If #A# is an invertible matrix, then we can write down the solution of an equation of the form #A\vec{x} = \vec{b}# in the unknown vector #\vec{x}# directly in terms of the determinants of submatrices that occur in the row or column expansion. These submatrices are the #((n-1)\times(n-1))#-matrices #A_{ij}# that can be obtained from #A# by deleting the #i#-th row and the #j#-th column. In addition, the following concept plays a central role.
The adjunct matrix (or adjugate matrix) of the #(n\times n)#-matrix #A# is the #(n\times n)#-matrix \(\text{adj}(A) \) whose #(i,j)#-entry is equal to #(-1)^{i+j}\det(A_{ji})#.
Thus, the #(i,j)#-entry of #\text{adj}(A)# is up to a sign, not the determinant of #A_{ij}# but the determinant of #A_{ji}#.
The adjunct matrix of #\matrix{a&b\\ c&d}# is #\matrix{d&-b\\ -c&a}#.
The adjunct matrix of #\matrix{a&b&c\\ d&e&f\\ g&h&i}# is
\[\matrix{e\,i-f\,h&c\,h-b\,i&b\,f-c\,e\\ f\,g-d\,i&a\,i-c\,g&c\,d-
a\,f\\ d\,h-e\,g&b\,g-a\,h&a\,e-b\,d }\]
The adjunct matrix #\text{adj}(A)# is equal to the zero matrix if and only if the rank of #A# is smaller than #n-1#.
To see this, we distinguish two cases:
- if #\text{adj}(A)\ne0#, then the determinant of an #((n-1)\times(n-1))#-submatrix of #A# is distinct from #0#, and so this submatrix is invertible (see theorem Invertibility in terms of determinant). The #n-1# columns of #A# with the same column numbers as those of this submatrix are therefore linearly independent (see theorem Invertibility and rank). This means that then #\text{rank}(A) \ge n-1#.
- If #\text{adj}(A)=0#, then the determinant of each #((n-1)\times(n-1))#-submatrix of #A# equals #0#. If the rank of #A# would be at least #n-1#, then #A# would have #n-1# linearly independent columns. By regarding the rows of the associated #(n\times(n-1))#-matrix as vectors of #\mathbb{R}^{n-1}#, we can apply the theorem Thinning, which states that there can be found #n-1# rows that are linearly independent. By deleting the dependent row from the above #(n\times(n-1))#-matrix we find an #((n-1)\times(n-1))#-submatrix of #A# rank #n-1#. According to theorem Invertibility and rank the determinant of the submatrix is not equal to #0#. This contradicts the assumption #\text{adj}(A)=0#. We conclude that #\text{rank}(A)\lt n-1#.
If the matrix #A# is invertible, then #\text{adj}(A)# is the inverse of #A# up to a scalar multiple as follows from the rule below.
Let #A# be an #(n\times n)#-matrix.
- \(\det(A)\cdot I_n=\text{adj}(A)\, A= A\, \text{adj}(A)\).
- If #\vec{b}# is a vector in #\mathbb{R}^n#, then each solution #\vec{x}# of the equation \(A\vec{x} =\vec{b}\) satisfies \(\det(A) \vec{x} =\text{adj}(A)\vec{b} \).
- Cramer's rule: If #\vec{b}# lies in the image of #A# and #\det(A)\ne0#, then the solution #\vec{x}# of the equation \(A\vec{x} =\vec{b}\) is equal to \[ \vec{x} =\dfrac{1}{\det(A)}\,\rv{\det (A_1(\vec{b} )) ,\ldots,\det (A_n(\vec{b} )) } \] where #A_j (\vec{b} )# is the matrix obtained from #A# by replacing the #j#-th column by #\vec{b}#.
By #\vec{k}_1,\ldots ,\vec{k}_n# we denote the columns of #A#. Let #\vec{b}# be any vector in #\mathbb{R}^n#. By #A_j (\vec{b} )# we indicate the matrix that can be obtained from #A# by replacing the #j#-th column by #\vec{b}#. Then we have \[ \det (A_j(\vec{k}_\ell))=\begin{cases}\det(A)&\text{if } \ell=j\\ 0&\text{otherwise}\end{cases}\]
1. Expanding the determinant \(\det (A_j(\vec{k}_\ell))\) with respect to the #j#-th column, we find \[\begin{array}{rcl}\det(A_j(\vec{k}_\ell)) &=& \sum_{i=1}^n (-1)^{i+j}a_{i\ell}\det (A_{ij})\\&&\phantom{xxx}\color{blue}{\text{the }j\text{-th column of }A_j(\vec{k}_\ell)\text{ is }\vec{k}_\ell=\cv{a_{1\ell}\\\vdots\\a_{n\ell}}}\\&=& \sum_{i=1}^n \left((-1)^{i+j}\det (A_{ij}) \right)\cdot a_{i\ell}\\&&\phantom{xxx}\color{blue}{\text{interchanged the order of the last two factors}}\\&=&\text{the } (j,\ell)\text{-entry of }\text{adj}(A) \, A\\&&\phantom{xxx}\color{blue}{(-1)^{i+j}\det (A_{ij})=\text{ the }(j,i)\text{-entry of }\text{adj}(A)} \end{array}\]In the last step, we used the definition of the matrix product.
Because \(\det(A_j(\vec{k}_\ell))=\det(A)\) if #\ell=j# and #0# otherwise, it follows that \(\text{adj}(A) \, A \) is equal to the matrix #\det(A)\cdot I_n#.
This proves the first equality. The second follows by transposing sufficiently often:
\[\begin{array}{rcl}\det(A^\top)\cdot I_n &=& \text{adj}(A^\top)\,A^\top\\&&\phantom{xx}\color{blue}{\text{first equality applied to }A^\top}\\ \det(A^\top)\cdot I_n^\top&=&A\,\left(\text{adj}(A^\top)\right)^\top\\&&\phantom{xx}\color{blue}{\text{both sides transposed}}\\\det(A)\cdot I_n&=&A\,\left(\text{adj}(A^\top)\right)^\top\\&&\phantom{xx}\color{blue}{\det(A^\top)=\det(A)\text{ and }I_n^\top=I_n\text{ on the left-hand side}}\\ \det(A)\cdot I_n&=&A\,\text{adj}(A)\\ &&\phantom{xx}\color{blue}{\left(\text{adj}(A^\top)\right)^\top=\left(\left(\text{adj}(A)\right)^\top\right)^\top=\text{adj}(A)} \end{array}\]
2. Apply #\text{adj}(A)# to the equation \(A\vec{x} =\vec{b}\):
\[\text{adj}(A)\,A\vec{x} =\text{adj}(A)\vec{b}\]
Because of 1, the left side equals \(\det(A) \vec{x} \). This establishes the equality of 2.
3. Each solution #\vec{x} = \rv{x_1,\ldots,x_n}# of the equation \(A\vec{x} =\vec{b}\) satisfies
\[
\vec{b}=x_1\vec{k}_1+x_2\vec{k}_2+\cdots +x_n\vec{k}_n
\] Therefore,
\[
\begin{array}{rcl}
\det \left(A_j(\vec{b} )\right) & = &\det \left(A_j\left(\sum_{i=1}^n x_i\,\vec{k}_i\right)\right)\\&&\phantom{xx}\color{blue}{\text{above expression for }\vec{b}}\\
& =& \sum_{i=1}^nx_i\det\left (A_j(\vec{k}_i)\right)\\&&\phantom{xx}\color{blue}{\text{linearity of}\det\text{in the }j\text{-th column and of }A}\\
& =& x_j\det (A) \\&&\phantom{xx}\color{blue}{\text{formula for}\det(A_j(\vec{k}_\ell))\text{ at the beginning of the proof}}\\
\end{array}
\] Thus we have found Cramer's Rule: if #\det(A)\ne0#, then the solution of the above system is
\[
x_j=\dfrac{\det (A_j(\vec{b}))}{\det (A)}\phantom{xxxx}\text{for }j=1,\ldots ,n
\]
Consider the equation #A\vec{x}=\vec{b}# for #n=2#: \[\underbrace{\matrix{a_{11}&a_{12}\\a_{21}&a_{22}}}_{A}\underbrace{\cv{x_1\\x_2}}_{\vec{x}}=\underbrace{\cv{b_1\\b_2}}_{\vec{b}}\] Cramer's rule immediately gives the solution \[
\vec{x}=\dfrac{1}{\det(A)}\,\cv{\det (A_1(\vec{b}))\\\det (A_2(\vec{b}))}=\dfrac{1}{\left|\begin{array}{cc}
a_{11} & a_{12}\\
a_{21} & a_{22}
\end{array}\right|}\cv{\left|\begin{array}{cc}
b_1 & a_{12}\\
b_2 & a_{22}
\end{array}\right|\\\left|\begin{array}{cc}
a_{11} & b_1\\
a_{21} & b_2
\end{array}\right|}=\dfrac{1}{a_{11}a_{22}-a_{12}a_{21}}\cv{a_{22}b_1-a_{12}b_2\\a_{11}b_2-a_{21}b_1}
\] provided #a_{11}a_{22}-a_{12}a_{21}\neq 0#. Notice that #\vec{x}=A^{-1}\vec{b}#, since\[A^{-1}=\frac{1}{a_{11}a_{22}-a_{12}a_{21}}\,\matrix{a_{22} & -a_{12} \\ -a_{21} & a_{11}}\] We may also obtain this well known formula for the inverse of #A# using Cramer's rule. For convenience, we rename the matrix elements as follows\[A=\matrix{a&b\\c&d}\] and indicate the columns of the inverse of #A# with column vectors #\vec{x}# and #\vec{y}#: \[A^{-1}=\matrix{\vec{x}&\vec{y}}=\matrix{x_1&y_1\\x_2&y_2}\] By definition, the inverse satisfies #A\,A^{-1}=I_2#, that is\[\matrix{a&b\\c&d}\matrix{x_1&y_1\\x_2&y_2}=\matrix{1&0\\0&1}\] We can split this equation into two parts\[\underbrace{\matrix{a&b\\c&d}}_{A}\underbrace{\cv{x_1\\x_2}}_{\vec{x}}=\underbrace{\cv{1\\0}}_{\vec{e}_1}\quad \hbox{and}\quad\underbrace{\matrix{a&b\\c&d}}_{A}\underbrace{\cv{y_1\\y_2}}_{\vec{y}}=\underbrace{\cv{0\\1}}_{\vec{e}_2}\]in which #\vec{e}_1# and #\vec{e}_2# are the standard basis vectors of #\mathbb{R}^2#. Cramer's rule gives\[\vec{x}=\dfrac{1}{\det(A)}\,\cv{\det (A_1(\vec{e}_1))\\\det (A_2(\vec{e}_1))}=\frac{1}{\left|\begin{array}{cc}a&b\\c&d\end{array}\right|}\cv{\left|\begin{array}{cc}1&b\\0&d\end{array}\right|\\\left|\begin{array}{cc}a&1\\c&0\end{array}\right|}=\frac{1}{ad-bc}\cv{d\\-c}\]\[\vec{y}=\dfrac{1}{\det(A)}\,\cv{\det (A_1(\vec{e}_2))\\\det (A_2(\vec{e}_2))}=\frac{1}{\left|\begin{array}{cc}a&b\\c&d\end{array}\right|}\cv{\left|\begin{array}{cc}0&b\\1&d\end{array}\right|\\\left|\begin{array}{cc}a&0\\c&1\end{array}\right|}=\frac{1}{ad-bc}\cv{-b\\a}\]from which the known result follows
\[
A^{-1}=\matrix{\vec{x}&\vec{y}}=\frac{1}{ad-bc}\,\matrix{d & -b \\ -c & a}
\]In one of the examples above we already remarked that #\matrix{d&-b\\ -c&a}# is the adjunct of #\matrix{a&b\\ c&d}#.
The system #A\vec{x} = \vec{b}# has exactly one solution if #\det (A)\neq 0#. The remarkable fact is that by using determinants this solution can even be written down explicitly thanks to rule 3. This rule is really of any practical significance only for #n=2#. For #n\geq 3#, directly solving the system is much faster than calculating all determinants.
By means of rule 1 we can find the inverse of an invertible #(n\times n)#-matrix as \[A^{-1} = \frac{1}{\det(A)}\text{adj}(A)\] Also here, for #n\geq 3#, this formula is mainly of theoretical importance and of little practical value. Calculating inverses using row reduction is almost always more practical.
Comparison of statements 2 and 3 shows that the #j#-th component #\left(\text{adj}(A)\vec{b}\right)_j# of the vector #\text{adj}(A)\vec{b}# is equal to #\det( A_j(\vec{b}))#. This can also be seen directly: if #\vec{b} = \sum_{i=1}^n\lambda_i\vec{e}_i#, then\[\begin{array}{rcl}\left(\text{adj}(A)\vec{b}\right)_j &=& \sum_{i=1}^n\lambda_i\left(\text{adj}(A)\vec{e}_i\right)_j\\ &&\phantom{xx}\color{blue}{\text{linearity}}\\ &=& \sum_{i=1}^n\lambda_i\left(\text{adj}(A)\right)_{ji}\\&&\phantom{xx}\color{blue}{\left(\text{adj}(A)\vec{e}_i\right)_j\text{ is the }(j,i)\text{-element of adj}(A)}\\&=& \sum_{i=1}^n\lambda_i(-1)^{i+j}\det(A_{ij})\\ &&\phantom{xx}\color{blue}{\text{definition adj}(A)}\\ &=& \sum_{i=1}^n\lambda_i\det(A_{j}(\vec{e}_i))\\&&\phantom{xx}\color{blue}{\text{expansion with respect to the }j\text{-th column of }A_{j}(\vec{e}_i)\text{ gives}}\\&&\phantom{xx}\color{blue}{\det(A_{j}(\vec{e}_i))=(-1)^{i+j}\det(A_{ij})}\\&=&\det(A_{j}(\vec{b}))\\ &&\phantom{xx}\color{blue}{\text{linearity}}\end{array}\]
Let \(A\) the following \((3\times3)\)-matrix: \[A=\matrix{0 & -1 & -1 \\ -1 & 1 & 1 \\ 1 & -1 & -1 \\ }\] Calculate the adjunct matrix of \(A\).
\(\text{adj}(A)={}\) \(\matrix{0 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -1 & -1 \\ }\)
After all, \[\begin{array}{rllr} (\text{adj}(A))_{11}\!\!\! & =(-1)^2\cdot (1\cdot-1 - 1\cdot-1)\!\!\! &=&\!\!\! 0\\
(\text{adj}(A))_{12}\!\!\! &=(-1)^3\cdot (-1\cdot-1+1\cdot-1)\!\!\! &=&\!\!\! 0\\
(\text{adj}(A))_{13}\!\!\! &=(-1)^4\cdot (-1\cdot1+1\cdot1)\!\!\! &=&\!\!\! 0\\
(\text{adj}(A))_{21}\!\!\! &=(-1)^3\cdot (-1\cdot-1-1\cdot1)\!\!\! &=&\!\!\! 0\\
(\text{adj}(A))_{22}\!\!\! &=(-1)^4\cdot(0\cdot-1+1\cdot1)\!\!\! &=&\!\!\! 1\\
(\text{adj}(A))_{23}\!\!\! &=(-1)^5\cdot (0\cdot1+1\cdot-1)\!\!\! &=&\!\!\! 1\\
(\text{adj}(A))_{31}\!\!\! &=(-1)^4\cdot (-1\cdot-1-1\cdot1)\!\!\! &=&\!\!\! 0\\
(\text{adj}(A))_{32}\!\!\! &=(-1)^5\cdot (0\cdot-1+1\cdot1)\!\!\! &=&\!\!\! -1\\
(\text{adj}(A))_{33}\!\!\! &=(-1)^6\cdot(0\cdot1+1\cdot-1)\!\!\! &=&\!\!\! -1
\end{array}\] so
\[\text{adj}(A)={}\matrix{0 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -1 & -1 \\ }\]
According to Cramer's rule, #A\,\text{adj}(A) = \det(A)\cdot I_3#. Multiplication of #A# by #\text{adj}(A)# gives the zero matrix, so #\det(A) =0#.