A square matrix is an upper triangular matrix if and only if all its entries below the entries in the main diagonal are equal to zero.
These are examples of upper triangular matrices.
Main diagonal entries are in red and all entries below them, in blue, are equal to zero.
A square matrix is a lower triangular matrix if and only if all its entries above the entries in the main diagonal are equal to zero.
These are examples of lower triangular matrices.
Main diagonal entries are in red and all entries above them, in blue, are equal to zero.
Some of the most important properties of the triangular matrices are given below.
Example 1
Which of the following matrices is an upper triangular matrix, a lower triangular matrix or none?
Solution
a) None: Matrix \( A \) is not square and therefore is not triangular.
b) Upper: Matrix \( B \) is a square matrix with the main diagonal entries \( \{ -2, 0, 1 \} \). It is an upper triangular matrix because all its entries below the main diagonal entries are equal to zero.
c) Lower: Matrix \( C \) is a square matrix with the main diagonal entries \( \{ -1, -7, 2, 3 \} \). It is a lower triangular matrix because all its entries above the main diagonal entries are equal to zero.
d) None: Matrix \( D \) is a square matrix with the main diagonal entries \( \{ 1, 1, 1, 0 \} \). It is not a triangular matrix because entries not equal to zero are above \( 9 \) and below \( -4 \) the main diagonal entries.
Example 2
Find the determinant of each of the following matrices
a) \( A =
\begin{bmatrix}
3 & 0 & 0 \\
1 & -1 & 0 \\
-2 & 3 & 2
\end{bmatrix} \) b) \( B =
\begin{bmatrix}
x & 0 & 1 \\
0 & x+1 & 1 \\
0 & 0 & x
\end{bmatrix} \) c) \( C =
\begin{bmatrix}
-1 & 0 & 0 & 0\\
5 & -9 & 0 & 0\\
5 & 4 & 7 & 0\\
5 & 0 & 0 & 0
\end{bmatrix} \)
Solution
a) Matrix \( A \) is a lower triangular matrix and therefore its determinant is equal to the product of its entries in the main diagonal. Hence
\( Det(A) = (3)(-1)(2) = -6 \)
b) Matrix \( B \) is an upper triangular matrix and its determinant is is equal to the product of its entries in the main diagonal. Hence
\( Det(B) = x(x+1)x = x^3 + x^2\)
c) Matrix \( C \) is a lower triangular matrix and its determinant is is equal to the product of its entries in the main diagonal. Hence
\( Det(C) = (-1)(-9)(7)(0) = 0\)
Example 3
Find all real values of \( x \) for which the matrix \( A = \begin{bmatrix}
x-1 & 0 & 0 & 0\\
5 & x+1 & 0 & 0\\
5 & 4 & x^2 + 5 & 0\\
5 & 0 & x+2 & 3
\end{bmatrix} \) is not invertible.
Solution
The given matrix is a lower triangular matrix and its determinant is is equal to the product of its entries in the main diagonal. Hence
\( Det(A) = (x-1)(x+1)(x^2+5)(3) \)
Matrix \( A \) is not invertible if its determinant is equal to zero. The determinant of matrix \( A \) is equal to zero if one of the factors of \( Det(A) \) is equal to zero.
\( x - 1 = 0 \) gives the solution \( x = 1 \)
\( x + 1 = 0 \) gives the solution \( x = - 1 \)
\( x^2 + 5 = 0 \) has no real solutions
The values of \( x \) for which the given matrix is not invertible are \( -1 \) and \( 1 \).
Example 4
Find the inverse of matrix \( A = \begin{bmatrix}
1 & 0\\
2 & -4
\end{bmatrix} \) and verify property 6 given above.
Solution
Use the formula of the inverse of a \( 2 \times 2 \) matrix \( \begin{bmatrix}
x & y \\
z & w \\
\end{bmatrix}^{-1} = \dfrac{1}{xw - yz}
\begin{bmatrix}
w & - y \\
- z & x \\
\end{bmatrix} \) to find \( A^{-1} \)
\( A^{-1} = \dfrac{1}{-4}
\begin{bmatrix}
-1 & 0 \\
-2 & 1 \\
\end{bmatrix} =
\begin{bmatrix}
\dfrac{1}{4} & 0 \\
\dfrac{1}{2} & -\dfrac{1}{4} \\
\end{bmatrix}
\)
Conclusion: Both \( A \) and its inverse are lower triangular matrices which verify property 6 above.
Example 5
Let \( A = \begin{bmatrix}
3 & 0 & 0\\
5 & -2 & 0 \\
9 & -2 & 4
\end{bmatrix} \), \( B = \begin{bmatrix}
-2 & 0 & 0\\
0 & -2 & 0 \\
-3 & 4 & -1
\end{bmatrix} \) and \( C = \begin{bmatrix}
7 & 0 & 0\\
-1 & 0 & 0 \\
12 & -2 & 3
\end{bmatrix} \).
Which of the following matrices \( A , \; B ,\; C ,\; A B ,\; A C,\; A + C \) is invertible?
Solution
A matrix with a determinant not equal to zero in invertible.
\( Det(A) = (3)(-2)(4) = -24 \) , hence matrix \( A \) is invertible.
\( Det(B) = (-2)(-2)(-1) = -4 \) , hence matrix \( B \) is invertible.
\( Det(C) = (7)(0)(3) = 0 \) , hence matrix \( C \) is NOT invertible.
The determinant of the product of two matrices is equal to the product of their determinants
\( Det(A B) = Det(A) Det(B) = (-24)(-4) = 96 \) , hence matrix \( A B \) is invertible.
\( Det(A C) = Det(A) Det(C) = (-24)(0) = 0 \) , hence matrix \( A C \) is NOT invertible.
\( Det(A+C) =
Det \left(\begin{bmatrix}
3 & 0 & 0\\
5 & -2 & 0 \\
9 & -2 & 4
\end{bmatrix} + \begin{bmatrix}
7 & 0 & 0\\
-1 & 0 & 0 \\
12 & -2 & 3
\end{bmatrix} \right)
\)
\(
\quad \quad =
Det \begin{bmatrix}
10 & 0 & 0\\
4 & -2 & 0 \\
21 & -4 & 7
\end{bmatrix} = (7) (-2) (7) = -98
\) , hence matrix \( A + C \) is invertible.