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}
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.
To determine whether a matrix is invertible or not, we need to check if its rank is equal to the size of the matrix. So, we need to find the rank of each matrix and compare it to the size of the matrix.
\mathbf{A}:
To find the rank of matrix \mathbf{A}, we can row reduce it:
\begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix}
Performing row operation R2 = R2 - 2R1:
\begin{pmatrix} 1 & 3 \\ 0 & 0 \end{pmatrix}
The matrix reduces to a matrix with a row of zeros, meaning the rank is 1. Since the matrix is a 2x2 matrix, the rank is not equal to the size of the matrix, so \mathbf{A} is not invertible.
\mathbf{B}:
To find the rank of matrix \mathbf{B}, we can row reduce it:
\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}
Performing row operation R2 = R2 - 2R1:
\begin{pmatrix} 1 & 2 \\ 0 & -3 \end{pmatrix}
The matrix is reduced to an upper triangular matrix, so the rank is 2. Since the matrix is a 2x2 matrix and the rank is equal to the size of the matrix, \mathbf{B} is invertible.
\mathbf{C}:
To find the rank of matrix \mathbf{C}, we can row reduce it:
\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}
Performing row operations R2 = R2 - R1, and R3 = R3 - R2:
\begin{pmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}
The matrix reduces to a matrix with two rows of zeros, meaning the rank is 2. Since the matrix is a 3x3 matrix and the rank is not equal to the size of the matrix, \mathbf{C} is not invertible.
\mathbf{D}:
To find the rank of matrix \mathbf{D}, we can row reduce it:
\begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{pmatrix}
Performing row operations R2 = R2 + 0.5R1, and R3 = R3 + 0.5R1:
\begin{pmatrix} 2 & -1 & -1 \\ 0 & 2.5 & -1.5 \\ 0 & -1.5 & 2.5 \end{pmatrix}
Performing row operations R3 = R3 + 0.6R2:
\begin{pmatrix} 2 & -1 & -1 \\ 0 & 2.5 & -1.5 \\ 0 & 0 & 0 \end{pmatrix}
The matrix reduces to a matrix with a row of zeros, meaning the rank is 2. Since the matrix is a 3x3 matrix and the rank is not equal to the size of the matrix, \mathbf{D} is not invertible.
To summarize:
- \mathbf{A} is not invertible.
- \mathbf{B} is invertible.
- \mathbf{C} is not invertible.
- \mathbf{D} is not invertible.