At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
As an example, let's invert the matrix
[tex]\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}[/tex]
We construct the augmented matrix,
[tex]\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 2 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right][/tex]
On this augmented matrix, we perform row operations in such a way as to transform the matrix on the left side into the identity matrix, and the matrix on the right will be the inverse that we want to find.
Now we can carry out Gaussian elimination.
• Eliminate the column 1 entry in row 2.
Combine 2 times row 1 with 3 times row 2 :
2 (-3, 2, 1, 1, 0, 0) + 3 (2, 1, 1, 0, 1, 0)
= (-6, 4, 2, 2, 0, 0) + (6, 3, 3, 0, 3, 0)
= (0, 7, 5, 2, 3, 0)
which changes the augmented matrix to
[tex]\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right][/tex]
• Eliminate the column 1 entry in row 3.
Using the new aug. matrix, combine row 1 and 3 times row 3 :
(-3, 2, 1, 1, 0, 0) + 3 (1, 1, 1, 0, 0, 1)
= (-3, 2, 1, 1, 0, 0) + (3, 3, 3, 0, 0, 3)
= (0, 5, 4, 1, 0, 3)
[tex]\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 5 & 4 & 1 & 0 & 3 \end{array} \right][/tex]
• Eliminate the column 2 entry in row 3.
Combine -5 times row 2 and 7 times row 3 :
-5 (0, 7, 5, 2, 3, 0) + 7 (0, 5, 4, 1, 0, 3)
= (0, -35, -25, -10, -15, 0) + (0, 35, 28, 7, 0, 21)
= (0, 0, 3, -3, -15, 21)
[tex]\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 3 & -3 & -15 & 21 \end{array} \right][/tex]
• Multiply row 3 by 1/3 :
[tex]\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right][/tex]
• Eliminate the column 3 entry in row 2.
Combine row 2 and -5 times row 3 :
(0, 7, 5, 2, 3, 0) - 5 (0, 0, 1, -1, -5, 7)
= (0, 7, 5, 2, 3, 0) + (0, 0, -5, 5, 25, -35)
= (0, 7, 0, 7, 28, -35)
[tex]\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 0 & 7 & 28 & -35 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right][/tex]
• Multiply row 2 by 1/7 :
[tex]\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right][/tex]
• Eliminate the column 2 and 3 entries in row 1.
Combine row 1, -2 times row 2, and -1 times row 3 :
(-3, 2, 1, 1, 0, 0) - 2 (0, 1, 0, 1, 4, -5) - (0, 0, 1, -1, -5, 7)
= (-3, 2, 1, 1, 0, 0) + (0, -2, 0, -2, -8, 10) + (0, 0, -1, 1, 5, -7)
= (-3, 0, 0, 0, -3, 3)
[tex]\left[ \begin{array}{ccc|ccc} -3 & 0 & 0 & 0 & -3 & 3 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right][/tex]
• Multiply row 1 by -1/3 :
[tex]\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 1 & -1 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right][/tex]
So, the inverse of our matrix is
[tex]\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}^{-1} = \begin{bmatrix}0&1&-1\\1&4&-5\\-1&-5&7\end{bmatrix}[/tex]
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
