Chapter 4: Q30. (page 199)

Find the inverse of each matrix, if it exists.30.
$[\begin{matrix} \frac{1}{2} & - \frac{3}{4} \\ \frac{1}{6} & \frac{1}{4} \end{matrix}]$

Short Answer

Expert verified

The inverse of the matrix is $[\begin{matrix} 1 & 3 \\ - \frac{2}{3} & 2 \end{matrix}]$

Step by step solution

- Define inverse of a matrix.

For the matrix A ,

$A = [\begin{matrix} a & b \\ c & d \end{matrix}]$

The inverse of matrix of the matrix A is: $A^{- 1} = \frac{1}{a d - b c} [\begin{matrix} d & - b \\ - c & a \end{matrix}]$

where $a d - b c \neq 0$ . $a d - b c$ is the determinant of thematrix A .

If $a d - b c = 0$ , the inverse of a matrix doesn not exist.

- Calculate the inverse.

Let A be the matrix $[\begin{matrix} \frac{1}{2} & - \frac{3}{4} \\ \frac{1}{6} & \frac{1}{4} \end{matrix}]$

That is, $A = [\begin{matrix} \frac{1}{2} & - \frac{3}{4} \\ \frac{1}{6} & \frac{1}{4} \end{matrix}]$

Comparing with the standard form $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ $a = \frac{1}{2}, b = - \frac{3}{4}, c = \frac{1}{6}, d = \frac{1}{4}$

Then, $A^{- 1}$ is:

$\begin{matrix} A^{- 1} = \frac{1}{a d - b c} [\begin{matrix} d & - b \\ - c & a \end{matrix}] \\ = \frac{1}{(\frac{1}{2} \times \frac{1}{4}) - (- \frac{3}{4} \times \frac{1}{6})} [\begin{matrix} \frac{1}{4} & - (- \frac{3}{4}) \\ - \frac{1}{6} & \frac{1}{2} \end{matrix}] \\ = \frac{1}{\frac{1}{8} + \frac{3}{24}} [\begin{matrix} \frac{1}{4} & \frac{3}{4} \\ - \frac{1}{6} & \frac{1}{2} \end{matrix}] \end{matrix}$

Here,

$\begin{matrix} a d - b c = \frac{1}{8} + \frac{3}{24} \\ = \frac{3 + 3}{24} \\ = \frac{6}{24} \\ = \frac{1}{4} \end{matrix}$

As $a d - b c \neq 0$ here, inverse exist.

Then, $A^{- 1}$ is:

$\begin{matrix} A^{- 1} = \frac{1}{\frac{1}{4}} [\begin{matrix} \frac{1}{4} & \frac{3}{4} \\ - \frac{1}{6} & \frac{1}{2} \end{matrix}] \\ = 4 [\begin{matrix} \frac{1}{4} & \frac{3}{4} \\ - \frac{1}{6} & \frac{1}{2} \end{matrix}] \\ = [\begin{matrix} \frac{1}{4} \times 4 & \frac{3}{4} \times 4 \\ - \frac{1}{6} \times 4 & \frac{1}{2} \times 4 \end{matrix}] \\ = [\begin{matrix} 1 & 3 \\ - \frac{2}{3} & 2 \end{matrix}] \end{matrix}$

- State the conclusion.

Therefore, the inverse of the matrix $[\begin{matrix} \frac{1}{2} & - \frac{3}{4} \\ \frac{1}{6} & \frac{1}{4} \end{matrix}]$ is $[\begin{matrix} 1 & 3 \\ - \frac{2}{3} & 2 \end{matrix}]$

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Find the inverse of each matrix, if it exists.30.
$[\begin{matrix} \frac{1}{2} & - \frac{3}{4} \\ \frac{1}{6} & \frac{1}{4} \end{matrix}]$

Short Answer

Step by step solution

- Define inverse of a matrix.

- Calculate the inverse.

- State the conclusion.

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Find the inverse of each matrix, if it exists.30.[12−341614]

Short Answer

Step by step solution

- Define inverse of a matrix.

- Calculate the inverse.

- State the conclusion.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Geometry

Calculus

Discrete Mathematics

Theoretical and Mathematical Physics

Statistics

Study anywhere. Anytime. Across all devices.

Find the inverse of each matrix, if it exists.30.
$[\begin{matrix} \frac{1}{2} & - \frac{3}{4} \\ \frac{1}{6} & \frac{1}{4} \end{matrix}]$