Question

Either compute the inverse of the given matrix, or else show that it is singular. \(\left(\begin{array}{lll}{2} & {1} & {0} \\ {0} & {2} & {1} \\ {0} & {0} & {2}\end{array}\right)\)

Short answer

Question: Compute the inverse of the given 3x3 matrix or show that it is singular: \(A=\left(\begin{array}{lll}{2} & {1} & {0} \\\ {0} & {2} & {1} \\\ {0} & {0} & {2}\end{array}\right)\) Answer: The inverse of the given matrix is \(A^{-1}=\left(\begin{array}{lll}{\frac{1}{2}} & {0} & {0} \\\ {-\frac{1}{8}} & {\frac{1}{2}} & {0} \\\ {0} & {-\frac{1}{8}} & {\frac{1}{2}}\end{array}\right)\)

Step-by-step solution

Compute the determinant of the given matrix

We are given the matrix: \(A=\left(\begin{array}{lll}{2} & {1} & {0} \\\ {0} & {2} & {1} \\\ {0} & {0} & {2}\end{array}\right)\) Since it is a triangular matrix, we can compute the determinant as the product of the diagonal elements: Det(A) = 2 * 2 * 2 = 8 Since the determinant is not zero, the matrix is not singular, and we can compute its inverse.

Compute the matrix of cofactors

First, we will compute the matrix of cofactors of matrix A: Cof(A) = \(\left(\begin{array}{lll}{C_{11}} & {C_{12}} & {C_{13}} \\\ {C_{21}} & {C_{22}} & {C_{23}} \\\ {C_{31}} & {C_{32}} & {C_{33}}\end{array}\right)\) For each element, we can calculate its cofactor as: \(C_{ij} = (-1)^{i+j} M_{ij}\) Where \(M_{ij}\) is the minor of element \(a_{ij}\). Since matrix A is a triangular matrix, the cofactors of the upper and lower triangle are easy to compute: Cof(A) = \(\left(\begin{array}{lll}{4} & {-1} & {0} \\\ {0} & {4} & {-1} \\\ {0} & {0} & {4}\end{array}\right)\)

Compute the adjugate of the matrix of cofactors

Now we need to compute the adjugate of the matrix of cofactors. The adjugate is the transpose of the matrix of cofactors: Adj(A) = \(\left(\begin{array}{lll}{4} & {0} & {0} \\\ {-1} & {4} & {0} \\\ {0} & {-1} & {4}\end{array}\right)\)

Compute the inverse of matrix A

Finally, we can compute the inverse of matrix A by multiplying the adjugate of the matrix of cofactors by the reciprocal of the determinant: \(A^{-1}=\frac{1}{\text{det}(A)} \cdot \text{Adj}(A) = \frac{1}{8} \left(\begin{array}{lll}{4} & {0} & {0} \\\ {-1} & {4} & {0} \\\ {0} & {-1} & {4}\end{array}\right)\) So the inverse of matrix A is: \(A^{-1}=\left(\begin{array}{lll}{\frac{1}{2}} & {0} & {0} \\\ {-\frac{1}{8}} & {\frac{1}{2}} & {0} \\\ {0} & {-\frac{1}{8}} & {\frac{1}{2}}\end{array}\right)\)