Math 511: Linear Algebra

3.3 Applications of the Determinant

---

3.3.1 The Adjugate Matrix

$$ \require{color} \definecolor{brightblue}{rgb}{.267, .298, .812} \definecolor{darkblue}{rgb}{0.0, 0.0, 1.0} \definecolor{palepink}{rgb}{1, .73, .8} \definecolor{softmagenta}{rgb}{.99,.34,.86} \definecolor{blueviolet}{rgb}{.537,.192,.937} \definecolor{jonquil}{rgb}{.949,.792,.098} \definecolor{shockingpink}{rgb}{1, 0, .741} \definecolor{royalblue}{rgb}{0, .341, .914} \definecolor{alien}{rgb}{.529,.914,.067} \definecolor{crimson}{rgb}{1, .094, .271} \def\ihat{\mathbf{\hat{\unicode{x0131}}}} \def\jhat{\mathbf{\hat{\unicode{x0237}}}} \def\khat{\mathrm{\hat{k}}} \def\tombstone{\unicode{x220E}} \def\contradiction{\unicode{x2A33}} $$

The Adjugate Matrix is the classical adjoint of an $n\times n$ matrix. The adjugate matrix is the transpose of the cofactor matrix of a $n\times n$ matrix. For $n\times n$ matrix $A$,

$$ \begin{align*} \text{adj}(A) &= \text{Cof}(A)^T \\ \\ &= \begin{bmatrix} \ A_{11}\ &\ A_{12}\ &\ \cdots\ &\ A_{1n}\ \\ \ A_{21}\ &\ A_{22}\ &\ \cdots\ &\ A_{2n}\ \\ \ \vdots\ &\ \vdots\ &\ \ddots\ &\ \vdots\ \\ \ A_{n1}\ &\ A_{n2}\ &\ \cdots\ &\ A_{nn}\ \end{bmatrix}^T \\ \\ &= \begin{bmatrix} \ A_{11}\ &\ A_{21}\ &\ \cdots\ &\ A_{n1}\ \\ \ A_{12}\ &\ A_{22}\ &\ \cdots\ &\ A_{n2}\ \\ \ \vdots\ &\ \vdots\ &\ \ddots\ &\ \vdots\ \\ \ A_{1n}\ &\ A_{2n}\ &\ \cdots\ &\ A_{nn}\ \end{bmatrix} \\ \\ \end{align*} $$

Recall that the cofactor of element $a_{ij}$ is

$$ A_{ij} = (-1)^{i+j}\left|M_{ij}\right| $$

---

3.3.2 Computing the Determinant using the Adjugate

If $A$ is and $n\times n$ matrix then recall the Laplace expansion of the determinant of $A$ using the $i^{\text{th}}$ row of matrix $A$.

$$ \begin{align*} |A| &= \displaystyle\sum_{j=1}^n a_{ij}A_{ij} \\ \\ &= a_{i1}A_{i1} + a_{i2}A_{i2} + \cdots + a_{in}A_{in} \\ \\ &= \begin{bmatrix} a_{i1}\ & a_{i2}\ & \cdots\ & a_{in}\ \end{bmatrix} \begin{bmatrix} A_{i1}\ \\ A_{i2}\ \\ \vdots\ \\ A_{in}\ \end{bmatrix} \\ \\ &= \mathbf{a}^i\left(\text{adj}\mathbf{(A)}\right)_i \end{align*} $$

The last expression is the $i^{\text{th}}$ row of matrix $A$ times the $i^{\text{th}}$ column of matrix adj$(A)$.

Similarly we can compute the determinant using the $j^{\text{th}}$ column of matrix $A$.

$$ \begin{align*} |A| &= \displaystyle\sum_{i=1}^n a_{ij}A_{ij} \\ \\ &= a_{1j}A_{1j} + a_{2j}A_{2j} + \cdots a_{nj}A_{nj} \\ \\ &= \begin{bmatrix} A_{1j} & A_{2j} & \cdots & A_{nj} \end{bmatrix} \begin{bmatrix} a_{1j} \\ a_{2j} \\ \vdots \\ a_{nj} \end{bmatrix} \\ &= \left(\text{adj}\mathbf{(A)}\right)^j\,\mathbf{a}_j \end{align*} $$

What happens if we multiply the $i^{\text{th}}$ row of matrix $A$ times another column of the adjugate matrix adj$(A)$?

Remember that when we compute the Laplace expansion of the determinant of a matrix,

• we use the factors $a_{ij}$ from a row or column,
• however all of the minors $M_{ij}$ used to compute the cofactors never use the elements of our selected row or column. We obtain the minor matrix $M_{ij}$ by removing the $i^{\text{th}}$ row and $j^{\text{th}}$ column.

The key here is that we never use the values in the $i^{\text{th}}$ row that we use to compute the Laplace expansion of the determinant so their values can be anything. So let us create a new matrix $B$ that has all the same rows as matrix $A$ except the $i^{\text{th}}$ one.

We will pick one of the other rows, row $k$, where $k\neq i$ and copy row $\mathbf{a}^k$ into the $i^{\text{th}}$ row of matrix $B$. Then we will have each row of $B$

$$ \mathbf{b}^j = \left\{\begin{array}{rcl} \mathbf{a}^j, & & i\neq j \\ \mathbf{a}^k, & & i=j \end{array}\right. $$

Recall we picked the $i^{\text{th}}$ row for our Laplace expansion so

$$ B = \begin{bmatrix} \mathbf{a}^1 \\ \mathbf{a}^2 \\ \vdots \\ \mathbf{b}^i = \mathbf{a}^k \\ \vdots \\ \mathbf{a}^n \end{bmatrix} $$

This means that matrix $B$ has two rows equal to $\mathbf{a}^k$, the $k^{\text{th}}$ and the $i^{\text{th}}$ one, so the determinant of matrix $B$ is zero as it has two equal rows. This also means that computing the product of the $k^{\text{th}}$ row of matrix $A$ times the $i^{\text{th}}$ column of $\text{adj}(A)$ when $i\neq k$ yields

$$ \begin{align*} \mathbf{a}^k\left(\text{adj}\mathbf{(A)}\right)_i &= \begin{bmatrix} a_{k1}\ & a_{k2}\ & \cdots\ & a_{kn}\ \end{bmatrix} \begin{bmatrix} A_{i1}\ \\ A_{i2}\ \\ \vdots\ \\ A_{in}\ \end{bmatrix} \\ \\ &= \begin{bmatrix} b_{k1}\ & b_{k2}\ & \cdots\ & b_{kn}\ \end{bmatrix} \begin{bmatrix} B_{i1}\ \\ B_{i2}\ \\ \vdots\ \\ B_{in}\ \end{bmatrix} \\ \\ &= \begin{bmatrix} b_{i1}\ & b_{i2}\ & \cdots\ & b_{in}\ \end{bmatrix} \begin{bmatrix} B_{i1}\ \\ B_{i2}\ \\ \vdots\ \\ B_{in}\ \end{bmatrix} \\ \\ &= \mathbf{b}^i\left(\text{adj}\mathbf{(B)}\right)_i = |B| = 0 \end{align*} $$

Since this is true for any for $k\neq i$ we have

$$ \mathbf{a}^i\left(\mathbf{\text{adj}(A)}\right)_k = \left\{\begin{array}{rcl} |A| &\ &\ i=k \\ 0 &\ &\ i\neq k \end{array}\right. $$

Theorem 3.3.1

The product of matrices $A$ and $\text{adj}(A)$ results in

$$ A\,\text{adj}(A) = \left[\,\mathbf{a}^i\left(\mathbf{\text{adj}(A)}\right)_k\,\right] = \left[\, |A|\delta_{ik}\,\right] = |A|I_n. $$

Notice that since a matrix is singular if and only if $\text{det}(A) = 0$, we have a nice corollary of this theorem.

Corollary 3.3.2

A matrix $A$ is singular if and only if the matrix product $ A\,\text{adj}(A) $ is the zero matrix.

---

3.3.3 Properties of the Adjugate Matrix

If matrix $A$ is nonsingular, then $|A|\neq 0$ so we have

$$ \begin{align*} A\,\text{adj}(A) = |A|\,I_n \\ \\ A\,\frac{1}{|A|}\,\text{adj}(A) = I_n \end{align*} $$

Theorem 3.3.2

If matrix $A$ is nonsingular, then

$$ A^{-1} = \dfrac{1}{|A|}\text{adj}(A). $$

We can derive several more identities involving the adjugate matrix,

1. The adjugate of the zero matrix is the zero matrix.
   
   $$ \text{adj}(\mathbf{0}) = \text{Cof}(\mathbf{0})^T = \mathbf{0}^T = \mathbf{0} $$
   
   

2. The adjugate of the $n\times n$ identity matrix is the $n\times n$ identity matrix.
   
   $$ \text{adj}(I_n) = \text{Cof}(I_n)^T = \left[ \delta_{ij} \right]^T = I_n $$
   
   

3. For every element $a_{ij}$ of $n\times n$ matrix $A$ and nonzero scalar $c$, the cofactor $\ ca_{ij}$ of matrix $cA$ is given by
   
   $$ \left(cA\right)_{ij} = (-1)^{i+j}\left|\,cM_{ij}\,\right| = c^{n-1}(-1)^{i+j}\left|\,M_{ij}\,\right| = c^{n-1}A_{ij} $$
   because $M_{ij}$ is an $(n-1)\times (n-1)$ matrix. Thus the cofactor matrix $\ \text{Cof}(cA) = \left[ c^{n-1}A_{ij} \right] = c^{n-1}\left[A_{ij}\right]$. Hence
   
   $$ \text{adj}(cA) = \text{Cof}(cA)^T = c^{n-1}\text{Cof}(A)^T = c^{n-1}\text{adj}(A) $$
   
   
   

4. Notice that if $B = A^T$, the cofactor of $a_{ij}$ equals the cofactor of $b_{ji}$ because $(-1)^{i+j} = (-1)^{j+i}$ and the determinant of the $(n-1)\times(n-1)$ submatrix equals the determinant of its transpose. So
   
   $$ \begin{align*} \text{adj}\left(A^T\right) &= \text{Cof}\left(A^T\right)^T \\ \\ &= \left[ \left(A^T\right)_{ij} \right]^T \\ \\ &= \left[ A_{ji} \right]^T \\ \\ &= \left[ A_{ij} \right] = \text{adj}(A)^T \end{align*} $$
   
   

5. If matrix $A$ is nonsingular, then $|A|\neq 0$ and
   
   $$ A^{-1}\,\text{adj}\left(A^{-1}\right) = \left|A^{-1}\right|\,I_n $$
   If one multiplies both sides of this equation on the left by matrix $A$,
   
   $$ \text{adj}\left(A^{-1}\right) = \dfrac{1}{\left|A\right|}\,A $$
   Moreover,
   
   $$ \text{adj}\left(A^{-1}\right)\text{adj}(A) = \dfrac{1}{\left|A\right|}\,A\,|A|\,A^{-1} = I_n $$
   So if matrix $A$ is invertible, then $\text{adj}(A)$ is invertible and
   
   $$ \left(\text{adj}(A)\right)^{-1} = \text{adj}\left(A^{-1}\right). $$
   
   
   
   
   

6. Notice also that since the cofactor $A_{ij} = (-1)^{i+j}\left|M_{ij}\right|$.
   
   $$ \begin{align*} \left(A_{ij}\right)^* &= \left((-1)^{i+j}\left| M_{ij} \right| \right)^* \\ \\ &= (-1)^{i+j}\left|\,M_{ij}^*\,\right| \\ \\ &= \left(A^*\right)_{ij} \end{align*} $$
   Thus the complex conjugate of the cofactor of element $\ a_{ij}\ $ in matrix $A$ is the cofactor of element $(i,j)$ of the conjugate of matrix $A$. Hence the adjugate of the conjugate of matrix $A$ is the conjugate of the adjugate matrix of $A$ as well
   
   $$ \text{adj}\left(A^*\right) = \left(\text{adj}(A)\right)^* $$
   
   
   

7. Combining properties 4 and 6 gives us that the adjugate matrix of the Hermitian of matrix $A$ is the Hermitian of the adjugate of matrix $A$.
   
   $$ \text{adj}\left(A^{\dagger}\right) = \text{adj}\left(A^H\right) = \text{adj}(A)^H = \text{adj}(A)^{\dagger} $$
   
   

8. Like the transpose, Hermitian or inverse of a product of two matrices, if $A$ and $B$ are nonsingular $n\times n$ matrices we have
   
   
   $$ \begin{align*} \text{adj}(AB) &= \text{det}(AB)\,(AB)^{-1} = |A||B|B^{-1}A^{-1} \\ \\ &= |B|B^{-1}\,|A|A^{-1} = \text{adj}(B)\,\text{adj}(A) \end{align*} $$
   We would need two results from vector calculus to conclude that for any $n\times n$ matrices $A$ and $B$ we have
   
   $$ \text{adj}(AB) = \text{adj}(B)\,\text{adj}(A). $$
   
   
   
   

9. From Property 8 we have
   
   $$ \text{adj}\left(A^k\right) = \left(\text{adj}(A)\right)^k $$
   
   

10. In general, for two $n\times n$ matrices, matrix multiplication is not commutative. But if two $n\times n$ matrices commute
    
    $$ AB = BA $$
    then if we multiply both sides of this equation by the matrix $\text{adj}(A)$ on the left and right one obtains
    
    $$ \begin{align*} \text{adj}(A)(AB)\text{adj}(A) &= \text{adj}(A)(BA)\text{adj}(A) \\ \\ \left(\text{adj}(A)\,A\right)\left(B\,\text{adj}(A)\right) &= \left(\text{adj}(A)\,B\right)\left(A\,\text{adj}(A)\right) \\ \\ \left(|A|\,I_n\right)\left(B\,\text{adj}(A)\right) &= \left(\text{adj}(A)\,B\right)\left(|A|\,I_n\right) \\ \\ |A|\,B\,\text{adj}(A) &= |A|\,\text{adj}(A)\,B \end{align*} $$
    If matrix $A$ is nonsingular then
    $$ B\,\text{adj}(A) = \text{adj}(A)\,B $$
    If one multiplies both sides of the resulting equation on the left and right by $\text{adj}(B)$ on obtains
    
    $$ \begin{align*} \text{adj}(B)\,\left(B\,\text{adj}(A)\right)\,\text{adj}(B) &= \text{adj}(B)\,\left(\text{adj}(A)\,B\right)\,\text{adj}(B) \\ \\ \left(\text{adj}(B)\,B\right)\,\left(\text{adj}(A)\,\text{adj}(B)\right) &= \left(\text{adj}(B)\,\text{adj}(A)\right)\,\left(B\,\text{adj}(B)\right) \\ \\ \left(|B|\,I_n\right)\,\left(\text{adj}(A)\,\text{adj}(B)\right) &= \left(\text{adj}(B)\,\text{adj}(A)\right)\,\left(|B|\,I_n\right) \\ \\ |B|\,\text{adj}(A)\,\text{adj}(B) &= |B|\,\text{adj}(B)\,\text{adj}(A) \end{align*} $$
    If matrix $B$ is also nonsingular
    $$ \text{adj}(A)\,\text{adj}(B) = \text{adj}(B)\,\text{adj}(A) $$
    
    
    
    

11. From all of these properties we have that if $n\times n$ matrix $A$ has any of the following attributes, then so does its adjugate:

    • Upper Triangular
    • Lower Triangular
    • Diagonal
    • Symmetric
    • Hermitian

---

3.3.4 Computing Inverses using the Adjugate

The adjugate of a $2\times 2$ matrix $A$ is simple to compute

$$ \begin{align*} \text{adj}(A) &= \text{adj}\left(\begin{bmatrix}\ a\ &\ b\ \\ \ c\ &\ d\ \end{bmatrix}\right) \\ \\ &= \text{Cof}\left(\begin{bmatrix}\ a\ &\ b\ \\ \ c\ &\ d\ \end{bmatrix}\right)^T \\ \\ &= \begin{bmatrix}\ d\ & -c\ \\ -b\ &\ a\ \end{bmatrix}^T \\ \\ &= \begin{bmatrix}\ d\ & -b\ \\ -c\ &\ a\ \end{bmatrix} \end{align*} $$

If $2\times 2$ matrix $A$ is nonsingular

$$ A^{-1} = \dfrac{1}{|A|}\text{adj}(A) = \dfrac{1}{|A|}\,\begin{bmatrix}\ d\ & -b\ \\ -c\ &\ a\ \end{bmatrix} $$

In order to compute the inverse of a $3\times 3$ matrix,

$$ A = \begin{bmatrix}\ a_{11}\ &\ a_{12}\ &\ a_{13}\ \\ \ a_{21}\ &\ a_{22}\ &\ a_{23}\ \\ \ a_{31}\ &\ a_{32}\ &\ a_{33}\ \end{bmatrix} $$

one has nine $2\times 2$ determinants to compute for the cofactors and a $3\times 3$ determinant for

$$ A^{-1} = \dfrac{1}{|A|}\begin{bmatrix}\ A_{11}\ &\ A_{21}\ &\ A_{31}\ \\ \ A_{12}\ &\ A_{22}\ &\ A_{32}\ \\ \ A_{13}\ &\ A_{23}\ &\ A_{33}\ \end{bmatrix} $$

The adjugate is not recommended for computing inverses of matrices larger than $3\times 3$.

Example 1

Compute the inverse of matrix $A = \begin{bmatrix} -1\ &\ 3\ \\ \ 3\ &\ 2\ \end{bmatrix}$

$$ \begin{align*} A^{-1} &= \dfrac{1}{|A|}\,\text{adj}(A) \\ \\ &= \dfrac{1}{(-1)(2) - (3)(3)}\begin{bmatrix}\ 2\ & -3\ \\ -3\ & -1\ \end{bmatrix} \\ \\ &= -\dfrac{1}{11}\begin{bmatrix}\ 2\ & -3\ \\ -3\ & -1\ \end{bmatrix} \\ \\ &= \begin{bmatrix} -\frac{2}{11}\ &\ \frac{3}{11}\ \\ \ \frac{3}{11}\ &\ \frac{1}{11} \end{bmatrix} \end{align*} $$

Example 2

Compute the inverse of matrix $A = \begin{bmatrix}\ 2\ &\ 1\ &\ 3\ \\ \ 2\ &\ 2\ &\ 1\ \\ \ 1\ &\ 3\ &\ 2\ \end{bmatrix}$

$$ \begin{align*} A^{-1} &= \dfrac{1}{|A|}\,\text{adj}(A) \\ \\ &= \dfrac{1}{11}\begin{bmatrix}\ \ \ \begin{vmatrix}\ 2\ &\ 1\ \\ \ 3\ &\ 2\ \end{vmatrix}\ & -\begin{vmatrix}\ 2\ &\ 1\ \\ \ 1\ &\ 2\ \end{vmatrix}\ &\ \ \ \begin{vmatrix}\ 2\ &\ 2\ \\ \ 1\ &\ 3\ \end{vmatrix}\ \\ -\begin{vmatrix}\ 1\ &\ 3\ \\ \ 3\ &\ 2\ \end{vmatrix}\ &\ \ \ \begin{vmatrix}\ 2\ &\ 3\ \\ \ 1\ &\ 2\ \end{vmatrix}\ & -\begin{vmatrix}\ 2\ &\ 1\ \\ \ 1\ &\ 3\ \end{vmatrix}\ \\ \ \ \ \begin{vmatrix}\ 1\ &\ 3\ \\ \ 2\ &\ 1\ \end{vmatrix}\ &\ -\begin{vmatrix}\ 2\ &\ 3\ \\ \ 2\ &\ 1\ \end{vmatrix}\ &\ \ \ \begin{vmatrix}\ 2\ &\ 1\ \\ \ 2\ &\ 2\ \end{vmatrix}\ \end{bmatrix}^T \\ \\ &= \dfrac{1}{11}\begin{bmatrix}\ 1\ & -3\ &\ 4\ \\ \ 7\ &\ 1\ & -5\ \\ -5\ &\ 4\ &\ 2\ \end{bmatrix}^T \\ \\ &= \begin{bmatrix}\ \frac{1}{11}\ &\ \frac{7}{11}\ & -\frac{5}{11}\ \\ -\frac{3}{11}\ &\ \frac{1}{11}\ &\ \frac{4}{11}\ \\ \ \frac{4}{11}\ & -\frac{5}{11}\ &\ \frac{2}{11}\ \end{bmatrix} \end{align*} $$

---

3.3.5 Cramer's Rule

Cramer's Rule

For Cramer's Rule we need some more notation. Recall that for $n\times n$ matrix

$$ A = \begin{bmatrix}\ a_{11}\ &\ a_{12}\ &\ \cdots\ &\ a_{1n}\ \\ \ a_{21}\ &\ a_{22}\ &\ \cdots\ &\ a_{2n}\ \\ \ \vdots\ &\ \vdots\ &\ \ddots\ &\ \vdots\ \\ \ a_{n1}\ &\ a_{n2}\ &\ \cdots\ &\ a_{nn}\ \end{bmatrix} $$

we have

• $a_{ij}$ is the element of matrix $A$ at position $(i,j)$
• $\mathbf{a}^i$ is the $i^{\text{th}}$ row of matrix $A$
• $\mathbf{a}_j$ is the $j^{\text{th}}$ column of matrix $A$
• $A_{ij}$ is the cofactor of element $a_{ij}$

Let us consider $n\times n$ nonsingular matrix $A$, $n\times 1$ vector $\mathbf{b}$ and the linear system

$$ A\mathbf{x} = \mathbf{b} $$

For Cramer's Rule we need to replace the $j^{\text{th}}$ column of matrix $A$ with vector $\mathbf{b}$. For each column $1\le j\le n$, denote by $A_j$, the matrix obtained by replacing the $j^{\text{th}}$ column of matrix $A$ with vector $\mathbf{b}$. Confusing, isn't it?

Cramer's Rule give us a method of computing the solution $\mathbf{x}$ by calculating each element of vector $\mathbf{x}$ as follows:

$$ x_j = \dfrac{\text{det}(A_j)}{\text{det}(A)} = \dfrac{|A_j|}{|A|} $$

Example 3

Solve the linear system $\begin{bmatrix}\ 1\ &\ 1\ \\ \ 2\ &\ 4\ \end{bmatrix}\mathbf{x} = \begin{bmatrix}\ 4\ \\ \ 2\ \end{bmatrix}$

$$ \begin{align*} x_1 &= \dfrac{|A_1|}{|A|} = \dfrac{\begin{vmatrix}\ 4\ &\ 1\ \\ \ 2\ &\ 4\ \end{vmatrix}}{\begin{vmatrix}\ 1\ &\ 1\ \\ \ 2\ &\ 4\ \end{vmatrix}} = \dfrac{14}{2} = 7 \\ \\ x_2 &= \dfrac{|A_2|}{|A|} = \dfrac{\begin{vmatrix}\ 1\ &\ 4\ \\ \ 2\ &\ 2\ \end{vmatrix}}{\begin{vmatrix}\ 1\ &\ 1\ \\ \ 2\ &\ 4\ \end{vmatrix}} = \dfrac{-6}{2} = -3 \end{align*} $$

The solution $\mathbf{x} = \begin{bmatrix}\ \ 7\ \\ -3\ \end{bmatrix}$.

---

Your use of this self-initiated mediated course material is subject to our Creative Commons License 4.0

---