4.2 3 by 3 Determinants
Definition 4.2.1. The determinant of the 3 by 3 matrix, \[ A= \begin {pmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\\ \end {pmatrix} \] is defined by \[\det A= a_{11}\det \begin {pmatrix} a_{22}&a_{23}\\ a_{32}&a_{33}\\ \end {pmatrix} -a_{12}\det \begin {pmatrix} a_{21}&a_{23}\\ a_{31}&a_{33} \end {pmatrix} +a_{13}\det \begin {pmatrix} a_{21}&a_{22}\\ a_{31}&a_{32}\\ \end {pmatrix} \]
Example 4.2.2. The determinant of \(A= \begin {pmatrix} 1&-1&3\\2&0&4\\-3&1&5\\ \end {pmatrix} \) is
\begin {align*} \det A &=\det \begin {pmatrix} 0&4\\1&5\\ \end {pmatrix} +\det \begin {pmatrix} 2&4\\-3&5\\ \end {pmatrix} +\det \begin {pmatrix} 2&0\\-3&1 \end {pmatrix}\\\\ &=-4+22+6\\ &=24.\\ \end {align*}
The expansion of \(\det A\) above is in terms of the first row. Any row or column can be used to obtain the expansion of \(\det A\). So there are six ways of expanding \(\det A\), as follows \[\det A=\det \begin {pmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\ \end {pmatrix} \]
- 1.
- (along the first row) \(\det A=a_{11}\det \begin {pmatrix} a_{22}&a_{23}\\a_{32}&a_{33}\\ \end {pmatrix} -a_{12}\det \begin {pmatrix} a_{21}&a_{23}\\a_{31}&a_{33}\\ \end {pmatrix} +a_{13}\det \begin {pmatrix} a_{21}&a_{22}\\a_{31}&a_{32}\\ \end {pmatrix} \)
- 2.
- \(\det A=-a_{21}\det \begin {pmatrix} a_{12}&a_{13}\\a_{32}&a_{33}\\ \end {pmatrix} +a_{22}\det \begin {pmatrix} a_{11}&a_{13}\\a_{31}&a_{33}\\ \end {pmatrix} -a_{23}\det \begin {pmatrix} a_{11}&a_{12}\\a_{31}&a_{32} \end {pmatrix} \)
- 3.
- \(\det A=a_{31}\det \begin {pmatrix} a_{12}&a_{13}\\a_{22}&a_{23}\\ \end {pmatrix} -a_{32}\det \begin {pmatrix} a_{11}&a_{13}\\a_{21}&a_{23}\\ \end {pmatrix} +a_{33}\det \begin {pmatrix} a_{11}&a_{12}\\a_{21}&a_{22}\\ \end {pmatrix} \)
- 4.
- (along the first column) \(\det A=a_{11}\det \begin {pmatrix} a_{22}&a_{23}\\a_{32}&a_{33}\\ \end {pmatrix} -a_{21}\det \begin {pmatrix} a_{12}&a_{13}\\a_{32}&a_{33}\\ \end {pmatrix} +a_{31}\det \begin {pmatrix} a_{12}&a_{13}\\a_{22}&a_{23}\\ \end {pmatrix} \)
- 5.
- (along the second column) \(\det A=-a_{12}\det \begin {pmatrix} a_{21}&a_{23}\\a_{31}&a_{33}\\ \end {pmatrix} +a_{22}\det \begin {pmatrix} a_{11}&a_{13}\\a_{31}&a_{33}\\ \end {pmatrix} -a_{32}\det \begin {pmatrix} a_{11}&a_{13}\\a_{21}&a_{23}\\ \end {pmatrix} \)
- 6.
- (along the third column) \(\det A=a_{13}\det \begin {pmatrix} a_{21}&a_{22}\\a_{31}&a_{32}\\ \end {pmatrix} -a_{23}\det \begin {pmatrix} a_{11}&a_{12}\\a_{31}&a_{32}\\ \end {pmatrix} +a_{33}\det \begin {pmatrix} a_{11}&a_{12}\\a_{21}&a_{22}\\ \end {pmatrix} \)
The signs follow the chequerboard pattern \(\begin {pmatrix}+&-&+\\-&+&-\\+&-&+\end {pmatrix}\): the sign attached to \(a_{ij}\) is \((-1)^{i+j}\). That all six expansions give the same number is the content of part (1) of the next theorem together with the fact that expanding along any row gives the same answer — it is not obvious, and it is what makes ”the” determinant well defined.
Like the case of 2 by 2 determinants the following properties of 3 by 3 determinants can be proved.
Theorem 4.2.3. \[\text {Let}\hspace {0.4cm} A= \begin {pmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\ \end {pmatrix} \hspace {0.4cm}\text {be a 3 by 3 matrix. Then:} \]
- 1.
- \(\det A=\det A^t\)
- 2.
- \(\det \begin {pmatrix} ka_{11}&ka_{12}&ka_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\ \end {pmatrix} =k\det \begin {pmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\ \end {pmatrix} \) for any real number \(k\).
- 3.
- Interchanging two rows (columns) changes the sign of the determinant.
- 4.
- If any two rows (columns) are identical, then \(\det A=0\).
- 5.
- If any row or columns is a scalar multiple of another row (column) the \(\det A=0\).
For elementary row operations and determinants we have the following properties.
- i.
- If \(B\) is the matrix obtained from \(A\) by applying the elementary row operation \(r_i\longrightarrow kr_i\), then the \(\det B=k\det A\).
- ii.
- If \(B\) is the matrix obtained from \(A\) by applying the elementary row operations \(r_i\longleftrightarrow r_j\), then \(\det B=-\det A\).
- iii.
- If \(B\) is the matrix obtained from \(A\) by applying the elementary row operations \(r_i\longrightarrow r_i+kr_j\), then \(\det B=\det A\).
If two or more determinants of a 3 by 3 matrix have a common row (column), then their sum can be expressed as a single determinant.
Proof. Each part can be verified by expanding both sides into the six products \(\pm a_{1i}a_{2j}a_{3k}\) and comparing, but the expansions are more illuminating than the arithmetic, so we indicate the reason in each case.
- 1.
- Expanding \(\det A\) along the first row and \(\det A^t\) along the first column produces the same three terms: transposing exchanges rows with columns, and the six expansions listed above already show that a row expansion and the corresponding column expansion agree.
- 2.
- Expanding along the row that has been multiplied, every term of the expansion contains exactly one entry from that row. Multiplying that row by \(k\) therefore multiplies every term by \(k\), so the whole determinant is multiplied by \(k\). Note this is a statement about one row: multiplying the entire \(3\) by \(3\) matrix by \(k\) multiplies the determinant by \(k^3\).
- 3.
- Each of the six products \(\pm a_{1i}a_{2j}a_{3k}\) contains exactly one entry from each row and one from each column. Interchanging two rows permutes these products among themselves but reverses the sign attached to each, so the sum changes sign.
- 4.
- If two rows are identical, interchanging them leaves the matrix unchanged but, by (3), changes the sign of the determinant. Hence \(\det A=-\det A\), so \(2\det A=0\) and \(\det A=0\).
- 5.
- Suppose row \(i\) equals \(k\) times row \(j\). If \(k=0\) the row is zero and every term of the expansion along it vanishes. Otherwise, by (2) we may take the factor \(k\) outside, leaving a determinant with two identical rows, which is zero by (4). So \(\det A=k\cdot 0=0\).
The three statements about elementary row operations now follow: (i) is (2); (ii) is (3); and for (iii), the determinant of the matrix obtained by \(r_i\rightarrow r_i+kr_j\) splits, by the additivity in row \(i\) recorded at the end of the theorem, into \(\det A\) plus \(k\) times a determinant with row \(i\) equal to row \(j\), and that second determinant is \(0\) by (4). □
Determinants of 3 by 3 matrices have the following additional properties.
- i.
- If \(A\) and \(B\) are any 3 by 3 matrices then \(\det AB=\det A\det B\).
- ii.
- If \(A\) is a non-singular 3 by 3 matrix then \(\det A^{-1}=\dfrac {1}{\det A}\).
- iii.
- For any 3 by 3 matrix \(A\) and any positive integer \(n\), \(\det A^n=(\det A)^n\).
Proof.
- i.
- First suppose \(A\) is an elementary matrix \(E\). The three elementary row operations have \(\det E=k\), \(\det E=-1\) and
\(\det E=1\) respectively (apply the previous theorem to \(I\)), and by Theorem 2.6.3(1) the matrix \(EB\) is \(B\)
with that same operation applied. The previous theorem then gives \(\det (EB)=\det E\cdot \det B\) in each of the three
cases.
Now take a general \(A\) and let \(R\) be its reduced echelon form, so \(A=E_1E_2\cdots E_k R\) by Theorem 2.6.3(4). Applying the elementary case repeatedly, \[\det (AB)=\det (E_1\cdots E_kRB)=\det E_1\cdots \det E_k\det (RB),\] \[\det A=\det E_1\cdots \det E_k \det R.\] If \(A\) is invertible then \(R=I\) and \(\det (RB)=\det B\), giving the result. If \(A\) is not invertible then \(R\) has a zero row, so both \(R B\) and \(R\) have a zero row, making \(\det (AB)=0=\det A\cdot \det B\).
- ii.
- Apply (i) to \(AA^{-1}=I\): since \(\det I=1\), \[\det A\cdot \det A^{-1}=1.\] As \(A\) is non-singular, \(\det A\neq 0\), so we may divide to get \(\det A^{-1}=1/\det A\).
- iii.
- Induction on \(n\). The case \(n=1\) is trivial. If \(\det A^{n}=(\det A)^{n}\), then by (i) \[\det A^{n+1}=\det (A^{n}A)=\det A^{n}\cdot \det A=(\det A)^{n}\det A =(\det A)^{n+1}.\]
Remark. Part (ii) explains the phrase “non-singular”: a matrix has an inverse exactly when its determinant is not zero, since \(\det A\cdot \det A^{-1}=1\) is impossible if \(\det A=0\).
- 1.
- Express \(\det \begin {pmatrix} x^2+y^2&x+y&x\\y^2+z^2&y+z&y\\x^2+z^2&x+y&z\\ \end {pmatrix} \) as a sum of four determinants.
- 2.
- Use elementary row operations to find the determinant of \(A= \begin {pmatrix} 1&-1&0\\3&2&-1\\3&4&1\\ \end {pmatrix} \).
- 3.
- Given that \(A= \begin {pmatrix} 1&-1&0\\3&2&-1\\3&-4&1\\ \end {pmatrix} \) verify that \(\det A^2=(\det A)^2\).
Solution. \begin {align*} 1.\hspace {0.5cm} \det \begin {pmatrix} x^2+y^2&x+y&z\\y^2+z^2&y+z&y\\x^2+z^2&x+y&z\\ \end {pmatrix} &=\det \begin {pmatrix} x^2&x+y&x\\y^2&y+z&y\\x^2&x+y&z \end {pmatrix} +\det \begin {pmatrix} y^2&x+y&x\\z^2+y+z&y\\z^2&x+y&z\\ \end {pmatrix}\\\\ &=\det \begin {pmatrix} x^2&x&x\\y^2&y&y\\x^2&x&z\\ \end {pmatrix} +\det \begin {pmatrix} x^2&y&x\\y^2&z&y\\x^2&y&z\\ \end {pmatrix} +\det \begin {pmatrix} y^2&x&x\\z^2&y&y\\z^2&x&z\\ \end {pmatrix}\\\\ & +\det \begin {pmatrix} y^2&y&x\\z^2&z&y\\z^2&y&z\\ \end {pmatrix}\\ \end {align*}
\begin {align*} 2.\hspace {0.5cm} \det A &=\det \begin {pmatrix} 1&-1&0\\3&2&-1\\3&4&1\\ \end {pmatrix}=\det \begin {pmatrix} 1&-1&0\\0&5&-1\\0&7&1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_2\rightarrow r_2-3r_1\\ \longrightarrow \\ r_3\rightarrow r_3-3r_1\\ \end {matrix}\\\\ &=\det \begin {pmatrix} 1&-1&0\\0&5&-1\\0&2&2\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_3\rightarrow r_3-r_2\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm}=2\det \begin {pmatrix} 1&-1&0\\0&5&-1\\0&1&1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_3\rightarrow \frac {1}{2}r_3 \end {matrix}\\\\ &=2\det \begin {pmatrix} 1&-1&0\\0&0&-6\\0&1&1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_2\rightarrow r_2-5r_3\\\\ \end {matrix} \hspace {0.3cm} =-12\det \begin {pmatrix} 1&-1&0\\0&0&1\\0&1&1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_2\rightarrow r_2-5r_3\\ \end {matrix}\\\\ &=-12\det \begin {pmatrix} 1&-1&0\\0&0&1\\0&1&0\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_3\rightarrow r_3-r_2\\ \end {matrix} \hspace {0.3cm} =-12\det \begin {pmatrix} 1&0&0\\ 0&0&1\\0&1&0\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_1\rightarrow r_1+r_3\\ \end {matrix}\\\\ &=12\det \begin {pmatrix} 1&0&0\\0&1&0\\0&0&1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_2\longleftrightarrow r_3 \end {matrix}\\\\ &=12\\ \end {align*}
\[3.\hspace {2cm} A^2= \begin {pmatrix} 1&-1&0\\3&2&-1\\3&-4&1\\ \end {pmatrix} \begin {pmatrix} 1&-1&0\\3&2&-1\\3&-4&1\\ \end {pmatrix} = \begin {pmatrix} -2&-3&1\\6&5&-3\\-6&-15&5\\ \end {pmatrix} \] \begin {align*} \therefore \hspace {0.3cm} \det A^2 &=\det \begin {pmatrix} -2&-3&1\\6&5&-3\\-6&-15&5\\ \end {pmatrix}\\\\ &=-2\det \begin {pmatrix} 5&-3\\-15&5\\ \end {pmatrix} +3\det \begin {pmatrix} 6&-3\\-6&5\\ \end {pmatrix} +\det \begin {pmatrix} 6&5\\-6&-15\\ \end {pmatrix}\\\\ &=-2(-20)+3(12)-60=40+36-60\\ &=16\\\\ \det A&=\det \begin {pmatrix} 1&-1&0\\3&2&-1\\3&-4&1 \end {pmatrix}=\det \begin {pmatrix} 2&-1\\-4&1\\ \end {pmatrix}+\det \begin {pmatrix} 3&-1\\3&1\\ \end {pmatrix} +0\det \begin {pmatrix} 3&2\\3&-4\\ \end {pmatrix} \\\\ &=-2+6=4\\\\ \therefore \hspace {0.4cm} (\det A)^2=4^2=16=\det A^2\\ \end {align*}
The determinant of a 3 by 3 matrix \(A= \begin {pmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\ \end {pmatrix} \) is given by \[\det A=a_{11}\det \begin {pmatrix} a_{22}&a_{23}\\a_{32}&a_{33}\\ \end {pmatrix} -a_{12}\det \begin {pmatrix} a_{21}&a_{23}\\a_{31}&a_{33}\\ \end {pmatrix} +a_{13}\det \begin {pmatrix} a_{21}&a_{22}\\a_{31}&a_{32}\\ \end {pmatrix} \] Each of the 2 by 2 determinants is obtained by deleting the row and the column of \(A\) that contains the entry the determinant is being multiplied by. For example the first term is \(a_{11}\) multiplied by the determinant of the 2 by 2 matrix obtained by deleting row 1 and column 1. Also the plus and minus signs alternate term by term in the expansion of \(A\).
Let \(A_{ij}\) denote the 2 by 2 matrix obtained by deleting the \(i^{\text {th}}\) row and \(j^{\text {th}}\) column of \(A\). Then the equation \begin {align*} \det A &=a_{11}\det \begin {pmatrix} a_{22}&a_{23}\\a_{32}&a_{33} \end {pmatrix} -a_{12}\det \begin {pmatrix} a_{21}&a_{23}\\a_{31}&a_{33}\\ \end {pmatrix} +a_{13}\det \begin {pmatrix} a_{21}&a_{22}\\a_{31}&a_{32}\\ \end {pmatrix}\\ \det A &=a_{11}\det A_{11}-a_{12}\det A_{12}+a_{13}\det A_{13}\\ \implies \hspace {0.3cm} \det A &=\sum ^3_{j=1}(-1)^{i+j}a_{ij}\det A_{ij} \end {align*}
The number \(m_{ij}=\det A_{ij}\) is called the \((i,j)-\) minor of \(A\).
Example 4.2.6. If \(A= \begin {pmatrix} 1&-1&0\\2&1&3\\-2&0&-1\\ \end {pmatrix} \), then the minors of \(A\) are \(M_{11}=\det \begin {pmatrix} 1&3\\0&-1\\ \end {pmatrix} =-1\),
\(M_{12}=\det \begin {pmatrix} 2&3\\-2&-1\\ \end {pmatrix} =4\), \(M_{13}=\det \begin {pmatrix} 2&1\\-2&0\\ \end {pmatrix} =2\), \(M_{21}=\det \begin {pmatrix} -1&0\\0&-1\\ \end {pmatrix} =1\),
\(M_{22}=\det \begin {pmatrix} 1&0\\-2&-1\\ \end {pmatrix} =-1\), \(M_{23}=\det \begin {pmatrix} 1&-1\\-2&0\\ \end {pmatrix} =-2\), \(M_{31}=\det \begin {pmatrix} -1&0\\1&3\\ \end {pmatrix} =-3\), \(M_{32}=\det \begin {pmatrix} 1&0\\2&3\\ \end {pmatrix} =3\), \(M_{33}=\det \begin {pmatrix} 1&-1\\2&1\\ \end {pmatrix} =3\).
The \((i,j)-\) cofactor of the matrix \(A= \begin {pmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\ \end {pmatrix} \) is defined by \(C_{ij}=(-1)^{i+j}M_{ij}\), where \(M_{ij}\) is the \(C_{ij}-\) minor of \(A\).
Example 4.2.7. From the previous example the minors of the matrix \(A= \begin {pmatrix} 1&-1&0\\2&1&3\\-2&0&1\\ \end {pmatrix} \) are \(M_{11}=-1\), \(M_{12}=4\), \(M_{13}=2\), \(M_{21}=1\), \(M_{22}=-1\), \(M_{23}=-2\), \(M_{31}=-3\), \(M_{32}=3\) and \(M_{33}=3\).
Therefore the cofactors of \(A\) are \(C_{11}=(-1)^{1+1}M_{11}=-1\), \(C_{12}=(-1)^{1+2}M_{12}=-4\),
\(C_{13}=(-1)^{1+3}M_{13}=2\), \(C_{21}=(-1)^{2+1}M_{21}=-1\), \(C_{22}=(-1)^{2+2}M_{22}=-1\),
\(C_{23}=(-1)^{2+3}M_{23}=2\), \(C_{31}=(-1)^{3+1}M_{31}=-3\), \(C_{32}=(-1)^{3+2}M_{32}=-3\),
\(C_{33}=(-1)^{3+3}M_{33}=3\).
Definition 4.2.8. The adjoint of the matrix \(A= \begin {pmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\ \end {pmatrix} \) is defined by \(\text {adj}A= \begin {pmatrix} C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{33}\\ \end {pmatrix}^t \) where \(C_{ij}\) is the \((i,j)\)-cofactor of \(A\).
Note that the adjoint of the matrix \(A\) is the transpose of the matrix of cofactors of \(A\).
Example 4.2.9. From Example 4.2.7 the adjoint of the matrix \(A= \begin {pmatrix} 1&-1&0\\2&1&3\\-2&0&-1\\ \end {pmatrix} \) is \[\text {adj} A= \begin {pmatrix} C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{33} \end {pmatrix}^t= \begin {pmatrix} -1&-4&2\\-1&-1&2\\-3&-3&3\\ \end {pmatrix}^t = \begin {pmatrix} -1&-1&-3\\-4&-1&-3\\2&2&3\\ \end {pmatrix} \]
Theorem 4.2.10. If \(A\) is a non-singular 3 by 3 matrix then \(A^{-1}=\dfrac {1}{\det A}\,\text {adj}\,A\).
Proof. Write \(C_{ij}\) for the cofactor of \(a_{ij}\), so that \(\text {adj}\,A\) is the matrix whose \((i,j)\) entry is \(C_{ji}\) — the transpose of the matrix of cofactors. Consider the \((i,j)\) entry of the product \(A\,\text {adj}\,A\): \[\left (A\,\text {adj}\,A\right )_{ij}=\sum _{k}a_{ik}\left (\text {adj}\,A\right )_{kj} =\sum _{k}a_{ik}C_{jk}.\] Two cases arise.
When \(i=j\)
The sum is \(\sum _k a_{ik}C_{ik}\), which is exactly the cofactor expansion of \(\det A\) along row \(i\). So the diagonal entries are all \(\det A\).
When \(i\neq j\)
The sum \(\sum _k a_{ik}C_{jk}\) pairs the entries of row \(i\) with the cofactors of row \(j\). That is the cofactor expansion, along row \(j\), of the matrix obtained from \(A\) by replacing row \(j\) with a copy of row \(i\) — the cofactors of row \(j\) do not involve row \(j\) at all, so they are unchanged by the replacement. That matrix has two equal rows, so its determinant is \(0\). Hence every off-diagonal entry is \(0\).
Therefore \[A\,\text {adj}\,A=(\det A)\,I.\] Since \(A\) is non-singular, \(\det A\neq 0\), and dividing by it gives \(A\left (\frac {1}{\det A}\text {adj}\,A\right )=I\). The same computation with columns in place of rows gives \(\left (\frac {1}{\det A}\text {adj}\,A\right )A=I\), so this matrix is the inverse. □
Example 4.2.11. Let \(A= \begin {pmatrix} 1&-1&0\\2&1&3\\-2&0&-1\\ \end {pmatrix} \). Then \(\det A=3\) from Example 4.2.9 \(\text {adj}A= \begin {pmatrix} -1&-1&-3\\-4&-1&-3\\2&2&3\\ \end {pmatrix} \), therefore \[A^{-1}=\frac {1}{\det A}\hspace {0.1cm}.\hspace {0.1cm}\text {adj}A=\frac {1}{3} \begin {pmatrix} -1&-1&-3\\-4&-1&-3\\2&2&3\\ \end {pmatrix} \]
Determinants and the inverse of a matrix can be used to solve systems of linear equations in which there are three variables and three equations. \begin {align*} \text {Let}\hspace {0.4cm} a_{11}x+a_{12}y+a_{13}z &=b_1\\ a_{21}x+a_{22}y+a_{23}z&=b_2\\ a_{31}x+a_{32}y+a_{33}z&=b_3 \end {align*}
be a system of 3 linear equations in 3 variables. If the coefficients matrix \(A= \begin {pmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\ \end {pmatrix} \) is non-singular, then \[ \begin {pmatrix} x\\y\\z\\ \end {pmatrix} = \begin {pmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\ \end {pmatrix}^{-1} \begin {pmatrix} b_1\\b_2\\b_3\\ \end {pmatrix} =\frac {1}{\det A}.\text {adj}A \begin {pmatrix} b_1\\b_2\\b_3\\ \end {pmatrix} \]
Example 4.2.12. Use the inverse matrix method to solve the system of linear equations \begin {align*} x-y&=1\\2x+y+3z&=-3\\-2x-z&=2\\ \end {align*}
Solution. The coefficient matrix \(A= \begin {pmatrix} 1&-1&0\\2&1&3\\-2&0&-1\\ \end {pmatrix} \) from Example 4.2.11 \begin {align*} \begin {pmatrix} x\\y\\z\\ \end {pmatrix} &=\frac {1}{\det A}\hspace {0.1cm}.\hspace {0.1cm}\text {adj}A \begin {pmatrix} 1\\-3\\2\\ \end {pmatrix} =\frac {1}{3} \begin {pmatrix} -1&-1&-3\\-4&-1&-3\\2&2&3\\ \end {pmatrix} \begin {pmatrix} 1\\-3\\2\\ \end {pmatrix}\\\\ &=\frac {1}{3} \begin {pmatrix} -4\\-7\\2\\ \end {pmatrix}\\\\ &= \begin {pmatrix} -4/3\\-7/3\\2/3\\ \end {pmatrix}\\ \end {align*}
Theorem 4.2.13 (Cramer’s Rule). \begin {align*} \text {Let}\hspace {0.5cm} a_{11}x+a_{12}y+a_{13}z&=b_1\\a_{21}x+a_{22}y+a_{23}z&=b_2\\a_{31}x+a_{32}y+a_{33}z&=b_3 \end {align*}
be a system of linear equations in the variables. If the coefficients matrix \(A= \begin {pmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\ \end {pmatrix} \) is non-singular, then the solution of the system is given by \[x=\frac { \begin {vmatrix} b_1&a_{12}&a_{13}\\b_2&a_{22}&a_{23}\\b_3&a_{32}&a_{33}\\ \end {vmatrix} }{ \begin {vmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\ \end {vmatrix} },\hspace {1cm} y=\frac { \begin {vmatrix} a_{11}&b_1&a_{13}\\a_{21}&b_2&a_{23}\\a_{31}&b_3&a_{33}\\ \end {vmatrix} }{ \begin {vmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end {vmatrix} } ,\hspace {1cm}z=\frac { \begin {vmatrix} a_{11}&a_{12}&b_1\\a_{21}&a_{22}&b_2\\a_{31}&a_{32}&b_3\\ \end {vmatrix} }{ \begin {vmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\ \end {vmatrix} } \]
Proof. Since \(A\) is non-singular the system \(A\underline {x}=\underline {b}\) has the unique solution \(\underline {x}=A^{-1}\underline {b}\), and by Theorem 4.2.10, \[\underline {x}=\frac {1}{\det A}\,\text {adj}\,A\,\underline {b}.\] Take the \(i\)-th component. Since \((\text {adj}\,A)_{ik}=C_{ki}\), \[x_i=\frac {1}{\det A}\sum _{k}C_{ki}b_k.\] Now let \(A_i\) be the matrix obtained from \(A\) by replacing its \(i\)-th column with \(\underline {b}\). Expanding \(\det A_i\) along that column gives \[\det A_i=\sum _{k}b_k\,C_{ki},\] because the cofactors of the \(i\)-th column do not involve the entries of that column, so replacing it leaves them unchanged. Comparing the two displays, \[x_i=\frac {\det A_i}{\det A},\] which for \(i=1,2,3\) is precisely the stated formula for \(x\), \(y\) and \(z\). □
Remark. Cramer’s rule is a formula, not a method: for anything larger than \(3\) by \(3\) it requires far more arithmetic than row reduction, since it computes \(n+1\) determinants instead of doing one elimination. Its value is theoretical — it shows the solution depends on the data as a ratio of polynomials in the entries, which is the fact used whenever the solution has to be differentiated or its sign tracked as the coefficients vary.
Example 4.2.14. Use Cramer’s rule to solve the system \begin {align*} x-y&=1\\2x+y+3z&=-3\\-2x-z&=2 \end {align*}
Solution. \begin {align*} x & =\frac { \begin {vmatrix} 1&-1&0\\-3&1&3\\2&0&-1\\ \end {vmatrix} }{ \begin {vmatrix} 1&-1&0\\2&1&3\\-2&0&-1\\ \end {vmatrix} }=\frac {-1-3}{-1+4}=\frac {-4}{3}\\\\ y & =\frac { \begin {vmatrix} 1&1&0\\2&-3&3\\-2&2&-1\\ \end {vmatrix} }{3}=\frac {-3}{3}=-1\\\\ z &=\frac { \begin {vmatrix} 1&-1&1\\2&1&-3\\-2&0&2\\ \end {vmatrix} }{3}=\frac {10}{3}\\\\ \end {align*}
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