Tag: applications of matrices and determinants

Quizzes (3) Questions (176)

Questions Related to applications of matrices and determinants

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

Let A=$\left( {\begin{array}{{20}{c}}{ - 5}&{ - 8}&{ - 7}\3&5&4\2&3&3\end{array}} \right),B = \left( {\begin{array}{{20}{c}}x\y\z\end{array}} \right)$. If AB is scalar $\left( { \ne 0} \right)$ multiple of B, then x+y=

  1. $z$
  2. $-z$
  3. $0$
  4. $2z$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$A=\begin{pmatrix} -5 & -8 & -7 \\ 3 & 5 & 4 \\ 2 & 3 & 3 \end{pmatrix}\quad B=\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
Given $AB=k\ B$
$AB\Rightarrow \begin{pmatrix} -5 & -8 & -7 \\ 3 & 5 & 4 \\ 2 & 3 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}\quad k=\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
$\Rightarrow \begin{pmatrix} -5x & -8y & -7z \\ 3x & +5y & +4z \\ x2 & +3y & +3z \end{pmatrix}=\quad k\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
On adding all the elements on left and right side 
$\Rightarrow \ (-5x-8y-7z)+(3x+5y+47)+(2x+3y+3z)$
$=k(x+y+z)$
$\Rightarrow \ D=k(x+y+z)$
$k\neq 0$
$\Rightarrow \ x+y+z=0$
$\Rightarrow \ x+y=-z$

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

If $A = \left[ {\begin{array}{*{20}{c}}1&2\3&4\end{array}} \right]$, then $8A^{-4}$ is equal to

  1. $145A^{-1}+27I$
  2. $145A^{-1}-27I$
  3. $27I - 145A^{-1}$
  4. $29A^{-1} +9I$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the Cayley-Hamilton theorem for matrix A = [[1, 2], [3, 4]], the characteristic equation is A^2 - 5A - 2I = 0. By manipulating this equation, one can express higher powers of A in terms of A and I, eventually leading to the expression 27I - 145A^-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 $A$ 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 maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

If $A=\begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix}$ and $B=\begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$, find the values of $\alpha$ for which $A^2=B$.

  1. $\pm 1$
  2. $4$
  3. $0$

  4. No value
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have,

$A^2=B$
$\begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix} \begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$

$\begin{bmatrix} \alpha^2 +0 & 0+0 \ \alpha +1 & 0+1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$

$\begin{bmatrix} \alpha^2 & 0 \ \alpha +1 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$

$\alpha^2=1$ and $\alpha +1=5$
$\alpha =\pm 1$ and $\alpha =4$, which is not possible.
Hence, there is no value of $\alpha$ for which $A^2=B$ is true.

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

Let $P=\begin{bmatrix} \cos { \dfrac { \pi  }{ 9 }  }  & \sin { \dfrac { \pi  }{ 9 }  }  \ -\sin { \dfrac { \pi  }{ 9 }  }  & \cos { \dfrac { \pi  }{ 9 }  }  \end{bmatrix}$ and $\alpha,\ \beta,\ \gamma$ be non-zero real numbers such that $\alpha P^{6}+\beta P^{3}+\gamma 1$ is the zero matrix. Then, $(\alpha^{2}+\beta^{2}+\gamma^{2})^{(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)}$ is

  1. $\pi$
  2. $\dfrac {\pi}{2}$
  3. $0$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Matrix P represents a rotation by pi/9. Raising P to the 6th and 3rd powers corresponds to rotations by 6pi/9 and 3pi/9 respectively. By analyzing the characteristic equation or evaluating the matrix powers, the coefficients alpha, beta, and gamma make the expression simplify such that the exponent product evaluates to 1.