Skip to main content

Section 4.5 The adjoint of a matrix and Cramer's rule

We have already used Definition 4.2.2 to define the cofactor matrix \(C\) of a matrix \(A\text{.}\) We use this to define the adjoint of a square matrix.

Definition 4.5.1. The adjoint of a matrix.

If a matrix \(A\) has \(C\) as a cofactor matrix then the adjoint of \(A\) is \(C^T\text{.}\) We write this as \(\adj(A)=C^T\text{.}\)

Example 4.5.2. The adjoint of a matrix.

Let

\begin{equation*} A= \begin{bmatrix} 1\amp2\amp1 \\ 3\amp1\amp1 \\ 1\amp2\amp2 \end{bmatrix}. \end{equation*}

Then

\begin{equation*} \det A = -5\\ M= \begin{bmatrix} 0\amp5\amp 5 \\ 2 \amp 1 \amp 0 \\ 1\amp -2\amp -5 \end{bmatrix} \\ C= \begin{bmatrix} 0\amp-5\amp 5 \\ -2 \amp 1 \amp 0 \\ 1\amp 2\amp -5 \end{bmatrix} \\ \adj(A)=C^T= \begin{bmatrix} 0\amp-2\amp1 \\ -5\amp1\amp 2\\ 5\amp0\amp-5 \end{bmatrix} \end{equation*}

and so

\begin{equation*} A^{-1}=-\frac 15 \begin{bmatrix} 0\amp-2\amp1 \\ -5\amp1\amp 2\\ 5\amp0\amp-5 \end{bmatrix} = \begin{bmatrix} 0\amp\frac25\amp-\frac15 \\ 1\amp-\frac15\amp -\frac25\\ -1\amp0\amp1 \end{bmatrix} \end{equation*}
\begin{equation*} A \adj A= \begin{bmatrix} a_{1,1} \amp a_{1,2} \amp \cdots \amp a_{1,n} \\ a_{2,1} \amp a_{2,2} \amp \cdots \amp a_{2,n} \\ \amp\amp\vdots\\ a_{n,1} \amp a_{n,2} \amp \cdots \amp a_{n,n} \\ \end{bmatrix} \begin{bmatrix} C_{1,1} \amp C_{2,1} \amp \cdots \amp C_{n,1} \\ C_{1,2} \amp C_{2,2} \amp \cdots \amp C_{n,2} \\ \amp\amp\vdots\\ C_{1,n} \amp C_{2,n} \amp \cdots \amp C_{n,n} \\ \end{bmatrix} \end{equation*}

Consider the \(i\)-\(j\) entry of \(A \adj A\text{,}\) which we write as

\begin{equation*} (A \adj A)_{i,j}=\sum_{k=1}^n A_{i,k}(\adj A)_{k,j} =\sum_{k=1}^n a_{i,k} C_{j,k}. \end{equation*}

There are two cases:

  • \(i=j\text{:}\) In this case, from the \(i\)-th row expansion of \(A\text{,}\)

    \begin{equation*} (A \adj A)_{i,i}=\sum_{k=1}^n a_{i,k} C_{i,k}=\det A \end{equation*}

  • \(i\not=j\text{:}\) For this case we use a new matrix \(B\) constructed from \(A\) by replacing \(R_j\) with \(R_i\text{,}\) that is \(R_j\gets R_i\) (this is not an elementary row operation). This means \(B_{j,k}=A_{i,k}=a_{i,k}\) for \(k=1,2,\ldots n\text{.}\) Since \(B\) has two identical rows, Theorem 4.4.12 tells us that \(\det B=0\text{.}\) On the other hand, by expanding on \(R_j\) of \(B\text{,}\) we have

    \begin{equation*} 0 =\sum_{k=1}^n B_{j,k} C_{j,k} =\sum_{k=1}^n a_{i,k} C_{j,k}= (A \adj A)_{i,j} \end{equation*}

    Combining the two cases.

    \begin{align*} A \adj A \amp = \begin{cases} \det A \amp \textrm{if } i=j\\ 0 \amp \textrm{otherwise} \end{cases}\\ \amp = (\det A)I \end{align*}

    Hence

    \begin{equation*} A\frac1{\det A} \adj A=I \end{equation*}

    and

    \begin{equation*} \frac1{\det A} \adj A=A^{-1}. \end{equation*}
Example 4.5.4. The inverse computed using the adjoint of \(A\).

Let

\begin{equation*} A= \begin{bmatrix} 1\amp 2\amp 1\\ 2\amp 3\amp 5 \\1\amp 2\amp 0 \end{bmatrix} \end{equation*}

First we compute \(\det A=1\)

Next we compute the minors:

\begin{equation*} \begin{array}{lll} M_{1,1}= -10 \amp M_{1,2}= -5 \amp M_{1,3}= 1 \\ M_{2,1}= -2 \amp M_{2,2}= -1 \amp M_{2,3}= 0 \\ M_{3,1}= 7 \amp M_{3,2}= 3 \amp M_{3,3}= -1 \end{array} \end{equation*}

from which we deduce

\begin{equation*} M= \begin{bmatrix} -10 \amp -5 \amp 1 \\ -2 \amp -1 \amp 0 \\ 7 \amp 3 \amp -1 \end{bmatrix} \qquad C= \begin{bmatrix} -10 \amp 5 \amp 1 \\ 2 \amp -1 \amp 0 \\ 7 \amp -3 \amp -1 \end{bmatrix} \end{equation*}

from which follows

\begin{equation*} A^{-1}=\frac1{\det A} \adj A=\frac 11 C^T = \begin{bmatrix} -10 \amp 2 \amp 7\\ 5 \amp -1 \amp -3 \\ 1\amp 0 \amp -1 \end{bmatrix} \end{equation*}

If \(\det A=\pm1\) then

\begin{equation*} A^{-1}=\frac1{\det A} \adj A = \pm C^T. \end{equation*}

where \(C\) is the cofactor matrix. But the entries of \(C\) are computed by taking the determinant of matrices with integer entries. Since this determinant is computed using products and sums of integers, \(C\) must have all integer entries, and hence so does \(A^{-1}\text{.}\)

Conversely, if both \(A\) and \(A^{-1}\) have all integer entries, the \(\det A\) and \(\det A^{-1}\) are both integers. But then

\begin{equation*} \det A\,\det A^{-1}= \det(AA^{-1})=\det I=1 \end{equation*}

Hence either \(\det A=\det A^{-1}=1\) or \(\det A=\det A^{-1}=-1\text{.}\)

There is a nice application of the adjoint to the solution of \(n\) equations in \(n\) unknowns. We can write such a system of linear equations as

\begin{equation*} Ax=b. \end{equation*}

If \(A\) is nonsingular, then this system has a unique solution \(x=A^{-1}b= \frac1{\det A} \adj Ab\text{.}\) We define new matrices \(A_1,A_2,\ldots,A_n\text{:}\) construct \(A_k\) by replacing the \(k\)-th column of \(A\) with \(b\text{.}\) More specifically, if the columns of \(A\) are \(C_1,C_2,\ldots,C_n\text{,}\) then

\begin{equation*} A_k= \begin{bmatrix} C_1 \cdots C_{k-1}\amp b\amp C_{k+1} \cdots C_n \end{bmatrix} \end{equation*}

Since \(A\) is invertible, we may use the cofactor matrix \(C\) to get

\begin{equation*} x=A^{-1}b=\frac1{\det A} (\adj A) b=\frac1{\det A} C^T b. \end{equation*}

Then

\begin{equation*} (C^Tb)_i=\sum_{k=1}^n C_{k,i}b_k =\sum_{k=1}^n b_kC_{k,i} =\det A_i \end{equation*}

since \(\sum_{k=1}^n b_kC_{k,i}\) is the \(i\)-th column expansion for the evaluation of \(\det A_i\text{.}\) Hence

\begin{equation*} x_i=\frac1{\det A} \det A_i=\frac{\det A_i}{\det A}. \end{equation*}
Example 4.5.7. Application of Cramer's rule.

Consider the system of linear equations

\begin{equation*} x_1+x_2+x_3=2\\ x_1-x_2+x_3=0\\ 2x_1-x_2+x_3=2 \end{equation*}

We have

\begin{equation*} A= \begin{bmatrix} 1 \amp 1\amp 1\\ 1 \amp -1 \amp 1 \\ 2 \amp -1 \amp 1 \end{bmatrix} \qquad \det A=2 \\ A_1= \begin{bmatrix} 2 \amp 1\amp 1\\ 0 \amp -1 \amp 1 \\ 2 \amp -1 \amp 1 \end{bmatrix} \qquad \det A_1=4 \\ A_2= \begin{bmatrix} 1 \amp 2\amp 1\\ 1 \amp 0 \amp 1 \\ 2 \amp 2 \amp 1 \end{bmatrix} \qquad \det A_2=2 \\ A_3= \begin{bmatrix} 1 \amp 1\amp 2\\ 1 \amp -1 \amp 0 \\ 2 \amp -1 \amp 2 \end{bmatrix} \qquad \det A_3=-2 \end{equation*}

and so

\begin{equation*} x_1=\frac42=2\\ x_2=\frac22=1\\ x_3=\frac{-2}2=-1 \end{equation*}