Question

An n \times n matrix \mathbf{A} is invertible if and only if \mathbf{A} has full rank, i.e. \mathrm{rank}(\mathbf{A}) = n.

Which of the following matrices are invertible? Choose all that apply.

\displaystyle \mathbf{A} \displaystyle = \begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix}
\displaystyle \mathbf{B} \displaystyle = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}
\displaystyle \mathbf{C} \displaystyle = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}
\displaystyle \mathbf{D} \displaystyle = \begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{pmatrix}


\mathbf{A}

\mathbf{B}

\mathbf{C}

\mathbf{D}

Answer 1

The matrix A is not invertible. We can see this because the second row is a multiple of the first row, which means the rank of A is 1, not 2.

The matrix B is invertible. We can see this because the determinant of B is not zero, which means it has full rank.

The matrix C is invertible. We can see this because it is a diagonal matrix with all non-zero entries, which means it has full rank.

The matrix D is invertible. We can see this because it is a symmetric matrix with all non-zero entries, which means it has full rank.

So the matrices B, C, and D are invertible.