Answered: 6. A =-2 -1 3 2

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Compute A⁻² two different ways and show that the results are equal.

Expert Solution

Step 1

The given matrix $A = [\begin{matrix} 6 & 0 & 4 \\ - 2 & 7 & - 1 \\ 3 & 1 & 2 \end{matrix}]$ .

We have to find the value of $A^{- 2}$ .

Step 2

Method $1$ :

The given matrix $A = [\begin{matrix} 6 & 0 & 4 \\ - 2 & 7 & - 1 \\ 3 & 1 & 2 \end{matrix}]$ .

To find the inverse matrix:

$A^{- 1} = \frac{1}{d e t (A)} a d j (A)$

$\begin{array}{rcl} d e t (A) & = & |\begin{matrix} 6 & 0 & 4 \\ - 2 & 7 & - 1 \\ 3 & 1 & 2 \end{matrix}| \\ = & 6 (14 + 1) - 0 + 4 (- 2 - 21) \\ = & 6 (15) + 4 (- 23) \\ = & 90 - 92 \\ d e t (A) & = & - 2 \end{array}$

$a d j (A) = [\begin{matrix} 15 & 4 & - 28 \\ 1 & 0 & - 2 \\ - 23 & - 6 & 42 \end{matrix}]$

$A^{- 1} = - \frac{1}{2} [\begin{matrix} 15 & 4 & - 28 \\ 1 & 0 & - 2 \\ - 23 & - 6 & 42 \end{matrix}]$

Now,

$\begin{array}{rcl} A^{- 2} & = & A^{- 1} \cdot A^{- 1} \\ = & - \frac{1}{2} [\begin{matrix} 15 & 4 & - 28 \\ 1 & 0 & - 2 \\ - 23 & - 6 & 42 \end{matrix}] \cdot - \frac{1}{2} [\begin{matrix} 15 & 4 & - 28 \\ 1 & 0 & - 2 \\ - 23 & - 6 & 42 \end{matrix}] \\ = & \frac{1}{4} [\begin{matrix} 15 & 4 & - 28 \\ 1 & 0 & - 2 \\ - 23 & - 6 & 42 \end{matrix}] \cdot [\begin{matrix} 15 & 4 & - 28 \\ 1 & 0 & - 2 \\ - 23 & - 6 & 42 \end{matrix}] \\ = & \frac{1}{4} [\begin{matrix} 225 + 4 + 644 & 60 + 0 + 168 & - 420 - 8 - 1176 \\ 15 + 0 + 46 & 4 + 0 + 12 & - 28 + 0 - 84 \\ - 345 - 6 - 966 & - 92 + 0 - 252 & 644 + 12 + 1764 \end{matrix}] \\ A^{- 2} & = & \frac{1}{4} [\begin{matrix} 873 & 228 & - 1604 \\ 61 & 16 & - 112 \\ - 1317 & - 344 & 2420 \end{matrix}] \end{array}$

The value is $A^{- 2} = \frac{1}{4} [\begin{matrix} 873 & 228 & - 1604 \\ 61 & 16 & - 112 \\ - 1317 & - 344 & 2420 \end{matrix}]$ .

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