In this section we introduce the determinant of a matrix and describe some of its properties. The determinant is defined through a recursive process, and as a result many of the theorems in this section involve proofs by mathematical induction. Since that technique is beyond the scope of this course we will omit most of the proofs in this section.
Subsection4.5.1Cofactor expansion
The definition we are about to give will seem quite strange and unmotivated. You might want to look ahead to Theorem 4.5.14 and Subsection 4.5.3 to see why determinants are important before you work through the details of the definition.
Definition4.5.1.
If \(A = \begin{bmatrix}a\end{bmatrix}\) is a \(1 \times 1\) matrix then \(\det(A) = a\text{.}\)
If \(A = \begin{bmatrix}a \amp b \\ c \amp d\end{bmatrix}\) is a \(2 \times 2\) matrix then \(\det(A) = ad-bc\text{.}\)
Notice that in both cases of the definition \(\det(A)\) is a number. There are formulas for determinants of larger matrices, but they are quite unpleasant, so instead we will describe a method for the computation of the determinant.
Definition4.5.2.
Let \(A\) be an \(n \times n\) matrix, and suppose that \(1 \leq i \leq n\) and \(1 \leq j \leq n\text{.}\) The \((i,j)\)-minor of \(A\) is the \((n-1) \times (n-1)\) matrix \(A_{i,j}\) defined by deleting row \(i\) and column \(j\) of \(A\text{.}\)
This method of computing \(\det(A)\) is referred to as cofactor expansion.
Note4.5.5.
Some authors use the symbol \(\abs{A}\) to refer to the determinant of \(A\text{.}\) Since we already have plenty of things denoted by vertical bars, we will always use the notation \(\det(A)\text{.}\)
As you can see, computing the determinant of a \(3 \times 3\) matrix involves computing \(3\) determinants of \(2 \times 2\) matrices. Similarly, computing the determinant of a \(4 \times 4\) matrix involves computing \(4\) determinants of \(3 \times 3\) matrices, for a total of \(4\cdot 3 = 12\) computations of \(2 \times 2\) determinants. For a \(5 \times 5\) matrix we need to compute \(5\) determinants of \(4 \times 4\) matrices, for a total of \(60\) determinants of \(2 \times 2\) matrices. This very quickly gets out of hand!
If we are lucky and \(a_{1,j} = 0\) for some \(j\) then we can skip computing \(\det(A_{1,j})\text{,}\) because in the cofactor expansion the term \(a_{1,j}\det(A_{1,j})\) will be \(0\text{.}\) Fortunately, it is possible to do cofactor expansion along any row, or indeed any column, of \(A\text{,}\) so we can "target" rows and columns that have many \(0\) entries.
Fact4.5.7.
The only difference in using a row or column other than row \(1\) is that the sign pattern in the cofactor expansion may change. Specifically, if we do cofactor expansion along an odd-numbered row or column then the sign of the first term of the expansion will be positive (and alternating signs thereafter), while if we do cofactor expansion along an even-numbered row or column then the sign of the first term of the expansion will be negative (and again, alternating signs thereafter).
While neither of these calculations are particularly appealing, the second one is certainly much shorter, as it only requires us to compute two \(3 \times 3\) determinants. Regardless of which method we choose we will get the same answer, \(\det(A) = 77\text{.}\)
Definition4.5.9.
An \(n \times n\) matrix \(A\) is called a lower triangular matrix if every entry above the main diagonal is \(0\) (that is, if \(a_{i,j} = 0\) whenever \(i \lt j\)). It is called an upper triangular matrix if every entry below the main diagonal is \(0\) (that is, if \(a_{i,j} = 0\) whenever \(i > j\)). In either case the matrix may also have \(0\) entries elsewhere.
A matrix is triangular if it is either upper triangular or lower triangular.
What we saw in Example 4.5.10 always happens for triangular matrices: If \(A\) is triangular, then \(\det(A)\) is the product of the diagonal entries of \(A\text{.}\)
Subsection4.5.2Properties of determinants
It is time to consider some properties of the determinant and how it interacts with other operations we perform on matrices. After seeing Example 4.5.10 you might hope that we could row-reduce our matrix to a triangular form (such as the reduced row echelon form) before computing the determinant. We can, but the row operations change the determinant, so we need to keep track of our row operations and how they affect the determinant.
Theorem4.5.12.
Suppose that \(A\) is an \(n \times n\) matrix, and \(B\) is obtained from \(A\) by a single elementary row operation. Then:
If \(A \to_{R_i \leftrightarrow R_j} B\) then \(\det(A) = -\det(B)\text{.}\)
If \(A \to_{kR_i} B\) then \(\det(A) = \frac{1}{k}\det(B)\text{.}\)
If \(A \to_{R_i+cR_j} B\) then \(\det(A) = \det(B)\text{.}\)
Let \(A = \begin{bmatrix}2 \amp 3 \amp 0 \\ 3 \amp 1 \amp 1 \\ 4 \amp 2 \amp -1\end{bmatrix}\text{.}\) We row-reduce \(A\text{,}\) keeping track of our row operations. To help clarify the determinant calculation at the end, we will give each intermediate matrix a name (one does not typically do this, it is just for exposition purposes).
Now we've reached a triangular matrix, so the determinant of \(B_7\) is the product of its diagonal entries, \(\det(B_7) = 15/4\text{.}\) To find the determinant of \(A\) we work backwards, using Theorem 4.5.12.
Thus \(\det(A) = 15\text{.}\) You can verify this by computing \(\det(A)\) directly using cofactor expansion.
As you can see from the example, row reduction can somewhat simplify the process of calculating determinants, but it remains a tedious process. In fact, in real-life applications of linear algebra (where matrices are often very large), calculating determinants simply isn't practical - it is too time-consuming and error-prone, even when done on a computer. Nevertheless, for certain problems the determinant can be very helpful. For our purposes, the most important property of determinants is that they provide a numerical way to detect invertibility of matrices; specifically, a square matrix is invertible if and only if it has non-zero determinant. We will make extensive use of this method for detecting invertibility in Section 5.1. For now, we get to add another statement to the Fundamental Theorem!
Theorem4.5.14.Fundamental Theorem - Version 3.
Let \(A\) be an \(n \times n\) matrix. The following are equivalent:
\(\RREF(A) = I_n\text{.}\)
\(A\) is invertible.
The system \([A|\vec{0}]\) has a unique solution.
The equation \(A\vec{x} = \vec{0}\) has a unique solution.
For every vector \(\vec{b}\) in \(\mathbb{R}^n\text{,}\) the system \([A|\vec{b}]\) has a unique solution.
For every vector \(\vec{b}\) in \(\mathbb{R}^n\text{,}\) the equation \(A\vec{x} = \vec{b}\) has a unique solution.
The columns of \(A\) are linearly independent.
The span of the columns of \(A\) is \(\mathbb{R}^n\text{.}\)
\(\rank(A) = n\text{.}\)
\(A\) can be written as a product of a finite collection of elementary matrices.
The key observation for both parts of this proof is that it follows from Theorem 4.5.12 that, although row operations can change determinants, they never change whether or not the determiant is zero.
\(1 \implies 11\text{:}\) Suppose that \(\RREF(A) = I_n\text{.}\) Then there is a sequence of row operations that transforms \(A\) into \(I_n\text{,}\) and \(\det(I_n) \neq 0\text{,}\) so \(\det(A) \neq 0\text{.}\)
\(11 \implies 1\text{:}\) Suppose that \(\det(A) \neq 0\text{.}\) Then \(\det(\RREF(A)) \neq 0\) as well. In particular, \(\RREF(A)\) does not have a row of all zero, so every row of \(\RREF(A)\) has a leading \(1\text{.}\) Thus \(A\) has \(n\) pivot columns. The only \(n \times n\) matrix in reduced row echelon form with \(n\) pivot columns is \(I_n\text{,}\) so \(\RREF(A) = I_n\text{.}\)
To wrap up this part of our discussion, here is how determinants interact with the other matrix operations we have seen.
Theorem4.5.15.
Suppose that \(A\) and \(B\) are \(n \times n\) matrices. Then:
For any scalar \(k\text{,}\)\(\det(kA) = k^n\det(A)\text{.}\)
\(\det(AB) = \det(A)\det(B)\text{.}\)
If \(A\) is invertible then \(\det(A^{-1}) = \frac{1}{\det(A)}\text{.}\)
\(\det(A^t) = \det(A)\text{.}\)
Note4.5.16.
Conspicuously absent from the theorem above is a nice formula for computing \(\det(A+B)\) from \(\det(A)\) and \(\det(B)\text{.}\) The reason for that omission is that no such formula exists: If all that you know about \(A\) and \(B\) is their determinants then you do not have enough information to compute \(\det(A+B)\text{.}\)
Subsection4.5.3A geometric interpretation
Recall that every \(n \times n\) matrix is associated with a linear transformation \(T_A : \mathbb{R}^n \to \mathbb{R}^n\text{,}\) where \(T_A(\vec{v}) = A\vec{v}\text{.}\) The number \(\det(A)\) has a geometric meaning in terms of the transformation \(T_A\text{.}\) The geometry is easiest to visualize in the case of \(2 \times 2\) matrices, so we start there.
Suppose that \(A = \begin{bmatrix}a \amp b \\ c \amp d\end{bmatrix}\text{.}\) Then \(T_A\left(\begin{bmatrix}1\\0\end{bmatrix}\right) = \begin{bmatrix}a\\c\end{bmatrix}\) and \(T_A\left(\begin{bmatrix}b\\d\end{bmatrix}\right)\text{.}\) Any point inside the square determined by \(\begin{bmatrix}1\\0\end{bmatrix}\) and \(\begin{bmatrix}0\\1\end{bmatrix}\) can be written as \(s\begin{bmatrix}1\\0\end{bmatrix}+t\begin{bmatrix}0\\1\end{bmatrix}\text{,}\) where \(0 \leq s \leq 1\) and \(0 \leq t \leq 1\text{.}\) Since \(T_A\) is a linear transformation, we have
That is, points inside the square determined by \(\begin{bmatrix}1\\0\end{bmatrix}\) and \(\begin{bmatrix}0\\1\end{bmatrix}\) are sent to points inside the parallelogram determined by \(\begin{bmatrix}a\\c\end{bmatrix}\) and \(\begin{bmatrix}b\\d\end{bmatrix}\).
Figure4.5.17.The action of \(T_A\) on the unit square determined by \(\vec{e_1}\) and \(\vec{e_2}\text{.}\)
In the setting described above, the geometric interpretation of the determinant is that the parallelogram determined by \(T_A(\vec{e_1})\) and \(T_A(\vec{e_2})\) (that is, determined by the columns of \(A\)) has area exactly \(\abs{\det(A)}\text{.}\) The sign of \(\det(A)\) is determined by whether \(T_A\) preserves or reverses the orientation between the sides of the square. The figure above shows the situation for a positive determinant. Here is what the image looks like if the determinant is negative - notice how the positions of \(T_A(\vec{e_1})\) and \(T_A(\vec{e_2})\) are reversed from those of \(\vec{e_1}\) and \(\vec{e_2}\text{.}\)
Figure4.5.18.A negative determinant.
In higher dimensions the geometry is similar: \(\abs{\det(A)}\) is the volume of the image of the parallelopiped determined by the standard basis of \(\mathbb{R}^n\text{,}\) and the sign of \(\det(A)\) is determined by the the orientation of the image.
In multivariable calculus you will see that determinants appear in the formula for change-of-variables in multiple integrals. The reason is fundamentally because of the geometric picture shown above - the determinant is giving the correction factor for how areas (or volumes, or their higher-dimensional analogues) have changed in applying a linear transformation to change coordinates. In fact, this use of determinants was discovered before the modern notions of "matrix" and "linear transformation"!
\begin{equation*}
\det\begin{bmatrix} 0 \amp a \amp 0 \\ b \amp c \amp d \\ 0 \amp e \amp 0 \end{vmatrix}
= -a\det\begin{bmatrix}b \amp d \\ 0 \amp 0 \end{bmatrix}
= a (-1) (0) =0 \text{.}
\end{equation*}
We wish to interchange two rows in order to get an upper triangular matrix. If we interchange two rows, then we must multiply the determinant of the resulting matrix by (-1) twice. Once we have obtained an upper triangular matrix, we multiply the diagonal elements to get the determinant.
\begin{equation*}
\det\begin{bmatrix} 0 \amp 0 \amp 0 \amp a \\ 0 \amp 0 \amp b \amp p \\ 0 \amp c \amp q \amp k \\ d \amp s \amp t \amp u \end{bmatrix} = (-1)^2 \begin{bmatrix} d \amp s \amp t \amp u \\ 0 \amp c \amp q \amp k \\ 0 \amp 0 \amp b \amp p \\ 0 \amp 0 \amp 0 \amp a \end{bmatrix} = abcd
\end{equation*}
8.
Evaluate by cursory inspection:
\(\det \begin{bmatrix} a \amp b \amp c \\ a+1 \amp b+1 \amp c+1 \\ a-1 \amp b-1 \amp c-1 \end{bmatrix}\)Answer.
We can subtract the first row from both the second and third row.
\begin{equation*}
\begin{bmatrix} a \amp b \amp c \\ a+1 \amp b+1 \amp c+1 \\ a-1 \amp b-1 \amp c-1 \end{bmatrix} \to \begin{bmatrix} a \amp b \amp c \\ 1 \amp 1 \amp 1 \\ -1 \amp -1 \amp -1 \end{bmatrix}
\end{equation*}
Then we multiply the third row by -1 to obtain a matrix which has two rows that are the same. Hence, the determinant of the original matrix would be (-1)(0) =0.
\begin{equation*}
\det\begin{bmatrix} a \amp b \amp c \\ a+1 \amp b+1 \amp c+1 \\ a-1 \amp b-1 \amp c-1 \end{bmatrix}=0
\end{equation*}
\(\det \begin{bmatrix} a \amp b \amp c \\ a+b \amp 2b \amp c+b \\ 2 \amp 2 \amp 2 \end{bmatrix} \)Answer.
\begin{equation*}
\begin{bmatrix} a \amp b \amp c \\ a+b \amp 2b \amp c+b \\ 2 \amp 2 \amp 2 \end{bmatrix} \to \begin{bmatrix} a \amp b \amp c \\ b \amp b \amp b\\ 2 \amp 2 \amp 2 \end{bmatrix}
\end{equation*}
Now if \(b \) is equal to zero, then we have a matrix with a row of zeros and hence the determinant must be zero. If \(b \) is not equal to zero, then we can multiple the second row by \(\frac{2}{b} \) to obtain a matrix which has two rows that are the same, and in this case the determinant must be zero as well.
\begin{equation*}
\begin{bmatrix} a \amp b \amp c \\ b \amp b \amp b\\ 2 \amp 2 \amp 2 \end{bmatrix} \to \begin{bmatrix} a \amp b \amp c \\ 2 \amp 2 \amp 2\\ 2 \amp 2 \amp 2 \end{bmatrix}
\end{equation*}
Therefore, the determinant of the matrix must be zero.
There are many ways to go about finding \(b\text{,}\) you may expand along various rows or columns. For this solution, let us expand along the first row.
As in the previous part (a), there are many ways to go about finding the answer by expanding along various rows or columns. In this solution, we will expand along the first column.
There exists matrices \(A\) and \(B\) such that \(\det(A+B) \neq \det(A) + \det(B)\text{.}\) Therefore, the statement of the exercise is false. Let us show this with a counterexample. Let \(A \) be the identity matrix, \(A = \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \end{bmatrix}\) and let \(B = \begin{bmatrix} -1 \amp 0 \\ 0 \amp -1 \end{bmatrix}\text{,}\) then
We know that the matrix is invertible if \(\det(A) \ne 0 \text{,}\) therefore we can compute the determinant in order to identify the values of \(c \) which make the determinant non-zero. We choose to expand along the first row because the zero in the second column of the first row will simplify our cofactor calculation.
We know that the matrix is invertible if \(\det(A) \ne 0 \text{,}\) therefore we can compute the determinant in order to identify the values of \(c \) which make the determinant non-zero. We choose to expand along the first row because the zero in the first column of the first row will simplify our cofactor calculation.
We know that the matrix is invertible if \(\det(A) \neq 0 \text{,}\) therefore we can compute the determinant in order to identify the values of \(c \) which make the determinant non-zero. We choose to expand along the first row because the zero in the third column of the first row will simplify our cofactor calculation.
\begin{align*}
\det(A) = c\det\begin{bmatrix}2 \amp c \\ c \amp 5 \end{bmatrix} + -1\det\begin{bmatrix} 0 \amp c \\ -1 \amp 5 \end{bmatrix} \amp \ne 0\\
c(10-c^2) -(0-c) \amp \ne 0\\
c(9-c^2) \amp \ne 0\\
c \amp \ne 0,-3,3
\end{align*}
The matrix will be invertile if \(c\) is not equal to 0, -3 or 3.
\(\begin{bmatrix} 4 \amp c \amp 3 \\ c \amp 2 \amp c \\ 5 \amp c \amp 4 \end{bmatrix}\)Answer.
Any real value \(c\) will make the matrix invertible because its determinant will always be equal to 2 regardless of the value of \(c \text{.}\)
We know that the matrix is invertible if \(\det(A) \ne 0 \text{,}\) therefore we can compute the determinant in order to identify the values of \(c \) which make the determinant non-zero. We choose to expand along the first column this time just to mix things up.
\begin{align*}
\det(A) = 4\det\begin{bmatrix}2 \amp c \\ c \amp 4 \end{bvmatrix} + -c\det\begin{bmatrix} c \amp 3 \\ c \amp 4 \end{bmatrix}+ 5\det\begin{bmatrix} c \amp 3 \\ 2 \amp c \end{bmatrix} \amp \ne 0\\
4(8-c^2) -(4c-3c) +5(c^2-6) \amp \ne 0\\
2 \amp \ne 0
\end{align*}
\(\begin{bmatrix} 1 \amp 2 \amp -1 \\ 0 \amp -1 \amp c \\ 2 \amp c \amp 1 \end{bmatrix} \)Answer.
The matrix will be invertible as long as \(c \) is not equal to neither 1 nor 3.
We know that the matrix is invertible if \(\det(A) \ne 0 \text{,}\) therefore we can compute the determinant in order to identify the values of \(c \) which make the determinant non-zero. We choose to expand along the first column because the zero in the second row of the first column will simplify our cofactor calculation.
We know that the matrix is invertible if \(\det(A) \ne 0 \text{,}\) therefore we can compute the determinant in order to identify the values of \(c \) which make the determinant non-zero. We choose to expand along the third row this time because the zero in the third row of the first column will simplify our cofactor calculation.
\begin{align*}
\det(A) = -\det\begin{bmatrix}-1 \amp c \\ c \amp 1 \end{bmatrix} + c\det\begin{bmatrix} 1 \amp c \\ c \amp 1 \end{bmatrix} \amp \ne 0\\
-1(1+c) + c(1-c^2) \amp \ne 0\\
c^3 \amp \ne -1\\
c\amp \ne 1
\end{align*}
