Tag: multiplicative inverse of a matrix

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Questions Related to multiplicative inverse of a matrix

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.

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

Consider three matrices $A=\begin{bmatrix} 2 & 1 \ 4 & 1 \end{bmatrix}, B=\begin{bmatrix} 3 & 4 \ 2 & 3 \end{bmatrix}$ and $C=\begin{bmatrix} 3 & -4 \ -2 & 3 \end{bmatrix}$. Then the value of the sum $tr(A)+tr\left(\dfrac{ABC}{2}\right)+tr\left(\dfrac{A(BC)^{2}}{4}\right)+tr\left(\dfrac{A(BC)^{3}}{8}\right)+....+\infty$ is 

  1. $6$
  2. $9$
  3. $12$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The given series involves traces of matrix products that form a geometric progression of matrices. By computing BC and evaluating the traces step-by-step, the sum converges to a total trace value of 6.