}{7}\ -\dfrac {1}{7} & \dfrac {2}{7}\end{bmatrix}$.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $A$ and $B$ are two square matrices of the same order and $m$ is a positive integer, then
$(A + B)^m =$ $^mC _0A^m +$ $^mC _1 A^{m -1} B + ^mC _2A^{m-2} B^2 + ... +$ $^mC _{m- 1} AB^{m-1}+$ $^mC _m B^m$ if

  1. $AB =BA$
  2. $AB + BA =0$
  3. $A^m = 0, \:B^m = 0$

  4. none of these.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Binomial theorem is applicable if and only if $AB=BA$.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

Let $A, : B : and : C$ be $2\times 2$ matrices with entries from the set of real numbers. Define $\ast $ as follows: $\displaystyle A\ast B=\frac{1}{2}(AB + BA)$, then

  1. $A\ast B=B\ast A$
  2. $A\ast A=A^2$
  3. $A\ast (B+C)=A\ast B+A\ast C$
  4. $A\ast I=A$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

$\displaystyle A*1=\frac { 1 }{ 2 } \left( AI+IA \right) =\frac { 1 }{ 2 } \left( A+A \right) =A$

$\therefore A*I=A,\therefore$ (a)holds.
$\displaystyle A*A=\frac { 1 }{ 2 } \left( AA+AA \right) =\frac { 1 }{ 2 } \left( { A }^{ 2 }+{ A }^{ 2 } \right) ={ A }^{ 2 }$
$\therefore$ (b) holds.
$\displaystyle \left( c \right) A*B=\frac { 1 }{ 2 } \left( AB+BA \right) B*A=\frac { 1 }{ 2 } \left( BA+AB \right) =\frac { 1 }{ 2 } \left( AB+BA \right) $
$[\because $ addition is commutative$]$
$\therefore A*B=B*A,\therefore$ (c)holds
$\displaystyle A*\left( B+C \right) =\frac { 1 }{ 2 } \left( A\left( B+C \right) +\left( B+C \right) A \right) =\frac { 1 }{ 2 } \left( AB+CA+BA+CA \right)$
$\displaystyle  =\frac { 1 }{ 2 } \left( AB+BA \right) +\frac { 1 }{ 2 } \left( AC+CA \right) =A*B+A*C$
$\therefore$ (d) holds.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $A^k=0$ for some value of $k$ and $B=1+A+A^2+...+A^{k-1},$ then $B^{-1}$ equal

  1. $I-A$
  2. $I+A$
  3. $I-A^{k-1}$

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

Let $\displaystyle B=I+A+{ A }^{ 2 }+{ A }^{ 3 }+...+{ A }^{ k-1 }$

$\displaystyle \therefore B\left( I-A \right) =\left( I+A+{ A }^{ 2 }+...+{ A }^{ k-1 } \right) \left( I-A \right) $
$\displaystyle =I-A+A-{ A }^{ 2 }+{ A }^{ 2 }+...+{ A }^{ k-1 }-{ A }^{ k }=I={ A }^{ k }=I\quad \quad \quad \left( \because { A }^{ k }=O \right) $
Hence, $\displaystyle { \left( I-A \right)  }^{ -1 }=I+A+{ A }^{ 2 }+...+{ A }^{ k }-1=B$
$\displaystyle \Rightarrow { B }^{ -1 }=I-A$