Determinants

According to the previous definition we have: \begin{eqnarray*} \left|\begin{array}{ccc} 2&3&-2\\ 5&3&-5\\ 4&1&7 \end{array}\right|&=& 2\left|\begin{array}{cc} 3&-5\\ 1&7 \end{array}\right|-3\cdot \left|\begin{array}{ccc} 5&-5\\ 4&7 \end{array}\right|+(-2)\cdot \left|\begin{array}{ccc} 5&3\\ 4&1 \end{array}\right| \\ &=& 2\cdot (3\cdot 7-1\cdot (-5))-3\cdot (5\cdot 7-4\cdot (-5))+(-2)\cdot (5\cdot 1-4\cdot 3)\\ &=&2\cdot 26-3\cdot 55-2\cdot (-7)=52-165+14=-99. \end{eqnarray*}

The proof is omitted for now. It will follow from Theorems 2-8.

Using the property (b) from Theorem 1 we obtain \begin{eqnarray*} \left|\begin{array}{ccccc} 5&10&0&0&0\\ 3&0&9&0&0\\ 1&1&0&0&3\\ 0&0&0&1&2\\ 2&4&6&8&10 \end{array}\right| &=& 5\left|\begin{array}{ccccc} 1&2&0&0&0\\ 3&0&9&0&0\\ 1&1&0&0&3\\ 0&0&0&1&2\\ 2&4&6&8&10 \end{array}\right|=5\cdot 3\cdot2\cdot \left|\begin{array}{ccccc} 1&2&0&0&0\\ 1&0&3&0&0\\ 1&1&0&0&3\\ 0&0&0&1&2\\ 1&2&3&4&5 \end{array}\right|. \end{eqnarray*} We will now multiply the first column by \(-2\) and add it to the second to obtain: \begin{eqnarray*} 30\cdot\left|\begin{array}{ccccc} 1&2&0&0&0\\ 1&0&3&0&0\\ 1&1&0&0&3\\ 0&0&0&1&2\\ 1&2&3&4&5 \end{array}\right| &=&30 \cdot\left|\begin{array}{ccccc} 1&0&0&0&0\\ 1&-2&3&0&0\\ 1&-1&0&0&3\\ 0&0&0&1&2\\ 1&0&3&4&5 \end{array}\right|. \end{eqnarray*} Using the recursive evaluation of the determinant we obtain that the last determinant is equal to: \begin{eqnarray*} 30 \cdot\left|\begin{array}{ccccc} 1&0&0&0&0\\ 1&-2&3&0&0\\ 1&-1&0&0&3\\ 0&0&0&1&2\\ 1&0&3&4&5 \end{array}\right|= 30 \cdot\left|\begin{array}{cccc} -2&3&0&0\\ -1&0&0&3\\ 0&0&1&2\\ 0&3&4&5 \end{array}\right| =-30 \cdot\left|\begin{array}{cccc} -1&0&0&3\\ -2&3&0&0\\ 0&0&1&2\\ 0&3&4&5 \end{array}\right|. \end{eqnarray*} In the last step we exchanged positions of the first two rows, which according to Theorem 1 c) results in multiplying the determinant by \(-1\). We can now multiply the first row by \(3\) and add it to the last to obtain: \begin{eqnarray*} -30 \cdot\left|\begin{array}{cccc} -1&0&0&3\\ -2&3&0&0\\ 0&0&1&2\\ 0&3&4&5 \end{array}\right|&=&-30 \cdot\left|\begin{array}{cccc} -1&0&0&0\\ -2&3&0&-6\\ 0&0&1&2\\ 0&3&4&5 \end{array}\right|=-15\cdot -1\cdot \left|\begin{array}{ccc} 3&0&-6\\ 0&1&2\\ 3&4&5 \end{array}\right|=30\cdot 3\cdot \left|\begin{array}{ccc} 1&0&-2\\ 0&1&2\\ 3&4&5 \end{array}\right|=90\cdot \left|\begin{array}{ccc} 1&0&0\\ 0&1&2\\ 3&4&11 \end{array}\right|\\ &=& 90\cdot \left|\begin{array}{cc} 1&2\\ 4&11 \end{array}\right|=90\cdot (1\cdot 11-2\cdot 4)=270. \end{eqnarray*}

We will prove by induction that the determinant is equal to \(\prod_{i < j}(\alpha_i-\alpha_j)\). The statement clearly holds for \(n=2\). Assume that it holds for \(n-1\).

We will first multiply the \(n-1\)-st column by \(\alpha_1^{n-2}\) and subtract it from the \(n\)-th. The original determinant becomes equal to: \[ \left|\begin{array}{ccccc} 1& \alpha_1&\alpha_1^2&\cdots&\alpha_1^{n-1}\\ 1& \alpha_2&\alpha_2^2&\cdots&\alpha_2^{n-1}\\ & & &\vdots& \\ 1& \alpha_n&\alpha_n^2&\cdots&\alpha_n^{n-1}\\ \end{array}\right| = \left|\begin{array}{cccccc} 1& \alpha_1&\alpha_1^2&\cdots&\alpha_1^{n-2}&0\\ 1& \alpha_2&\alpha_2^2&\cdots&\alpha_2^{n-2}&\alpha_2^{n-2}(\alpha_2-\alpha_1)\\ & & &\vdots&& \\ 1& \alpha_n&\alpha_n^2&\cdots&\alpha_n^{n-2}&\alpha_n^{n-2}(\alpha_n-\alpha_1)\\ \end{array}\right|. \] We now multiply the \(n-2\)-nd column by \(\alpha_1^{n-3}\) and subtract it from the \(n-1\)st. The determinant becomes equal to: \[ \left|\begin{array}{cccccc} 1& \alpha_1&\alpha_1^2&\cdots&\alpha_1^{n-2}&0\\ 1& \alpha_2&\alpha_2^2&\cdots&\alpha_2^{n-2}&\alpha_2^{n-2}(\alpha_2-\alpha_1)\\ & & &\vdots&& \\ 1& \alpha_n&\alpha_n^2&\cdots&\alpha_n^{n-2}&\alpha_n^{n-2}(\alpha_n-\alpha_1)\\ \end{array}\right|= \left|\begin{array}{cccccc} 1& \alpha_1&\alpha_1^2&\cdots&0&0\\ 1& \alpha_2&\alpha_2^2&\cdots&\alpha_2^{n-3}(\alpha_2-\alpha_1)&\alpha_2^{n-2}(\alpha_2-\alpha_1)\\ & & &\vdots&& \\ 1& \alpha_n&\alpha_n^2&\cdots&\alpha_n^{n-3}(\alpha_n-\alpha_1)&\alpha_n^{n-2}(\alpha_n-\alpha_1)\\ \end{array}\right|. \] Repeating the same procedure with the remaining columns, and using the part (b) of Theorem 1 we obtain: \begin{eqnarray*} \left|\begin{array}{ccccc} 1& \alpha_1&\alpha_1^2&\cdots&\alpha_1^{n-1}\\ 1& \alpha_2&\alpha_2^2&\cdots&\alpha_2^{n-1}\\ & & &\vdots& \\ 1& \alpha_n&\alpha_n^2&\cdots&\alpha_n^{n-1}\\ \end{array}\right| &=& \left|\begin{array}{cccccc} 1& 0&0&\cdots&0&0\\ 1& \alpha_2-\alpha_1&\alpha_2(\alpha_2-\alpha_1)&\cdots&\alpha_2^{n-3}(\alpha_2-\alpha_1)&\alpha_2^{n-2}(\alpha_2-\alpha_1)\\ & & &\vdots&& \\ 1& \alpha_n-\alpha_1&\alpha_n(\alpha_n-\alpha_1)&\cdots&\alpha_n^{n-3}(\alpha_n-\alpha_1)&\alpha_n^{n-2}(\alpha_n-\alpha_1)\\ \end{array}\right| \\&=& (\alpha_2-\alpha_1)(\alpha_3-\alpha_1)\cdots(\alpha_n-\alpha_1) \left|\begin{array}{cccccc} 1& 0&0&\cdots&0&0\\ 1& 1&\alpha_2 &\cdots&\alpha_2^{n-3} &\alpha_2^{n-2} \\ & & &\vdots&& \\ 1& 1&\alpha_n &\cdots&\alpha_n^{n-3} &\alpha_n^{n-2} \\ \end{array}\right|\\ &=& (\alpha_2-\alpha_1)(\alpha_3-\alpha_1)\cdots(\alpha_n-\alpha_1) \left|\begin{array}{ccccc} 1&\alpha_2 &\cdots&\alpha_2^{n-3} &\alpha_2^{n-2} \\ & &\vdots&& \\ 1&\alpha_n &\cdots&\alpha_n^{n-3} &\alpha_n^{n-2} \\ \end{array}\right|. \end{eqnarray*} We have reduced the problem to the case \(n-1\) and we may use inductional hypothesis to conclude the result.

By the property of the alternating forms we have for any three vectors \(v_1\), \(v_2\), and \(v_3\) the following equality: \(\phi(v_1,v_2,v_3)=-\phi(v_3,v_2,v_1)\) with \(v_1=v_3=e_1\) and \(v_2=e_3\), we obtain \(\phi(e_1,e_3,e_1)=-\phi(e_1,e_3,e_1)\) which is equivalent to \(\phi(e_1,e_3,e_1)=0\).

\(\phi(e_1,e_3,e_2)=-\phi(e_1,e_2,e_3)=-1\).

\(\phi(e_3,e_1,e_2)=-\phi(e_1,e_3,e_2)=-(-1)=1\).

We will use the induction on \(n\) to prove that every permutation can be expressed as a composition of transpositions. The statement clearly holds for the set \(\{1,2\}\). Assume that it holds for \(n-1\), where \(n\geq 3\) is an integer. Assume that \(\sigma=(\sigma_1, \dots, \sigma_n)\) is a permutation of \(\{1,2,\dots, n\}\). Let \(k\) be an integer such that \(\sigma_k=n\). Let \(\tau\) be the transposition \((1,2,\dots, k-1,n,k,\dots, n)\). Let \(\sigma^{\prime}=(\sigma_1,\dots, \sigma_{k-1},\sigma_n,\sigma_{k+1},\dots, \sigma_{n-1},n)\). Then we have \(\sigma=\sigma^{\prime}\circ \tau\). The permutation \(\sigma^{\prime}\) leaves the last coordinate \(n\) fixed, hence the inductional hypothesis can be applied to conclude that there are transpositions \(\tau_1\), \(\dots\), \(\tau_j\) such that \(\sigma^{\prime}=\tau_1\circ\cdots\circ \tau_j\). Thus \[\sigma=\sigma^{\prime}\circ\tau=\tau_1\circ\cdots\circ \tau_j\circ \tau,\] which completes the proof of the existence of transpositions.

The pair \((i,j)\) of indices is called inversion for the permutation \(\sigma\) if \(\sigma_i > \sigma_j\). Each transposition has an odd number of inversions. Indeed if \(p < q\), the transposition \((1,2,\dots,p-1, q,p+1, \dots, q-1,\dots, p, q+1,\dots, n)\) has exactly \(2(q-p)-1\) inversions.

Assume that \(\phi\) is a permutation and \(\tau\) a transposition. Assume that \(\phi\) has \(u\) inversions, and that \(\phi\circ \tau\) has \(v\) inversions. We will prove that \(u-v\) is odd. Let \(\tau=(1,2,\dots,p-1, q,p+1, \dots, q-1,\dots, p, q+1,\dots, n)\), where \(p < q\). If \(i,j\in\{1,2,\dots, n\}\setminus\{p,q\}\), then \((i,j)\) is an inversion for \(\phi\circ \tau\) if and only if it is an inversion for \(\phi\). Let \begin{eqnarray*}A&=&\{i\in\{p+1,\dots, q-1\}: \phi_p < \phi_i\},\\ B&=&\{i\in\{p+1,\dots, q-1\}: \phi_p > \phi_i\},\\ C&=&\{i\in\{p+1,\dots, q-1\}: \phi_q < \phi_i\}, \\D&=&\{i\in\{p+1,\dots, q-1\}: \phi_q > \phi_i\}. \end{eqnarray*} We will now use \(|A|\), \(|B|\), \(|C|\), and \(|D|\) to express the difference \(u-v\) between the number of inversions in \(\phi\) and \(\phi\circ \tau\). If \(\phi_p > \phi_q\), then \(u-v=|B|+|C|-|A|-|D|+1\). If \(|phi_p < \phi_q\), then \(u-v=|B|+|C|-|A|-|D|-1\). In either case we have \(u-v\equiv |B|+|C|-|A|-|D|+1\) (mod 2). Thus \[u-v\equiv |B|+|C|-|A|-|D|+1\equiv |B|+|C|+|A|+|D|+1\quad\quad (\mbox{mod }2).\] Since \(|A|+|B|=|C|+|D|=q-p-1\), we conclude that \(u-v\) is odd.

The expression \(\sigma=\tau_1\circ \tau_2\circ \cdots \circ\tau_m\) implies that the number of inversions in \(\sigma\) is congruent to \((-1)^m\) modulo \(2\). Similarly, from \(\sigma=\mu_1\circ \mu_2\circ\cdots\circ \mu_l\) we conclude that the number of inversions in \(\sigma\) is congruent to \((-1)^l\) modulo \(2\). Thus \(k-l\) is divisible by \(2\).

Let \(\tau_1\), \(\dots\), \(\tau_m\) be the sequence of transpositions such that \(\sigma=\tau_1\circ\tau_2\circ\cdots\circ \tau_k\). Then \begin{eqnarray*}\phi(e_1, \dots, e_k)&=&-\phi(e_{\tau_m(1)},\dots, e_{\tau_m(k)})=\phi(e_{\tau_{m-1}\circ\tau_m(1)},\dots, e_{\tau_{m-1}\circ\tau_m(k)}) \\&=&-\phi(e_{\tau_{m-2}\circ\tau_{m-1}\circ\tau_m(1)},\dots, e_{\tau_{m-2}\circ\tau_{m-1}\circ\tau_m(k)})\\ &=&(-1)^k\phi(e_{\sigma_1},\dots, e_{\sigma_k})\\ &=&\mbox{sgn }(\sigma)\cdot \phi(e_{\sigma_1},\dots, e_{\sigma_k}). \end{eqnarray*}

Assume that \(v_1\), \(\dots\), \(v_k\) are vectors from \(\mathbb R^k\). Assume that \((\alpha_{ij})_{1\leq i,j\leq k}\) are real numbers such that for each \(i\in\{1,2,\dots, k\}\) we have \(v_i=\alpha_{1i}e_1+\cdots+\alpha_{ki}e_k\). Then we have \begin{eqnarray*} L(v_1,\dots, v_k)&=&L\left(\sum_{j=1}^k \alpha_{j1}e_j,\dots, \sum_{j=1}^k\alpha_{jk}e_j\right)\\ &=&\sum_{j=1}^k \alpha_{j1}L\left(e_j,\sum_{m=1}^k\alpha_{m2}e_m,\dots, \sum_{m=1}^k\alpha_{mk}e_m\right)\\ &=&\cdots\\ &=&\sum_{i_1,\dots, i_k=1}^k \alpha_{i_11}\cdots\alpha_{i_kk}L\left(e_{i_1},\dots, e_{i_k}\right). \end{eqnarray*} In the last sum the term \(L\left(e_{i_1},\dots, e_{i_k}\right)\) is non-zero if and only if \(\{i_1,\dots, i_k\}=\{1,2,\dots, k\}\), i.e. if and only if \((i_1,\dots, i_k)\) is a permutation of \(\{1,2,\dots, k\}\). Thus \[L(v_1,\dots, v_k)=\sum_{\pi}\alpha_{\pi_11}\cdots \alpha_{\pi_kk}L\left(e_{\pi_1},\dots, e_{\pi_k}\right),\] where the last sum is taken over all permutations \(\pi=(\pi_1,\dots, \pi_k)\) of \(\{1,2,\dots, k\}\). The value \(L\left(e_{\pi_1},\dots, e_{\pi_k}\right)\) is either \(1\) or \(-1\), and it depends on whether the number of transpositions in the permutation \((\pi_1,\dots, \pi_k)\) is odd or even. Therefore \[L(v_1,\dots, v_k)=\sum_{\pi}\alpha_{\pi_11}\cdots \alpha_{\pi_kk}\cdot \mbox{sgn }(\pi)\cdot L\left(e_1,\dots, e_k\right)=\sum_{\pi}\alpha_{\pi_11}\cdots \alpha_{\pi_kk}\cdot \mbox{sgn }(\pi). \] which means that \(L(v_1, \dots, v_k)\) is uniquely determined if we require \(L(e_1,\dots, e_k)=1\).

We will use the formula from the proof of Theorem 4. If we denote \(\displaystyle v_i=\sum_{j=1}^3 \alpha_{ij}\) then \[\phi(v_1,v_2,v_3)=\sum_{\pi}\alpha_{1\pi_1}\alpha_{2\pi_2} \alpha_{3\pi_3}\phi\left(e_{\pi_1},e_{\pi_2}, e_{\pi_2}\right)= \sum_{\pi}\alpha_{1\pi_1}\alpha_{2\pi_2} \alpha_{3\pi_3}\mbox{sgn }(\pi) ,\] where the summation is over all permutations \(\pi=(\pi_1,\pi_2,\pi_3)\) of \(\{1,2,3\}\). Thus: \begin{eqnarray*}\phi(v_1, v_2, v_3)&=&1\cdot1\cdot 1-1\cdot (-1)\cdot (-1)-0\cdot 0\cdot 1+0\cdot (-1)\cdot 2+2\cdot 0\cdot (-1)-2\cdot 1\cdot 2\\ &=&-4. \end{eqnarray*}

This theorem follows directly from the representations \[\det (A)=\sum_{\pi} a_{\pi_11}\cdots a_{\pi_nn}\mbox{sgn }(\pi)\quad\quad\quad \mbox{and}\quad\quad\quad \det(A^T)=\sum_{\pi} a_{1\pi_i}\cdots a_{n\pi_n}\mbox{sgn }(\pi),\] and the fact that each permutation has the same sign as its inverse.

We will prove the first equality. The second one is analogous. \begin{eqnarray*} \det (A)&=&\sum_{\pi}a_{\pi_11}\cdots a_{\pi_nn} \mbox{sgn }(\pi)=\sum_{k=1}^n\sum_{\pi:\pi_k=i}a_{\pi_11}\cdots a_{\pi_nn} \mbox{sgn }(\pi)\\ &=&\sum_{k=1}^n a_{ik}\cdot \sum_{\pi:\pi_k=i}a_{\pi_11}\cdots a_{\pi_{k-1},k-1}\cdot a_{\pi_{k+1},k+1}\cdots a_{\pi_nn} \mbox{sgn }(\pi). \end{eqnarray*} It remains to prove that \[(-1)^{i+k}\hat A_{ik}=\sum_{\pi:\pi_k=i}a_{\pi_11}\cdots a_{\pi_{k-1},k-1}\cdot a_{\pi_{k+1},k+1}\cdots a_{\pi_nn} \mbox{sgn }(\pi).\] Consider the permutation \(\pi=(\pi_1,\dots, \pi_n)\) such that \(\pi_k=i\). Consider the permutation \(\hat \pi\) on the set \(\{1,2,\dots, n\}\setminus\{i\}\) obtained when the entry \(\pi_k=i\) is removed from \(\pi\). Let us determine the number of inversions in \(\pi\) that are not present in \(\hat\pi\). Assume that \(u\) of the numbers \(\{\hat\pi_1,\dots, \hat\pi_{k-1}\}\) are smaller than \(i\). Then \(k-1-u\) of the numbers \(\{\hat \pi_1,\dots, \hat\pi_{k-1}\}\) are larger than \(i\), and they will generate inversions in \(\pi\) that are not present in \(\hat \pi\). We can also conclude that \(i-1-u\) of the numbers \(\{\hat\pi_{k},\dots, \hat\pi_n\}\) are smaller than \(i\), and they will form inversion in \(\pi\) that are not in \(\hat \pi\). Hence the difference between the number of inversions in \(\pi\) and \(\hat \pi\) is equal to \(k-1-u+i-1-u=k+i-2u-2\equiv k+i\) (mod \(2\)). Therefore \(\mbox{sgn }(\pi)=(-1)^{k+i}\mbox{sgn }(\pi)\), hence: \[\sum_{\pi:\pi_k=i}a_{\pi_11}\cdots a_{\pi_{k-1},k-1}\cdot a_{\pi_{k+1},k+1}\cdots a_{\pi_nn} \mbox{sgn }(\pi)=(-1)^{k+i}\sum_{\pi:\pi_k=i} a_{\pi_11}\cdots a_{\pi_{k-1},k-1}\cdot a_{\pi_{k+1},k+1}\cdots a_{\pi_nn} \mbox{sgn }(\hat\pi)=(-1)^{k+i} \hat A_{ik}. \] This completes the proof of the theorem.

Let \(C=AB\), and let \(c_{ij}\) be the entry on the position \((i,j)\) of the matrix \(C\). We have \[c_{ii}=\sum_{k=1}^n a_{ik}b_{ki}=\sum_{k=1}^n a_{ik}\cdot (-1)^{i+k}\hat A_{ik}=\det (A),\] according to Theorem 6. We will now prove that \(c_{ji}=0\) for \(i\neq j\). Consider the matrix \(M\) that is obtained from matrix \(A\) by replacing the \(i\)-th row by the \(j\)-th row. Then \(\det(M)=\det(M^T)=0\), since \(\det(M^T)\) is the alternating multilinear form evaluated on \(n\) vectors where \(i\)-th and \(j\)-th vectors are identical. Using Theorem 6 we obtain \[\det (M)=0=\sum_{k=1}^n a_{ik}(-1)^{i+k}\hat A_{ik}=\sum_{k=1}^n a_{jk}(-1)^{i+k}\hat A_{ik}=\sum_{k=1}^n a_{jk}b_{ki}=c_{ji}.\]

Assume that \(a_{ij}\) (\(1\leq i,j\leq n\)) are the entries of \(A\), and \(b_{ij}\), \(1\leq i,j\leq n\), are the entries of \(B\). Let us denote by \(b_1\), \(\dots\), \(b_n\) the column vectors of \(B\), i.e. for \(i\in\{1,2,\dots, n\}\) we have \[\displaystyle b_i=\left[\begin{array}{c}b_{1i}\\b_{2i}\\\vdots\\b_{ni}\end{array}\right].\] Let us denote by \(e_1\), \(\dots\), \(e_n\) the standard basis of \(\mathbb R^n\), and let us denote by \(\phi\) the multilinear form corresponding to the determinant. Then we have \(\displaystyle b_i=\sum_{j=1}^n b_{ji}e_j\) and \begin{eqnarray*} \mbox{det }(AB)&=&\phi(Ab_1,\dots, Ab_n)=\phi\left(\sum_{j=1}^n Ab_{j1}e_j,\dots, \sum_{j=1}^nAb_{jn}e_j\right)\\ &=&\sum_{j_1,\dots, j_n=1}^n \phi\left(b_{j_11}Ae_{j_1}, b_{j_22}Ae_{j_2},\dots, b_{j_nn}Ae_{j_n}\right)\\ &=&\sum_{j_1,\dots, j_n=1}^n b_{j_11}b_{j_22}\cdots b_{j_nn}\phi\left(Ae_{j_1}, Ae_{j_2},\dots,Ae_{j_n}\right). \end{eqnarray*} In the last summation the term \(\phi\left(Ae_{j_1}, Ae_{j_2},\dots,Ae_{j_n}\right)\) is zero unless \((j_1,\dots, j_n)\) is a permutation of \(\{1,2,\dots, n\}\). Therefore \begin{eqnarray*}\mbox{det }(AB)&=&\sum_{\pi} b_{\pi_11}b_{\pi_22}\cdots b_{\pi_nn}\phi\left(Ae_{\pi_1},\dots, Ae_{\pi_n}\right) \\&=&\sum_{\pi} b_{\pi_11}b_{\pi_22}\cdots b_{\pi_nn}\cdot \mbox{sgn }(\pi)\cdot \phi\left(Ae_1,\dots, Ae_n\right). \end{eqnarray*} The summations are taken over all permutations \(\pi\) of \(\{1,2,\dots, n\}\). Notice that the last factor \(\phi\left(Ae_1,\dots, Ae_n\right)\) is equal to \(\mbox{det }(A)\), hence \begin{eqnarray*}\mbox{det }(AB)&=&\mbox{det }(A)\cdot \sum_{\pi} b_{\pi_11}b_{\pi_22}\cdots b_{\pi_nn}\cdot \mbox{sgn }(\pi)=\mbox{det }(A)\cdot \mbox{det }(B).\end{eqnarray*}

The formula for evaluation of \(2\times 2\) determinant implies: \begin{eqnarray*} \left|\begin{array}{cc} {a_{11}} & {a_{12}} \\ {a_{21}} & {a_{22}} \end{array} \right|&=&{a_{11}}\cdot {a_{22}} {ma_{21}}\cdot {a_{21}}={answer}. \end{eqnarray*}

According to the theorem we have that \(\displaystyle b_{23}=\frac{(-1)^{2+3}}{\det(A)}\cdot \left|\begin{array}{cc} 4 & 2\\ 0 & 3\end{array}\right|\). We have \begin{eqnarray*} \det(A)&=&\left|\begin{array}{ccc} 4 & 3 & 2\\ 0 & 4 & 3\\ 2 & 4 & 5 \end{array}\right|=4\cdot \left|\begin{array}{cc} 4 & 3\\ 4 & 5\end{array}\right|-3\cdot \left|\begin{array}{cc} 0 & 3\\ 2 & 5\end{array}\right| + 2\cdot \left|\begin{array}{cc} 0 & 4\\ 2 & 4\end{array}\right|\\ &=&34 \end{eqnarray*} and \begin{eqnarray*} \left|\begin{array}{cc} 4 & 2\\ 0 & 3\end{array}\right|=12. \end{eqnarray*} Thus, \(\displaystyle b_{23}=-\frac{6}{17}\).

We first transform the determinant into equal one by multiplying the first column by \(-5\) and adding to the fourth. After that it is easy to reduce the determinant \(4\times 4\) to the determinant \(3\times 3\) and finish the evaluation: \begin{eqnarray*} \left|\begin{array}{cccc} 1&0&0&5\\ 0&3&0&2\\ 6&4&9&6\\ 5&0&7&5\end{array}\right| &=& \left|\begin{array}{cccc} 1&0&0&0\\ 0&3&0&2\\ 6&4&9&-24\\ 5&0&7&-20\end{array}\right| = \left|\begin{array}{ccc} 3&0&2\\ 4&9&6\\ 0&7&5\end{array}\right| =3\cdot \left|\begin{array}{cc}9&-24\\ 7&-20\end{array}\right| 2\cdot\left|\begin{array}{cc}4&9\\ 0&7\end{array}\right|=20. \end{eqnarray*}

The set of inversions of the permutation \(\pi\) is \(\{(i,j): 1\leq i < j\leq 7: \pi_i > \pi_j\}\subseteq \{(i,j): 1\leq i < j\leq 7\}\).

Let us count the number of ordered pairs \((i,j)\) of positive integers smaller than or equal to \(7\) that satisfy \(i < j\). There is a bijection between such pairs and the sequences of \(5\) zeroes and \(2\) ones where ones are at positions \(i\) and \(j\) and zeroes are at all other positions in the sequence. The number of such sequences is \(\binom{7}{2}=21\), hence the maximal number of inversions is smaller than or equal to \(21\).

The permutation \((7,6,5,\dots, 2,1)\) has exactly \(21\) inversions, hence the answer to the problem is \(21\).

Let us denote the given permutation by \(\sigma=(\sigma_1,\sigma_2,\sigma_3,\sigma_4,\sigma_5)\), where \(\sigma_1=2\), \(\sigma_2=4\), \(\sigma_3=1\), \(\sigma_4=5\), and \(\sigma_5=3\). The number of inversions is the number of pairs \((i,j)\) with \(1\leq i < j\leq 5\) such that \(\sigma_i > \sigma_j\). Since the set of such pairs is \[\{{ }(1,3), (2,3), (2,5), (4,5)\},\] the number of inversions is equal to 4.