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 \(A=\matrix{-2 & 2 & 0 \\ -4 & 5 & 0 \\ -2 & 3 & 1 \\ }\).

\(\mathrm{adj}(A)={}\) \(\matrix{5 & -2 & 0 \\ 4 & -2 & 0 \\ -2 & 2 & -2 \\ }\)

Consider the matrix \(A=\matrix{-2 & 2 & 0 \\ -4 & 5 & 0 \\ -2 & 3 & 1 \\ }\). The cofactors of the nine elements of \(A\) are: \[\small\begin{array}{rclcrrclcrrclcr}
A_{11}&=&{+}\begin{vmatrix}5 & 0 \\ 3 & 1 \end{vmatrix}&=&5,
&A_{12}&=&{-}\begin{vmatrix}-4 & 0 \\ -2 & 1 \end{vmatrix}&=&4,
&A_{13}&=&{+}\begin{vmatrix}-4 & 5 \\ -2 & 3 \end{vmatrix}&=&-2\\
A_{21}&=&{-}\begin{vmatrix}-4 & 5 \\ -2 & 3 \end{vmatrix}&=&-2,
&A_{22}&=&{+}\begin{vmatrix}-2 & 0 \\ -2 & 1 \end{vmatrix}&=&-2,
&A_{23}&=&{-}\begin{vmatrix}-2 & 2 \\ -2 & 3 \end{vmatrix}&=&2\\
A_{31}&=&{+}\begin{vmatrix}2 & 0 \\ 5 & 0 \end{vmatrix}&=&0,
&A_{32}&=&{-}\begin{vmatrix}-2 & 0 \\ -4 & 0 \end{vmatrix}&=&0,
&A_{33}&=&{+}\begin{vmatrix}-2 & 2 \\ -4 & 5 \end{vmatrix}&=&2
\end{array} \]
We take the transform of the matrix of cofactors to obtain the adjoint matrix \[\mathrm{adj}(A)=\matrix{5 & -2 & 0 \\ 4 & -2 & 0 \\ -2 & 2 & -2 \\ }\]

We can calculate the determinant of the matrix \(A\) via Laplace expansion; for example \[\begin{aligned} |A| & =a_{11}\cdot A_{11}+a_{12}\cdot A_{12}+a_{13}\cdot A_{13} \\[0.25cm] &= -2\times 5 +2\times 4+0\times -2\\[0.25cm] &= -2\end{aligned}\] Note that \[\begin{aligned}A\cdot \mathrm{adj}(A)&=\matrix{-2 & 2 & 0 \\ -4 & 5 & 0 \\ -2 & 3 & 1 \\ }\cdot \matrix{5 & -2 & 0 \\ 4 & -2 & 0 \\ -2 & 2 & -2 \\ }\\[0.25cm] &= \matrix{ -2 & 0 & 0\\ 0 & -2 & 0\\ 0 & 0 & -2}\\[0.25cm] &=-2 \matrix{ 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1}\\[0.25cm] &=|A|\,I\end{aligned}\] where \(I\) is the \((3\times 3)\) identity matrix.

So: \[A^{-1}=\frac{1}{|A|}\,\mathrm{adj}(A)=\matrix{-{{5}\over{2}} & 1 & 0 \\ -2 & 1 & 0 \\ 1 & -1 & 1 \\ }\]

New example

Cramer's Rule 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: \[\lineqs{-3 x + 4 y &=& 2\cr -4 x +8 y &=& -3\cr} \]

\(\cv{x\\y} = {} \) \(\cv{-\frac{7}{2}\\-\frac{17}{8}}\)

The corresponding matrix \(A\) is equal to \[\left(\begin{array}{cc}
-3 &4 \\
-4 &8
\end{array}\right)\]
The determinant \(D\) of \(A\) can easily be computed: \[D=(-3)\times (8) - (4)\times (-4)=-8\]
Replace the first column in \(A\) by \(\cv{2\\-3}\): \(A_1\left(\cv{2\\-3}\right)=\left(\begin{array}{cc} 2 &4 \\ -3 &8
\end{array}\right)\)
So: \(D_1=\left|\begin{array}{cc} 2 &4 \\ 2 &8
\end{array}\right|=(2)\times (8) - (4)\times (-3)=28\)

Replace the second column in \(A\) by \(\cv{2\\-3}\): \(A_2\left(\cv{2\\-3}\right)=\left(\begin{array}{cc} -3 &2 \\ -4 &-3
\end{array}\right)\)
Thus: \(D_2=\left|\begin{array}{cc} -3 &2 \\ -4 &-3
\end{array}\right|=(-3)\times (-3) - (2)\times (-4)=17\)

According to Cramer's Rule: \[x=\frac{D_1}{D}=\frac{28}{-8}=-{{7}\over{2}}\quad\text{and}\quad y=\frac{D_2}{D}=\frac{17}{-8}\] So the solution is \[\cv{x\\y} = \cv{-\frac{7}{2}\\-\frac{17}{8}}\]

New example