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Questions Related to business maths

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

If f (x)= $\left[ {\begin{array}{{20}{c}}  {\cos \,x}&{ - \sin \,x}&0 \   {\sin \,x}&{\cos \,x\,}&0 \   0&0&1 \end{array}} \right]$ and $\left[ {\begin{array}{{20}{c}}  {\cos \,x}&0&{\sin \,x} \   0&1&0 \   { - \sin \,x}&0&{\cos \,x} \end{array}} \right]$
then ${\left[ {f\left( x \right)g\left( y \right)} \right]^{ - 1}}$ is equal to 

  1. $f\left( { - x} \right)g\left( { - y} \right)$
  2. $f\left( {{x^{ - 1}}} \right)g\left( {{y^{ - 1}}} \right)$
  3. $g\left( { - y} \right)f\left( { - x} \right)$
  4. $g\left( {{y^{ - 1}}} \right)f\left( {{x^{ - 1}}} \right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Letting the given matrices be f(x) and g(y), we note that their transposes equal their inverses with negative arguments, specifically f(x)^(-1) = f(-x) and g(y)^(-1) = g(-y). Using the reverse order law for inverses, [f(x)g(y)]^(-1) becomes g(y)^(-1)f(x)^(-1). Substituting the inverse forms yields g(-y)f(-x).

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

Let $A$ be a square matrix such that $A^2 = A$ and $|A| \neq 0$, then choose the correct option.

(A' represents transpose of matrix A)

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

$A$ is a square matrix such that $A^2 = A$ and $|A| \neq 0$
$\Rightarrow A^{-1}A^2=A^{-1}A=I$
$\Rightarrow A=I$
$\therefore A=A'=I$
Hence, option A.

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

For two matrices $A$ and $B$, if $AB=0$, then

  1. $A=0$ and $B=0$
  2. $A=0$ or $B=0$
  3. it is not necessary that $A=0$ or $B=0$

  4. all above are false
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

It is not necessary that, if $AB=0$ then $A=0$ or $B=0$


Take example, $A=\begin{bmatrix}0&0\0&1 \end{bmatrix}, B=\begin{bmatrix}0&1\0&0 \end{bmatrix}$

Clearly $AB=\begin{bmatrix}0&0\0&0 \end{bmatrix}$ but $A,B\neq 0$

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

If $A=\begin{bmatrix} \cos { \alpha  }  & -\sin { \alpha  }  \ \sin { \alpha  }  & \cos { \alpha  }  \end{bmatrix}$, $B=\begin{bmatrix} \cos { 2\beta  }  & \sin { 2\beta  }  \ \sin { 2\beta  }  & -\cos { 2\beta  }  \end{bmatrix}$, where 0 < $\beta$ < ${ \pi  }/{ 2 }$, then prove that $BAB=$ ${ A }^{ -1 }$.

  1. True

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

By performing the matrix multiplication BAB and comparing it with the inverse of A (A^-1), one can verify the identity. Given the definitions of A and B, the statement is mathematically true.

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

Let $A$ be a $3\times 2$ matrix with real entries. Let $H = A(A^{T}A)^{-1}A^{T}$ where $A^{T}$ is the transpose of $A$ and let $I$ be the identity matrix of order $3\times 3$. Then

  1. $H^{2} = I$
  2. $H^{2} = -I$
  3. $H^{2} = H$
  4. $H^{2} = -H$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that from the propeties of inverse matrices,

${ \left( A.B \right)  }^{ -1 }={ B }^{ -1 }.A^{ -1 }$
$\Longrightarrow H=A{ A }^{ -1 }{ \left( { A }^{ T } \right)  }^{ -1 }{ A }^{ T }=I\ \Longrightarrow H.H={ H }^{ 2 }=I.H=H$

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

If $ \begin{bmatrix}
              2 & 1 \[0.3em]
              3 & 2
              \end{bmatrix} \ A \begin{bmatrix}
              -3 & 2 \[0.3em]
              5 & -3
              \end{bmatrix} = \begin{bmatrix}
              1 & 0 \[0.3em]
              0 & 1
              \end{bmatrix}$ then the matrix A is equal to

  1. $\begin{bmatrix}

    1 & 1 \\[0.3em]

    1 & 0

    \end{bmatrix}$
  2. $\begin{bmatrix}

    1 & 1 \\[0.3em]

    0 & 1

    \end{bmatrix}$
  3. $\begin{bmatrix}

    1 & 0 \\[0.3em]

    1 & 1

    \end{bmatrix}$
  4. $\begin{bmatrix}

    0 & 1 \\[0.3em]

    1 & 1

    \end{bmatrix}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{bmatrix} 2 & 1 \ 3 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$

Pre multiplying by inverse of $1^{st}$ matrix on $LHS$
$A \begin{bmatrix} -3 & 2 \ 5 & -3 \end{bmatrix} = \begin{bmatrix} 2 & -3 \ -1 & 2 \end{bmatrix} \, \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -3 \ -1 & 2 \end{bmatrix}$
$A \begin{bmatrix} -3 & 2 \ 5 & -3 \end{bmatrix} = \begin{bmatrix} 2 & -3 \ -1 & 2 \end{bmatrix}$
Post multiplying by $2^{nd}$ matrix.
$A = \begin{bmatrix} 2 & -3 \ -1 & 2 \end{bmatrix} \begin{bmatrix} -3 & -5 \ -2 & -3 \end{bmatrix}$
$= \begin{bmatrix} 1 &  1 \ 1 & 0 \end{bmatrix}$

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

Find the number of all possible ordered sets of two $(n\times n)$ matrices A and B for which $AB-BA=$$I$.

  1. Infinite
  2. $n^2$
  3. $n!$

  4. Zero
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$AB-BA=I$

Let a $n\times n$ matrix. Take trace of both sides. trace $(AB-BA)=trace \ (I)$

trace $(AB)-$trace $(BA)=n$

$\Rightarrow \ n=0$

so zero possible ordered let of two $(n\times n)$ matrix.
Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $A=|a _{ij}| _{2\times 2}$, where $a _{ij}=\left{\begin{matrix} i+j, & if & i\neq j\ i^2-2j, & if & i=j\end{matrix}\right.$, then $A^{-1}=?$

  1. $\dfrac{1}{9}\begin{bmatrix} 0 & 3\\ 3 & 1\end{bmatrix}$
  2. $\dfrac{1}{9}\begin{bmatrix} 0 & -3 \\ -3 & -1\end{bmatrix}$
  3. $\dfrac{1}{9}\begin{bmatrix} 4 & 1\\ -1 & 2\end{bmatrix}$
  4. $\dfrac{-1}{9}\begin{bmatrix} 4 & 1\\ -1 & 2\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} A={ \left| { aij } \right| _{ 2\times 2 } } \ { a _{ 11 } }=1-2=-1 \ { a _{ 12 } }=3 \ { a _{ 21 } }=3 \ { a _{ 22 } }=4-2\times 2=0 \ A=\left[ \begin{array}{l} -1 & 3 \ 3 & D \end{array} \right]  \ { A^{ -1 } }=\frac { 1 }{ { \left| A \right|  } } adjA \ =\frac { { -1 } }{ { 09 } } \left[ \begin{array}{l} 0 & -3 \ -3 & -1 \end{array} \right]  \ =\frac { 1 }{ 9 } \left[ \begin{array}{l} 0 & 3 \ 3 & 1 \end{array} \right]  \end{array}$