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Linear Algebra : The Inverse

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Example Question #341 : Linear Algebra

$A= \begin{bmatrix} 0. 3 & 1. 7 \\ 0.5 & 2.3 \end{bmatrix}$

Find $A^{-1}$ .

Possible Answers:

$\begin{bmatrix} -14.375 &-10.625 \\- 3.125 &-1.875 \end{bmatrix}$

$\begin{bmatrix} -1.875 &-10.625 \\- 3.125 & -14.375 \end{bmatrix}$

$\begin{bmatrix} -14.375 &10.625 \\ 3.125 &-1.875 \end{bmatrix}$

$\begin{bmatrix} 14.375 &-10.625 \\- 3.125 & 1.875 \end{bmatrix}$

$\begin{bmatrix} -1.875 &10.625 \\ 3.125 & -14.375 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -14.375 &10.625 \\ 3.125 &-1.875 \end{bmatrix}$

Explanation:

The inverse of a two-by-two matrix

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

is the matrix

$A^{-1} =\frac{1}{D} \begin{bmatrix} a_{22} & -a_{12} \\ -a_{21}& a_{11} \end{bmatrix}$ .

is the determinant of the matrix, which is the product of the main diagonal elements minus the product of the other two:

$D = a_{11}a_{22}- a_{12}a_{21}$

= 0.3(2.3)- 1.7(0.5)

= 0.69 - 0.85

= -0.16

Therefore,

$A^{-1}=\frac{1}{-0.16} \begin{bmatrix} 2. 3 &- 1. 7 \\ -0.5 & 0.3 \end{bmatrix}$

$=\begin{bmatrix} \frac{1}{-0.16} \cdot 2. 3 & \frac{1}{-0.16} \cdot(- 1. 7) \\ \\ \frac{1}{-0.16} \cdot(-0.5) & \frac{1}{-0.16} \cdot0.3 \end{bmatrix}$

$=\begin{bmatrix} -14.375 &10.625 \\ 3.125 &-1.875 \end{bmatrix}$

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Linear Algebra : The Inverse

Study concepts, example questions & explanations for Linear Algebra

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Example Question #341 : Linear Algebra

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