Tag: business maths

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Questions Related to business maths

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

If $A$ and $B$ are square matrices such that $B=-A^{-1}BA$, then 

  1. $AB+BA=0$
  2. $(A+B)^{o}=A^{2}+B^{2}$
  3. $(A+B)^{2}=A^{2}+2AB+B^{2}$
  4. $(A+B)^{2}=A+B$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given B = -A^-1BA, multiply both sides by A on the right to get BA = -A^-1BA^2, or simply rearrange to AB = -BA. Adding BA to both sides yields AB + BA = 0.

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

If $A$ is a $2\times 2$ matrix such that $A^{2}-4A+3I=0$, then the inverse of $A+3I$ is equal to

  1. $\dfrac{1}{24}S-\dfrac{7}{24}I$
  2. $\dfrac{1}{21} A-\dfrac{7}{21}I$
  3. $\dfrac{7}{24}I+\dfrac{1}{24}A$
  4. $A-3I$`
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From A^2 - 4A + 3I = 0, we can write A^2 - 4A = -3I. Factoring gives A(A - 4I) = -3I, so A(4I - A) = 3I. The inverse of (A + 3I) is found by manipulating the characteristic equation; the result is 1/24 A - 7/24 I.

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

If $A=\left[ \begin{matrix} 1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{matrix} \right] $ and $10B=\left[ \begin{matrix} 4 & 2 & 2 \ -5 & 0 & \alpha  \ 1 & -2 & 3 \end{matrix} \right] $ where $B=A^{-1}$ then $\alpha$ is equal to-

  1. $2$
  2. $-1$
  3. $-2$
  4. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since B = A^-1, we have AB = I. Multiplying the first row of A by the second column of B must equal 0. Calculating this gives 1(2) + (-1)(0) + 1(-2) = 0, which is consistent. Solving for the element at (2,3) of the product AB = I yields alpha = 2.

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

The inverse of the matrix  $\left[ \begin{array} { c c c } { 1 } & { 0 } & { 0 } \ { 3 } & { 3 } & { 0 } \ { 5 } & { 2 } & { - 1 } \end{array} \right]$  is

  1. $- \dfrac { 1 } { 3 } \left[ \begin{array} { c c c } { - 3 } & { 0 } & { 0 } \\ { 3 } & { 1 } & { 0 } \\ { 9 } & { 2 } & { - 3 } \end{array} \right]$
  2. $- \dfrac { 1 } { 3 } \left[ \begin{array} { c c c } { - 3 } & { 0 } & { 0 } \\ { 3 } & { - 1 } & { 0 } \\ { - 9 } & { - 2 } & { 3 } \end{array} \right]$
  3. $- \dfrac { 1 } { 3 } \left[ \begin{array} { c c c } { 3 } & { 0 } & { 0 } \\ { 3 } & { - 1 } & { 0 } \\ { - 9 } & { - 2 } & { 3 } \end{array} \right]$
  4. $- \dfrac { 1 } { 3 } \left[ \begin{array} { c c c } { - 3 } & { 0 } & { 0 } \\ { - 3 } & { - 1 } & { 0 } \\ { - 9 } & { - 2 } & { 3 } \end{array} \right]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The inverse of a matrix M is (1/det(M)) * adj(M). The determinant of the given matrix is 1(3*-1 - 0) = -3. Calculating the adjugate matrix and multiplying by -1/3 yields the correct option.

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

If $A=\left[ \begin{matrix} 1 & 0 & -1 \ 3 & 4 & 5 \ 0 & 6 & 7 \end{matrix} \right]$ and $A^{-1}=[\alpha _{ij}] _{3\times 3}$ then $\alpha _{23}=$

  1. $-1/5$
  2. $1/5$
  3. $-2/5$
  4. $2/5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The element alpha_23 of the inverse matrix A^-1 is given by (-1)^(2+3) times the minor of the element at row 3, column 2 of matrix A, divided by the determinant of A. Computing the determinant of A and the appropriate cofactor yields 2/5.

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

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

  1. $\begin{bmatrix} 2/13 & -5/13 \\ 3/13 & -1/13 \end{bmatrix}$
  2. $\begin{bmatrix} -2/13 & 5/13 \\ -3/13 & 1/13 \end{bmatrix}$
  3. $\begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix}$
  4. $Cannot\ be\ determined$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a 2x2 matrix [[a, b], [c, d]], the inverse is (1/(ad-bc)) * [[d, -b], [-c, a]]. Here, det = (-1)(2) - (5)(-3) = -2 + 15 = 13. The inverse is (1/13) * [[2, -5], [3, -1]].

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

If $I=I=\left[ \begin{matrix} 1 \ 0 \end{matrix}\begin{matrix} 0 \ 1 \end{matrix} \right] ,j=\left[ \begin{matrix} 0 \ -1 \end{matrix}\begin{matrix} 1 \ 0 \end{matrix} \right] and B=\left[ \begin{matrix} cos\theta  \ -sin\theta  \end{matrix}\begin{matrix} sin\theta  \ cos\theta  \end{matrix} \right] ,$ then B =

  1. $Icos\theta +Jsin\theta $
  2. $Icos\theta -Jsin\theta $
  3. $Isin\theta +Jcos\theta $
  4. $-Icos\theta +Jsin\theta $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given, $I=\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}, J=\begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix}$

and $B=\begin{bmatrix} \cos \theta &\sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$

$=\cos\theta\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix} +\sin\theta \begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix}$

$=I\cos\theta +J\sin\theta$.
Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

Let p be a non-singular matrix, $1+p+p^{2}+....+p^{n}=0$ (0 denotes the null matrix) then $p^{-1}=$

  1. $p^{n}$
  2. -$p^{n}$
  3. -(1+p+...+$p^{n}$)

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

Given I + P + P^2 + ... + P^n = 0, we have P + P^2 + ... + P^n = -I. Factoring out P gives P(I + P + ... + P^(n-1)) = -I. This is a standard identity for geometric series of matrices.

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

Let A be a $3 \times 3$  matrix such that is: $A\left[ \begin{matrix} 1 & 2 & 3 \ 0 & 2 & 3 \ 0 & 1 & 1 \end{matrix} \right]=\left[ \begin{matrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \end{matrix} \right]  $Then $A^{-1}$ is

  1. $\left[ \begin{matrix} 0 & 1 & 3 \\ 0 & 2 & 3 \\ 1 & 1 & 1 \end{matrix} \right] $
  2. $\left[ \begin{matrix} 3 & 2 & 1 \\ 3 & 2 & 0 \\ 1 & 1 & 0 \end{matrix} \right] $
  3. $\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 2 & 3 \end{matrix} \right] $
  4. $\left[ \begin{matrix} 3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1 \end{matrix} \right] $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If AX = B, then A^-1 = XB^-1. Here X is the matrix on the left and B is the matrix on the right. Calculating B^-1 and multiplying by X yields the inverse.