00:01
In this question, we have to use matrix inversion to solve each collection of systems of linear equations.
00:08
This question consists of three parts.
00:10
We will start with part a.
00:12
In this part, the given equations are minus x, minus 4y plus 2 z equals to 4, x plus 2y minus z equals to 3, x plus y minus z equals to 8.
00:35
Now these equations can be written in the form as ax equals to b.
00:43
A matrix is equals to minus 1, minus 4, 1, 2, minus 1, 1, 1, 1, 1, x matrix equals to x, y, z, and b matrix equals to 4, 3, 8.
01:15
Now a matrix is invertible so a inverse can be calculated using row operations.
01:30
So we can write the matrix as minus 1 minus 4 2, 1 2 minus 1, 1, 1 minus 1 and on the right side identity matrix of 4 by 3 .1, 010, 0 .1.
01:54
1.
01:56
Now we will apply the operation r2 scales to r2 plus r1 and r3 scales to r3 plus r1.
02:14
On applying these operations we get the matrix.
02:21
Minus 1 minus 4, 2, 0 minus 2 1, 0 minus 3 1 and on the right side, 1 0 0, 110 1 0, 1 0, 1 0.
02:35
1 .1.
02:46
Now we will apply the operations.
02:52
R1 scales to r1 minus 2 r2, r2 scales 2 minus 1 by 2 r2 and r3 scales to r3 plus 3 times r2.
03:18
On applying these operations, we get the matrix minus 1 0 0 01 minus 1 minus 1 by 2.
03:33
0 minus 1 by 2 and on the right side we get minus 1 minus 2 0 minus 1 by 2 minus 1 by 2 0 minus 1 by 2 minus 3 by 2 1 now again we will apply the operations r 2 2 x r2 x r1 scales 2 minus r 3 r 1 scales 2 minus r 1 times r1 and 1 and r3 scales 2, minus 2 times, r3 on applying these operations, we get the matrix.
04:37
1 010, 010, 0 ,0, and on the right side, 120, 1 ,0 minus 1, 1 3 minus 1, 13 minus 2.
04:58
From here we get a inverse matrix which is 120 -0 -1 minus 1 1 -13 minus 2.
05:15
Therefore we can write it as x equals to a inverse b a -inverse matrix is 1 -20 -0 -1 -1 -13 -2 multiplied by b matrix which is 4 b 8 now we will multiply these matrix in order to get the desired matrix they will multiply first row by first column so we can write it as 1 multiplied by 4 plus 2 multiplied by 3 plus 0 multiplied by 8 when solving it we get 10 so we can write this element in the desired matrix 10 similarly, we will multiply this row by this column and this row by this column.
06:30
So our desired matrix is 10 minus 5, minus 3.
06:38
So we get the value of x y x y, x equals to 10, y equals to minus 5, and z equals to minus 3...