Exercises: Determinants

Solution. \[\begin{aligned} \det (A) &= 1 \cdot \det \left[\begin{array}{cc} 1 & 1 \\ 0 & 2 \end{array}\right] - 2 \cdot \det \left[\begin{array}{cc} 0 & 1 \\ 1 & 2 \end{array}\right] + 1 \cdot \det \left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right] \\[6pt] &= 1 \times 2 - 2 \times (-1) + 1 \times (-1) = 3. \end{aligned}\]

Solution. \[\det ( M_\theta ) = \cos^2 \theta -(-\sin ^2 \theta ) = 1 .\]

Solution. Let \(A^j\) be the \(j\)th column of \(A,\) that is, \[A = (A^1 ,\, A^2 ,\, A^3 ,\, \cdots ,\, A^n ).\] Then \[\begin{aligned} \det (\lambda A) &= \det((\lambda A^1 ,\, \lambda A^2 ,\, \lambda A^3 ,\, \cdots ,\, \lambda A^n )) \\[6pt] &= \lambda \det(( A^1 ,\, \lambda A^2 ,\, \lambda A^3 ,\, \cdots ,\, \lambda A^n )) \\[6pt] &= \lambda ^2 \det(( A^1 ,\, A^2 ,\, \lambda A^3 ,\, \cdots ,\, \lambda A^n )) \\[6pt] & \quad \quad \vdots \\[6pt] &= \lambda ^n \det(( A^1 ,\, A^2 ,\, A^3 ,\, \cdots ,\, A^n )) \\[6pt] &= \lambda ^n \det(A). \end{aligned}\]

Solution. Let \(A = (a_{ij})\in M_n (K)\) and suppose that \(p\)th row consists only of \(0,\) that is, \(a_{pj}=0\) for \(j=1,\,2,\,\cdots,\,n.\) Calculating the determinant of \(A\) along the \(p\)th row, we have \[\begin{aligned} \det(A) &= \sum_{j=1}^n (-1)^{p+j} a_{pj} \det(\partial_{pj} A) \\[4pt] &= \sum_{j=1}^n (-1)^{p+j} \cdot 0 \cdot \det(\partial_{pj} A) =0. \end{aligned}\]

Solution. Let \(A = (a_{ij})\in M_n (K),\) and let \(A^j\) be the \(j\)th column of \(A,\) that is, \[A = (A^1 ,\, A^2 ,\, \cdots ,\, A^n ).\] Consider the operation \[kA^p + A^q \,\rightarrow\, A_q \quad (p\ne q)\] on \(A\) and let \(B\) be the result. Define \[b_{ij} = \begin{cases} k & \text{if} \, (i,\,j) = (p,\,q) \\[5pt] 0 & \text{otherwise,} \end{cases}\] and take \[R = (r_{ij})_{n\times n}= I_n + (b_{ij})_{n\times n}.\] Then we have \[AR = B.\] But for \[\det(\partial_{pq} R)=0\] and \[\begin{aligned} \det(R) &= \sum_{i=1}^n (-1)^{i+q} r_{iq} \det(\partial_{iq} R) \\[6pt] &= \sum_{i=1}^n (-1)^{i+q} \delta_{iq} \det(\partial_{iq} R) \\[6pt] &= \det(I_n) = 1, \end{aligned}\] we have \[\det(AR) = \det(A)\det(R) = 1\cdot \det(A) = \det(A).\] As an example, let \(n=3\) and consider \(A = (a_{ij})_{3\times 3}.\) Suppose that \(k\) is a nonzero real number and we are given the operation \[kA^2 + A^3 \,\rightarrow\, A^3 ,\] that is, the result of the operation is \[B = [A^1 ,\, A^2 ,\, kA^2 + A^3 ].\] Observe that \[AR=\left[\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right] \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & k \\ 0 & 0 & 1 \end{array}\right] = [A^1 ,\, A^2 ,\, kA^2 + A^3 ].\] Since \[\begin{aligned} \det(R) &= \det\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & k \\ 0 & 0 & 1 \end{array}\right] \\[6pt] &= 0 \cdot \det\left[\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right] - k\cdot\det\left[\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right] + 1\cdot\det\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] \\[6pt] &= 0 + 0 + \det\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] = 1 = \det(I_3 ), \end{aligned}\] we have \[\det(B) = \det(AR) = \det(A)\det(R) = \det(A)\cdot 1 = \det(A).\]

Solution. Let \(A^j\) be the \(j\)th column of \(A.\)

If \(A^2,\) \(A^3\) are linearly dependent, then \(\det(A)=0\) for any \((x_1 ,\,x_2 ,\,x_3 ).\) Hence \[\left\{ (x_1 ,\,x_2 ,\,x_3 ) \,\vert\, \det(A) =0\right\} = \mathbb{R}^3.\] We now consider the case that \(A^1,\) \(A^2\) are linearly independent. Then \[\det(A)=0\] if and only if \[(x_1 ,\,x_2 ,\,x_3) = \lambda_2 A^2 + \lambda_3 A^3\] for some scalars \(\lambda_2\) and \(\lambda_3 .\) That is, \[\left\{ (x_1 ,\,x_2 ,\,x_3 )\,\vert\,\det(A)=0\right\} = \operatorname{Span}(A^2 ,\,A^3 ).\] Hence this set constitutes a \(2\)-dimensional subspace of \(\mathbb{R}^3.\)

Solution. \(A\) is expressible by \[A = [ {\mathbf{e}}_4 ,\, {\mathbf{e}}_1 ,\, {\mathbf{e}}_2 ,\, {\mathbf{e}}_5 ,\, {\mathbf{e}}_3 ].\] Since \[\left(\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 4 & 1 & 2 & 5 & 3 \end{array}\right)\] is an even permutation, we need even number of column-exchange operations to make \(A\) as the identity matrix. Hence we have \(\det(A) =1.\)

Solution. Along the \(4\)th row, we have \[\begin{aligned} \det(A) &= -0 \cdot \det (\partial_{41} A) + 2 \cdot \det(\partial_{42} A) - 0 \cdot \det(\partial_{43} A) + 0\cdot \det(\partial_{44} A) \\[6pt] &= 2\cdot \det\left[\begin{array}{crc} 1 & -1 & 2 \\ 0 & 2 & 0 \\ 4 & 3 & 1 \end{array}\right] \\[6pt] &= 2 \times 2 \times \det\left[\begin{array}{cc} 1 & 2 \\ 4 & 1 \end{array}\right] \\[6pt] &= 2 \times 2 \times (-7) = -28. \end{aligned}\]

Solution. Along the \(2\)nd column, we have \[\begin{aligned} \det(A) &= -1 \cdot \det\left[\begin{array}{ccr} 1 & 2 & 0 \\ 4 & 2 & -1 \\ 3 & 0 & 4 \end{array}\right] +1 \cdot \det\left[\begin{array}{crr} 0 & -1 & 2 \\ 4 & 2 & -1 \\ 3 & 0 & 4 \end{array}\right] +2 \cdot \det\left[\begin{array}{crr} 0 & -1 & 2 \\ 1 & 2 & 0 \\ 4 & 2 & -1 \end{array}\right] \\[6pt] &= -1 \times (8-6-32) +1 \times (3+16-12) + 2\times (4-1-16) \\[6pt] &= -1 \times (-3) + 1\times 7 + 2\times (-13) = 11. \end{aligned}\]

Solution. Using the rule of Sarrus, we have \[\begin{aligned} \det(A) &= 1 \times 0 \times 0 + 1 \times 2 \times 4 + 5 \times 2 \times 0 \\[6pt] &\quad\quad -1 \times 2 \times 0 - 1 \times 2 \times 0 - 5 \times 0 \times 4 \\[6pt] &=8. \end{aligned}\]

Solution. \[\begin{aligned} \det(A) &= -1 \times \det(\partial_{14} A) \\[6pt] &= -1 \times \det\left[\begin{array}{ccc} 0 & 2 & 2 \\ 3 & 0 & 1 \\ 0 & 0 & 2 \end{array}\right]\\[6pt] &= -1 \times 2 \times \det\left[\begin{array}{cc} 0 & 2 \\ 3 & 0 \end{array}\right] \\[6pt] &= -1 \times 2 \times (-6) = 12. \end{aligned}\]

Solution. \[\begin{aligned} \det(A) &=-\det \left[\begin{array}{ccccc} 2 & 6 & 1 & 0 & 8 \\ 0 & 4 & 5 & 1 & 7 \\ 0 & 0 & 4 & 7 & 9 \\ 0 & 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] \\[6pt] &= -2 \times 4 \times 4 \times 5 \times 1 = -160. \end{aligned}\]

Solution. We want to find the number of column-exchange operations to make \(A\) as an upper-triangular matrix. If \(n=1,\) then there is nothing to consider, hence assume that \(n\ge 2.\) If \(n=2,\) only a single operation \[A^1 \,\leftrightarrow\,A^2\] is needed. If \(n=3,\) then a single operation \[A^1 \,\leftrightarrow A^3\] is sufficient. If \(n=4\) or \(n=5,\) then two operations \[A^1 \,\leftrightarrow \, A^n \quad\text{and}\quad A^2 \,\leftrightarrow\, A^{n-1}\] are needed.

In general, if \(n=4k\) or \(n=4k+1\) where \(k\) is a positive integer, then \(2k\) operations are needed to make \(A\) as an upper-triangular matrix, and if \(n=4k+2\) or \(n=4k+3,\) then \(2k+1\) operations are needed. Hence if \(n=4k\) or \(n=4k+1,\) then \[\det(A) = (\text{product of minor diagonal entries of } A),\] and if \(n=4k+2\) or \(n=4k+3,\) then \[\det(A) = -(\text{product of minor diagonal entries of } A).\]

Solution. \[\begin{aligned} \det(A) &= \det(LU)\\[6pt] &= \det(L)\cdot\det(U) \\[6pt] &= (\text{product of diagonal entries of }L) \times \\[6pt] &\quad\quad(\text{product of diagonal entries of }U). \end{aligned}\]

Solution. Since \(\det(A) =0,\) we have \[\det(AB) = \det(A)\det(B) = 0\times \det(B) =0.\]

Solution. Let \(\det(A) =p,\) \(\det(A^{-1})=q.\) Since \(A\) and \(A^{-1}\) are integer matrices, both \(p\) and \(q\) are integers. Besides, since \[pq = \det(A)\det(A^{-1}) = \det(AA^{-1}) = \det(I_n) = 1,\] we have \(p=\pm 1.\)

Solution. (i) Interchanging two rows changes the sign of the determinant.

(ii) Multiplication of a row by a scalar \(k\) multiplies the determinant by \(k.\)

(iii) Addition of a scalar multiple of one row to another changes nothing of the determinant. (See Problem 5.)

Solution. \[\begin{aligned} \det(AA^T) &= \det(A) \det(A^T) \\[6pt] &= \det(A) \det(A) \\[6pt] &= (\det(A))^2 \ge 0. \end{aligned}\]

Solution. By the definition of similarity, there is an invertible matrix \(P\) such that \[A = P^{-1}BP.\] Hence \[\begin{aligned} \det(A) &= \det(P^{-1}BP) \\[6pt] &= \det(P^{-1})\det(B)\det(P) \\[6pt] &= (\det(P))^{-1} \det(B) \det(P) \\[6pt] &= \det(B). \end{aligned}\]

Solution. Since \[\det(A) = 6 \ne 56 = \det(B),\] two matrices are not similar. (By the preceding problem.)

Solution. Let \[A=\left[\begin{array}{rcr} 1 & 6 & 4 \\ 2 & 0 & -4 \\ -1 & 2 & 2 \end{array}\right].\] Then \(\det(A) = 24 \ne 0.\) Hence the family of column vectors of \(A\) is linearly independent, that is, the given vectors constitute a linearly independent family.

Solution. We immediately give the formula for the general case. Let \[A = \left[\begin{array}{ccccc} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1} \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-1} \\ 1 & x_3 & x_3^2 & \cdots & x_3^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1} \end{array}\right].\] As a polynomial of \(x_1,\) \(x_2,\) \(\cdots,\) \(x_n,\) the degree of \(\det(A)\) is \[1+2+3+\cdots + (n-1) = \frac{n(n-1)}{2}.\] IF \(x_i = x_j\) for \(i > j,\) then \(det(A) =0;\) Hence \(\det(A)\) has the factor \((x_i - x_j )\) for all \(i,\) \(j\) such that \(n \ge i > j \ge 1.\) Consider \[p(x_1 ,\,x_2 ,\, \cdots ,\,x_n ) = \prod_{i > j}(x_i - x_j).\] Then the degree of \(p\) is \(n(n-1)/2\) which coincides the degree of \(\det(A),\) and \(p\) has the factor \((x_i - x_j )\) for all \(i,\) \(j\) such that \(n \ge i > j \ge 1.\) Hence \[\det(A) = kp(x_1 ,\,x_2 ,\, \cdots ,\,x_n )\] for some \(k.\)

If we take the first term of each factor of \(p,\) then their product becomes \[x_2 \, x_3 ^2 \, x_4^3 \cdots x_n^{n-1}\] which is the product of diagonal entries of \(A.\) This term is unique among the expansion of \(p,\) so is in \(\det(A).\) Hence \(k=1.\)

We conclude that the family of column vectors \(A^1 ,\) \(A^2 ,\) \(\cdots ,\) \(A^n\) is linearly independent if and only if all \(x_j ' \text{s}\) are distinct.

Solution. Since the family \(f_1 ,\) \(f_2 ,\) \(\cdots ,\) \(f_n\) is linearly dependent, there exist scalars \(\lambda_j ' \text{s}\) such that \[\lambda_1 f_1 + \lambda_2 f_2 + \cdots + \lambda_n f_n = 0\] and not all \(\lambda_j ' \text{s}\) are zeros. Let \[A=\left[\begin{array}{ccccc} f_1 & f_2 & f_3 & \cdots & f_n \\ f_1 ^{(1)} & f_2 ^{(1)} & f_3 ^{(1)} & \cdots & f_n ^{(1)} \\ f_1 ^{(2)} & f_2 ^{(2)} & f_3 ^{(2)} & \cdots & f_n ^{(2)} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ f_1 ^{(n-1)} & f_2 ^{(n-1)} & f_3 ^{(n-1)} & \cdots & f_n ^{(n-1)} \end{array}\right].\] Let \(A^j\) be the \(j\)th column of \(A.\) Then \[\lambda_1 A^1 + \lambda_2 A^2 + \cdots + \lambda_n A^n = 0.\] Hence the family of column vectors of \(A\) is linearly dependent. Consequently we have \(\det(A) =0.\)

Solution. \[\begin{aligned} \det \left[\begin{array}{ccc} e^x & e^{2x} & e^{3x} \\ e^x & 2e^{2x} & 3e^{3x} \\ e^x & 4e^{2x} & 9e^{3x} \end{array}\right] &= \det \left( e^x e^{2x} e^{3x} \left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2^2 & 3^2 \end{array}\right] \right) \\[6pt] &= e^x e^{2x} e^{3x} \det \left[ \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2^2 & 3^2 \end{array}\right] \ne 0. \end{aligned}\]

Solution. If \(\lambda_i = \lambda_j\) for some \(i\ne j,\) then the family of given functions is linearly dependent.

Suppose now that \(\lambda_j ' \text{s}\) are all distinct. Let \[A = \left[\begin{array}{ccccc} e^{\lambda_1 x} & e^{\lambda_2 x} & e^{\lambda_3 x} & \cdots & e^{\lambda_n x} \\ \lambda_1 e^{\lambda_1 x} & \lambda_2 e^{\lambda_2 x} & \lambda_3 e^{\lambda_3 x} & \cdots & \lambda_n e^{\lambda_n x} \\ \lambda_1 ^2 e^{\lambda_1 x} & \lambda_2 ^2 e^{\lambda_2 x} & \lambda_3 ^2 e^{\lambda_3 x} & \cdots & \lambda_n ^2 e^{\lambda_n x} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \lambda_1 ^{n-1} e^{\lambda_1 x} & \lambda_2 ^{n-1} e^{\lambda_2 x} & \lambda_3 ^{n-1} e^{\lambda_3 x} & \cdots & \lambda_n ^{n-1} e^{\lambda_n x} \\ \end{array}\right].\] Then \[\det(A) = e^{\lambda_1 x} e^{\lambda_2 x} \cdots e^{\lambda_n x} \det\left[\begin{array}{ccccc} 1 & 1 & 1 & \cdots & 1 \\ \lambda_1 & \lambda_2 & \lambda_3 & \cdots & \lambda_n \\ \lambda_1 ^2 & \lambda_2 ^2 & \lambda_3 ^2 & \cdots & \lambda_n ^2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \lambda_1 ^{n-1} & \lambda_2 ^{n-1} & \lambda_3 ^{n-1} & \cdots & \lambda_n ^{n-1} \\ \end{array}\right] \ne 0\] by the Problem 22. Hence by Problem 23, the family of \(e^{\lambda_j x}\)'s is linearly independent.

Solution. If \(A^j\) is the \(j\)th column vector of \(A,\) then \[2A^1 + A^2 = 2A^3,\] that is, the family of column vectors of \(A\) is not linearly independent. Hence \(\det(A) =0\) and \(A^{-1}\) does not exist.

Solution. If \(A,\) \(B\in \operatorname{SL}_n (K),\) then \[\det(AB) = \det(A)\det(B) = 1\times 1 = 1,\] hence \(AB\in\operatorname{SL}_n (K).\)

If \(A\in\operatorname{SL}_n (K),\) then \(\det(A) =1.\) Hence \(A^{-1}\) exists and \[\det(A^{-1}) = (\det(A))^{-1} = 1^{-1} = 1,\] that is, \(A^{-1} \in \operatorname{SL}_n (K).\)

Therefore \(\operatorname{SL}_n(K)\) is a subgroup of \(\operatorname{GL}_n(K).\)

Solution. Keep calm and sketch a picture to derive \[\left\lvert \det(A) \right\rvert.\]