Tag: business maths

Quizzes (44) Questions (670)

Questions Related to business maths

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

The inverse of the matrix $\begin{bmatrix}3 & 5 & 7 \ 2 & -3 & 1 \ 1 & 1 & 2\end{bmatrix}$ is $\begin{bmatrix}7 & -3 & 26 \ 3 & 1 & 11 \ -5 & -2 & 0\end{bmatrix}$.
State true or false.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To check if the given matrix is the inverse, multiply it with the original matrix. The product of a matrix and its true inverse must yield the identity matrix. Since the resulting product is not the identity matrix, the statement is false.

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

 In matrices $AB = O$ does not necessarily mean that 

  1. $A=0$
  2. $B=0$
  3. Both $ A = 0$ and $B=0$

  4. all of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $\quad A = \begin{bmatrix}1 & -1 & 1 \ -3 & 2 & -1

\ -2 & 1 & 0\end{bmatrix}$ and $B = \begin{bmatrix}1&2

& 3\2&4&6 \ 1&2 &3\end{bmatrix}$

$\therefore\quad

AB = \begin{bmatrix}1 & -1 & 1 \ -3 & 2 & -1 \ -2

& 1 & 0\end{bmatrix}\times\begin{bmatrix}1&2 &

3\2&4&6 \ 1&2 &3\end{bmatrix}$

$\quad                   = \begin{bmatrix}0&0&0 \ 0&0&0 \ 0&0&0\end{bmatrix} = O$

$\therefore \quad AB = O$
But neither $A = O$ nor $B = O$.

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

If inverse of $A=\left[ \begin{matrix} 1 & 1 & 1 \ 2 & -1 & -1 \ 1 & -1 & 1 \end{matrix} \right] $ is $\cfrac { -1 }{ 6 } \left[ \begin{matrix} -2 & -2 & 0 \ -3 & 0 & \alpha  \ -1 & 2 & -3 \end{matrix} \right] $ then $\alpha=$

  1. $0$
  2. $-3$
  3. $3$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given,

$A=\begin{bmatrix}1&1&1\\ 2&-1&-1\\ 1&-1&1\end{bmatrix}$

$A^{-1}=\begin{bmatrix}1&1&1\\ 2&-1&-1\\ 1&-1&1\end{bmatrix}^{-1}$

$=\begin{bmatrix}1&1&1&\mid \:&1&0&0\\ 2&-1&-1&\mid \:&0&1&0\\ 1&-1&1&\mid \:&0&0&1\end{bmatrix}$

$\:R _1\:\leftrightarrow \:R _2$

$=\begin{bmatrix}2&-1&-1&\mid \:&0&1&0\\ 1&1&1&\mid \:&1&0&0\\ 1&-1&1&\mid \:&0&0&1\end{bmatrix}$

$R _2\:\leftarrow \:R _2-\frac{1}{2}\cdot \:R _1$

$R _3\:\leftarrow \:R _3-\frac{1}{2}\cdot \:R _1$

$=\begin{bmatrix}2&-1&-1&\mid \:&0&1&0\\ 0&\frac{3}{2}&\frac{3}{2}&\mid \:&1&-\frac{1}{2}&0\\ 0&-\frac{1}{2}&\frac{3}{2}&\mid \:&0&-\frac{1}{2}&1\end{bmatrix}$

$R _3\:\leftarrow \:R _3+\frac{1}{3}\cdot \:R _2$

$=\begin{bmatrix}2&-1&-1&\mid \:&0&1&0\\ 0&\frac{3}{2}&\frac{3}{2}&\mid \:&1&-\frac{1}{2}&0\\ 0&0&2&\mid \:&\frac{1}{3}&-\frac{2}{3}&1\end{bmatrix}$

$R _3\:\leftarrow \frac{1}{2}\cdot \:R _3$

$R _2\:\leftarrow \:R _2-\frac{3}{2}\cdot \:R _3$

$=\begin{bmatrix}2&-1&-1&\mid \:&0&1&0\\ 0&\frac{3}{2}&0&\mid \:&\frac{3}{4}&0&-\frac{3}{4}\\ 0&0&1&\mid \:&\frac{1}{6}&-\frac{1}{3}&\frac{1}{2}\end{bmatrix}$

$R _1\:\leftarrow \:R _1+1\cdot \:R _3$

$R _2\:\leftarrow \frac{2}{3}\cdot \:R _2$

$=\begin{bmatrix}2&-1&0&\mid \:&\frac{1}{6}&\frac{2}{3}&\frac{1}{2}\\ 0&1&0&\mid \:&\frac{1}{2}&0&-\frac{1}{2}\\ 0&0&1&\mid \:&\frac{1}{6}&-\frac{1}{3}&\frac{1}{2}\end{bmatrix}$

$R _1\:\leftarrow \:R _1+1\cdot \:R _2$

$=\begin{bmatrix}2&0&0&\mid \:&\frac{2}{3}&\frac{2}{3}&0\\ 0&1&0&\mid \:&\frac{1}{2}&0&-\frac{1}{2}\\ 0&0&1&\mid \:&\frac{1}{6}&-\frac{1}{3}&\frac{1}{2}\end{bmatrix}$

$R _1\:\leftarrow \frac{1}{2}\cdot \:R _1$

$=\begin{bmatrix}1&0&0&\mid \:&\frac{1}{3}&\frac{1}{3}&0\\ 0&1&0&\mid \:&\frac{1}{2}&0&-\frac{1}{2}\\ 0&0&1&\mid \:&\frac{1}{6}&-\frac{1}{3}&\frac{1}{2}\end{bmatrix}$

$=\begin{bmatrix}\frac{1}{3}&\tfrac{1}{3}&0\\ \tfrac{1}{2}&0&-\tfrac{1}{2}\\ \tfrac{1}{6}&-\tfrac{1}{3}&\tfrac{1}{2}\end{bmatrix}$

$=-\dfrac{1}{6}\begin{bmatrix}-2 &-2  &0 \\  -3& 0 &3 \\  -1& 2 &-3 \end{bmatrix}$

$\therefore \alpha =3$
Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

Inverse of $\begin{bmatrix}3& 1\5&2\end{bmatrix}$ is:

  1. $\begin{bmatrix}3&-1 \\-5 &-3\end{bmatrix}$
  2. $\begin{bmatrix}2&-1 \\-5 &3\end{bmatrix}$
  3. $\begin{bmatrix}-3&5 \\1 &-2\end{bmatrix}$
  4. $\begin{bmatrix}-2&5 \\1 &-3\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let $A=\begin{bmatrix}3& 1\\5&2\end{bmatrix}$
$\left|A\right|=6-5=1\neq 0$
$\therefore {A}^{-1}$ exists.
${C} _{11}={\left(-1\right)}^{1+1}{M} _{11}={\left(-1\right)}^{2}2=2$
${C} _{12}={\left(-1\right)}^{1+2}{M} _{12}={\left(-1\right)}^{3}5=-5$
${C} _{13}={\left(-1\right)}^{1+3}{M} _{13}={\left(-1\right)}^{4}1=1$
${C} _{14}={\left(-1\right)}^{1+4}{M} _{14}={\left(-1\right)}^{5}3=-3$
${C} _{ij}=\begin{bmatrix}2& -5\\-1 & 3\end{bmatrix}$
Adj$\left(A\right)={\begin{bmatrix}2& -5\\-1 & 3\end{bmatrix}}^{T}$
$=\begin{bmatrix}2& -1\\-5 & 3\end{bmatrix}$
${A}^{-1}=\dfrac{adj\left(A\right)}{\left|A\right|}=\dfrac{1}{1}\begin{bmatrix}2& -1\\-5 & 3\end{bmatrix}$
$\therefore {A}^{-1}=\begin{bmatrix}2& -1\\-5 & 3\end{bmatrix}$

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

Let $\displaystyle A=\begin{pmatrix}1 &2 \3  &4
\end{pmatrix}$ and $\displaystyle B=\begin{pmatrix}a &0 \0  &b \end{pmatrix} a,b \epsilon N.$Then

  1. there cannot exist any B such that $\displaystyle AB = BA $
  2. there exist more than one but finite number of B's such that $\displaystyle AB = BA$
  3. there exists exactly One B such that $\displaystyle AB = BA$
  4. there exist infinitely many B's such that $\displaystyle AB = BA.$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$A=\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$ and $B=\begin{bmatrix} a & 0 \ 0 & b \end{bmatrix}$

$AB = \begin{bmatrix} a & 2b \ 3a & 4b \end{bmatrix}$

$BA = \begin{bmatrix} a & 2a \ 3b & 4b \end{bmatrix}$

$AB\quad =\quad BA \Rightarrow a=b$

$\therefore$ there exist infinitely many  $B's$  such that $AB=BA$.

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

If matrix $A = [a _{ij}] _{2\times 2}$, where $a _{ij} = \left{\begin{matrix} 1,& \ \text{if}\ &i\neq j \ 0, & \ \text{if}\ & i + j\end{matrix}\right.$, then $A^{2}$ is equal to

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

$A=\begin{bmatrix}  a _{11}& a _{12} \  a _{21}&  a _{22}\end{bmatrix}$


$a _{11}:1\,\,\,a _{12}:0$

$a _{21}:0\,\,\,a _{22}:1$

$A=\begin{bmatrix}  1&  0\ 0 &  1\end{bmatrix}$

$A^2=\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \ 0 &  0\end{bmatrix}=\begin{bmatrix}  1& 0 \  0&  1\end{bmatrix}=I$

$\boxed{Hence\,A^2=I}$

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

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