Let A = eginbmatrix 1 3 -2 1 2 -1 0 1...

Let A = egin{bmatrix} 1 & 3 & -2 \ 1 & 2 & -1 \ 0 & 1 & 3 end{bmatrix} and b = egin{bmatrix} 1 \ 2 \ 3 end{bmatrix} are given, find the solution to the system Ax = b, by using Adjoint Matrix A (Adj (A)) for computing A^{-1}.

Transcript

00:01 In this problem, we have been given the matrix, which are containing the elements as shown here, and b is also another matrix.

00:10 And using these two metrics, we have to determine the matrix x here.

00:16 Provided that, here we will use the idea that on multiplying this equation on both sides with a inverse, we're going to get identity matrix times x is equal to b times a inverse.

00:31 Or it will be a inverse times b because this is a 3 by 3 matrix and this is a 3 by 1 matrix and the resulting matrix that we will get after multiplying inverse of a with matrix b it will be a 3 by 1 matrix as well so first let's get the inverse of matrix a so that can be obtained when we divide the adjoint of this matrix with the determinant of this matrix so first let's get the determinant so that will be 1 into the determinant of 2 1 minus 1 3 minus 3 times the determinant of 1 0 minus 1 3 plus minus 2 that's minus 2 times determinant of 1 0 to 1 and simplifying this we're gonna get 1 times 3 into 2 that's 6 plus 1 minus 3 times 3 minus 2 times 1 and this comes out to be 7 minus 9 minus 2 and that's minus 4 and here we observe that the determinant it's not 0 so basically here the matrix is inverse it exists let's get the adjoint of this matrix so this adjoint that's the transpose of the co -factor metrics so we get the co -factor of this matrix a so that will be one times the determinant of 2 -1 minus 1 3 that's 7 and then we determine and we determine here the co -factor that's minus 3 times the determinant of 1 -0 minus 1 3 so similarly we perform the operation and we can get the co -factor matrix as minus 3 1 and in the second row we get minus 11 3 1 and in the third row we're going to get here 1 minus 1 and the last element that will be minus 1 so we take the transpose of this matrix and that's going to be 7 minus 3 1 minus 11 3 minus 1 and 1 minus 11 minus 1 and 1 minus 11 minus 1 and now we can just divide this adjoint of metrics with the determinant of a so we can get the a inverse as minus 7 by 4 11 by 4 minus 1 by 4 3 by 4 minus 3 by 4 1 by 4 minus 1 by 4 minus 1 by 4 1 by 4 and 1 by 4 and once we have got this a inverse let's multiply this with matrix b that's 1 2 3 so when we multiply this, we're going to get the resultant metrics coming out to be, let's say, a, b, c, these are the three elements...

Question

Let A = egin{bmatrix} 1 & 3 & -2 \ 1 & 2 & -1 \ 0 & 1 & 3 end{bmatrix} and b = egin{bmatrix} 1 \ 2 \ 3 end{bmatrix} are given, find the solution to the system Ax = b, by using Adjoint Matrix A (Adj (A)) for computing A^{-1}.

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Let A = egin{bmatrix} 1 & 3 & -2 \ 1 & 2 & -1 \ 0 & 1 & 3 end{bmatrix} and b = egin{bmatrix} 1 \ 2 \ 3 end{bmatrix} are given, find the solution to the system Ax = b, by using Adjoint Matrix A (Adj (A)) for computing A^{-1}.