The adjoint matrix and Cramer's Rule

Matrices: Matrices

The adjoint matrix and Cramer's Rule

Let $A$ be an ( $n\times n$ ) matrix. For any index $(i,j)$ you can create the ( $(n-1)\times(n-1)$ ) matrix $M_{ij}$ by deleting the $i$ -th row and $j$ -th column in $A$ . The determinant $\det(M_{ij})$ is called the ( $i,j$ ) minor of the element $a_{ij}$ of $A$ . The cofactor of the element $a_{ij}$ , denoted $A_{ij}$ , is the ( $i,j$ ) minor with a sign: $A_{ij}=(-1)^{i+j}\cdot\det(M_{ij})$ If we combine these cofactors in a matrix and take the transpose of this matrix, we get the adjoint matrix $\mathrm{adj}(A)=\matrix{A_{11} & A_{21} & \cdots & A_{n1} \\ A_{12} & A_{22} & \cdots & A_{n2}\\ \vdots & \vdots &\ddots & \vdots \\ A_{1n} & A_{2n} & \cdots & A_{nn}}$

Compute the adjoint of the matrix $A=\matrix{2 & -3 & -2 \\ 1 & -2 & -1 \\ 1 & -2 & 0 \\ }$ .

$\mathrm{adj}(A)={}$ $\matrix{-2 & 4 & -1 \\ -1 & 2 & 0 \\ 0 & 1 & -1 \\ }$

Consider the matrix $A=\matrix{2 & -3 & -2 \\ 1 & -2 & -1 \\ 1 & -2 & 0 \\ }$ . The cofactors of the nine elements of $A$ are: $\small\begin{array}{rclcrrclcrrclcr} A_{11}&=&{+}\begin{vmatrix}-2 & -1 \\ -2 & 0 \end{vmatrix}&=&-2, &A_{12}&=&{-}\begin{vmatrix}1 & -1 \\ 1 & 0 \end{vmatrix}&=&-1, &A_{13}&=&{+}\begin{vmatrix}1 & -2 \\ 1 & -2 \end{vmatrix}&=&0\\ A_{21}&=&{-}\begin{vmatrix}1 & -2 \\ 1 & -2 \end{vmatrix}&=&4, &A_{22}&=&{+}\begin{vmatrix}2 & -2 \\ 1 & 0 \end{vmatrix}&=&2, &A_{23}&=&{-}\begin{vmatrix}2 & -3 \\ 1 & -2 \end{vmatrix}&=&1\\ A_{31}&=&{+}\begin{vmatrix}-3 & -2 \\ -2 & -1 \end{vmatrix}&=&-1, &A_{32}&=&{-}\begin{vmatrix}2 & -2 \\ 1 & -1 \end{vmatrix}&=&0, &A_{33}&=&{+}\begin{vmatrix}2 & -3 \\ 1 & -2 \end{vmatrix}&=&1 \end{array}$
We take the transform of the matrix of cofactors to obtain the adjoint matrix $\mathrm{adj}(A)=\matrix{-2 & 4 & -1 \\ -1 & 2 & 0 \\ 0 & 1 & -1 \\ }$

We can calculate the determinant of the matrix $A$ via Laplace expansion; for example $\begin{aligned} |A| & =a_{11}\cdot A_{11}+a_{12}\cdot A_{12}+a_{13}\cdot A_{13} \\[0.25cm] &= 2\times -2 -3\times -1-2\times 0\\[0.25cm] &= -1\end{aligned}$ Note that $\begin{aligned}A\cdot \mathrm{adj}(A)&=\matrix{2 & -3 & -2 \\ 1 & -2 & -1 \\ 1 & -2 & 0 \\ }\cdot \matrix{-2 & 4 & -1 \\ -1 & 2 & 0 \\ 0 & 1 & -1 \\ }\\[0.25cm] &= \matrix{ -1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & -1}\\[0.25cm] &=-1 \matrix{ 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1}\\[0.25cm] &=|A|\,I\end{aligned}$ where $I$ is the $(3\times 3)$ identity matrix.

So: $A^{-1}=\frac{1}{|A|}\,\mathrm{adj}(A)=\matrix{2 & -4 & 1 \\ 1 & -2 & 0 \\ 0 & -1 & 1 \\ }$

New example

We consider a system of $n$ linear equations in $n$ unknowns: $\left\{\;\begin{array}{llllllllll} a_{11}x_1 \!\!\!\!&+&\!\! a_{12}x_2 \!\!\!\!&+&\!\!\!\! \cdots \!\!\!\!&+&\!\!\!\! a_{1n}x_n\!\!\!\!&=&\!\!\!\!b_1\\ a_{21}x_1 \!\!&+&\!\! a_{22}x_2 \!\!&+&\!\! \cdots \!\!&+&\!\!\!\! a_{2n}x_n\!\!\!\!&=&\!\!\!\!b_2\\ \vdots &&\vdots &&&& \vdots&&\!\!\!\!\vdots\\ a_{n1}x_1 \!\!\!\!&+&\!\!\!\! a_{n2}x_2 \!\!\!\!&+&\!\!\!\! \cdots \!\!\!\!&+&\!\!\!\! a_{nn}x_n\!\!\!\!&=&\!\!\!\!b_n\end{array}\right.$ This can also be written using matrix-vector multiplication as: $\begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & & \vdots \\ a_{n1} & \cdots & a_{nn} \end{pmatrix}\!\!\begin{pmatrix}x_1\\ \vdots \\ x_n\end{pmatrix}= \begin{pmatrix}b_1\\ \vdots \\ b_n\end{pmatrix}$ or simply as $A\vec{x}=\vec{b}$ with $A=\begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & & \vdots \\ a_{n1} & \cdots & a_{nn} \end{pmatrix}, \qquad \vec{x}=\begin{pmatrix}x_1\\ \vdots \\ x_n\end{pmatrix},\qquad \vec{b}=\begin{pmatrix}b_1\\ \vdots \\ b_n\end{pmatrix}$ We assume that the determinant of $A$ , denoted by $D=|A|$ , is unequal to is zero so that the matrix $A$ is invertible and there is exactly one solution of the system of equation, namely $\vec{x}=A^{-1}\vec{b}=\frac{1}{D}\mathrm{adj}(A)\,\vec{b}$ Now suppose that $A_j(\vec{b})$ is the matrix obtained from $A$ by replacing the $j$ -th column of $A$ with the column vector $\vec{b}$ . Suppose $D_j$ is the determinant of $A_j(\vec{b})$ , that is, $D_j=|A_j(\vec{b})|$ . Then the unique solution of the system of linear equations is given by $x_1=\frac{D_1}{D},\quad x_2=\frac{D_2}{D}, \quad\ldots\quad x_n=\frac{D_n}{D}$

Let $I$ be the $(n\times n)$ identity matrix with column vectors $\vec{e_1}, \dots, \vec{e_n}$ . If $A\vec{x}=\vec{b}$ , then $A\cdot I_j(\vec{x})= A\cdot (\vec{e_1}, \ldots, \vec{e_{j_1}}, \vec{x}, \vec{e_{j+1}}, \ldots, \vec{e_n}) = (A\vec{e_1}, \ldots, A\vec{e_{j_1}}, A\vec{x}, A\vec{e_{j+1}}, \ldots, A\vec{e_n})=A_j(\vec{b})$ Then: $\det(A)\cdot \det(I_j\vec{x}))=\det(A_j(\vec{b}))$ Then: $\det(I_j)(\vec{x})= \begin{vmatrix} 1 & 0 & \cdots & x_1 & \cdots 0 & 0\\ 0 & 1 & \cdots & x_2 & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & & \vdots & \vdots \\ 0 & 0 & \cdots & x_j & \cdots & 0 & 0\\ \vdots & \vdots & & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & x_{n-1} & \cdots & 1 & 0\\ 0 & 0 & \cdots & x_{n} & \cdots & 0 & 1 \end{vmatrix} = x_j$ So: $x_j = \frac{D_j}{D}\qquad\text{for all }j$ \]

Determine the solution $\cv{x\\y}$ of the following system of equations via Cramer's Rule: $\lineqs{-4 x -3 y &=& -1\cr -3 x +2 y &=& -3\cr}$

$\cv{x\\y} = {}$ $\cv{\frac{11}{17}\\-\frac{9}{17}}$

The corresponding matrix $A$ is equal to $\left(\begin{array}{cc} -4 &-3 \\ -3 &2 \end{array}\right)$
The determinant $D$ of $A$ can easily be computed: $D=(-4)\times (2) - (-3)\times (-3)=-17$
Replace the first column in $A$ by $\cv{-1\\-3}$ : $A_1\left(\cv{-1\\-3}\right)=\left(\begin{array}{cc} -1 &-3 \\ -3 &2 \end{array}\right)$
So: $D_1=\left|\begin{array}{cc} -1 &-3 \\ -1 &2 \end{array}\right|=(-1)\times (2) - (-3)\times (-3)=-11$

Replace the second column in $A$ by $\cv{-1\\-3}$ : $A_2\left(\cv{-1\\-3}\right)=\left(\begin{array}{cc} -4 &-1 \\ -3 &-3 \end{array}\right)$
Thus: $D_2=\left|\begin{array}{cc} -4 &-1 \\ -3 &-3 \end{array}\right|=(-4)\times (-3) - (-1)\times (-3)=9$

According to Cramer's Rule: $x=\frac{D_1}{D}=\frac{-11}{-17}\quad\text{and}\quad y=\frac{D_2}{D}=\frac{9}{-17}$ So the solution is $\cv{x\\y} = \cv{\frac{11}{17}\\-\frac{9}{17}}$

New example