Perform the row operation on the given augmented matrix.
[tex]\left[\begin{array}{rrr|r}
1 & 8 & -6 & -6 \\
4 & 9 & 5 & 7 \\
-2 & 1 & 2 & 9
\end{array}\right] \quad R_2=-4 R_1+R_2[/tex]
What is the resultant matrix?
Answer (2)
Multiply the first row by -4: − 4 R 1 = [ − 4 , − 32 , 24 , 24 ]
Add the result to the second row: R 2 = [ 4 , 9 , 5 , 7 ] + [ − 4 , − 32 , 24 , 24 ] = [ 0 , − 23 , 29 , 31 ]
Replace the second row with the new row, keeping the first and third rows unchanged.
The resultant matrix is: [ 1 8 − 6 − 6 0 − 23 29 31 − 2 1 2 9 ]
Explanation
Understanding the Row Operation We are given the augmented matrix [ 1 8 − 6 − 6 4 9 5 7 − 2 1 2 9 ] and we want to perform the row operation R 2 = − 4 R 1 + R 2 . This means we will replace the second row with the result of multiplying the first row by -4 and adding it to the second row.
Performing the Calculation First, we multiply the first row by -4: − 4 R 1 = − 4 × [ 1 \t 8 \t − 6 \t − 6 ] = [ − 4 \t − 32 \t 24 \t 24 ] Next, we add this result to the second row: R 2 + ( − 4 R 1 ) = [ 4 \t 9 \t 5 \t 7 ] + [ − 4 \t − 32 \t 24 \t 24 ] = [ 0 \t − 23 \t 29 \t 31 ] So the new second row is [ 0 \t − 23 \t 29 \t 31 ] .
Resultant Matrix The first and third rows remain unchanged. Therefore, the resultant matrix is [ 1 8 − 6 − 6 0 − 23 29 31 − 2 1 2 9 ]
Examples
Row operations on matrices are fundamental in solving systems of linear equations. For example, consider a scenario where you want to optimize the cost of producing a blend of fruit juices. Each juice type has different costs and nutritional values. By setting up a system of linear equations representing the cost and nutritional constraints, you can use row operations to find the optimal blend that minimizes cost while meeting the required nutritional standards. This technique is widely used in operations research and linear programming to solve real-world optimization problems.
By performing the row operation R 2 = − 4 R 1 + R 2 , we multiply the first row by -4 and add it to the second row. The new second row becomes [ 0 , − 23 , 29 , 31 ] , resulting in the final augmented matrix. Therefore, the resultant matrix is 1 0 − 2 8 − 23 1 − 6 29 2 − 6 31 9
;
