{"id":5231,"date":"2020-09-09T11:20:13","date_gmt":"2020-09-09T02:20:13","guid":{"rendered":"https:\/\/sasamath.com\/blog\/?p=5231"},"modified":"2021-05-24T21:08:57","modified_gmt":"2021-05-24T12:08:57","slug":"exercises-determinants","status":"publish","type":"post","link":"https:\/\/sasamath.com\/blog\/articles\/exercises-determinants\/","title":{"rendered":"Exercises: Determinants"},"content":{"rendered":"
\n

This set of exercises is retrieved from the eighth chapter of Linear Algebra by Robert J. Valenza. Note that these solutions are not fully elaborated; You have to fill the descriptions by yourself.<\/p>\n<\/div>\n

<\/p>\n

Problem 8.1<\/span>
\nUsing the recursive definition given in the proof of the existence of determinant, systematically evaluate the determinant of the following matrix:
\n\\[A=\\begin{pmatrix}1&2&1\\\\0&1&1\\\\1&0&2\\end{pmatrix}.\\]\n<\/p>\n

\n

Solution.<\/span>
\n\\[\\begin{aligned}
\n\\det (A)
\n&= 1 \\cdot \\det \\left[\\begin{array}{cc} 1 & 1 \\\\ 0 & 2 \\end{array}\\right]
\n– 2 \\cdot \\det \\left[\\begin{array}{cc} 0 & 1 \\\\ 1 & 2 \\end{array}\\right]
\n+ 1 \\cdot \\det \\left[\\begin{array}{cc} 0 & 1 \\\\ 1 & 0 \\end{array}\\right] \\\\[6pt]
\n&= 1 \\times 2 – 2 \\times (-1) + 1 \\times (-1) = 3.
\n\\end{aligned}\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.2<\/span>
\nFor any angle \\(\\theta,\\) evaluate the determinant of the matrix
\n\\[M_\\theta = \\left(
\n\\begin{array}{cr} \\cos\\theta & -\\sin \\theta \\\\ \\sin \\theta & \\cos \\theta \\end{array}
\n\\right).\\]
\nRecall that this is the matrix of rotation about the origin by the angle \\(\\theta.\\)\n<\/p>\n

\n

Solution.<\/span>
\n\\[\\det ( M_\\theta ) = \\cos^2 \\theta -(-\\sin ^2 \\theta ) = 1 .\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.3<\/span>
\nShow that for \\(A\\in M_n (K)\\) and \\(\\lambda \\in K,\\) \\(\\det (\\lambda A) = \\lambda^n \\det(A).\\)\n<\/p>\n

\n

Solution.<\/span>
\nLet \\(A^j\\) be the \\(j\\)th column of \\(A,\\) that is,
\n\\[A = (A^1 ,\\, A^2 ,\\, A^3 ,\\, \\cdots ,\\, A^n ).\\]
\nThen
\n\\[\\begin{aligned}
\n\\det (\\lambda A)
\n&= \\det((\\lambda A^1 ,\\, \\lambda A^2 ,\\, \\lambda A^3 ,\\, \\cdots ,\\, \\lambda A^n )) \\\\[6pt]
\n&= \\lambda \\det(( A^1 ,\\, \\lambda A^2 ,\\, \\lambda A^3 ,\\, \\cdots ,\\, \\lambda A^n )) \\\\[6pt]
\n&= \\lambda ^2 \\det(( A^1 ,\\, A^2 ,\\, \\lambda A^3 ,\\, \\cdots ,\\, \\lambda A^n )) \\\\[6pt]
\n& \\quad \\quad \\vdots \\\\[6pt]
\n&= \\lambda ^n \\det(( A^1 ,\\, A^2 ,\\, A^3 ,\\, \\cdots ,\\, A^n )) \\\\[6pt]
\n&= \\lambda ^n \\det(A).
\n\\end{aligned}\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.4<\/span>
\nShow that if a matrix has a row or column of \\(0 \\text{‘s,}\\) then its determinant is \\(0.\\)\n<\/p>\n

\n

Solution.<\/span>
\nLet \\(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
\n\\[\\begin{aligned}
\n\\det(A)
\n&= \\sum_{j=1}^n (-1)^{p+j} a_{pj} \\det(\\partial_{pj} A) \\\\[4pt]
\n&= \\sum_{j=1}^n (-1)^{p+j} \\cdot 0 \\cdot \\det(\\partial_{pj} A) =0.
\n\\end{aligned}\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.5<\/span>
\nShow that the determinant of a matrix is unchanged if we add a scalar multiple of one column to another.\n<\/p>\n

\n

Solution.<\/span>
\nLet \\(A = (a_{ij})\\in M_n (K),\\) and let \\(A^j\\) be the \\(j\\)th column of \\(A,\\) that is,
\n\\[A = (A^1 ,\\, A^2 ,\\, \\cdots ,\\, A^n ).\\]
\nConsider the operation
\n\\[kA^p + A^q \\,\\rightarrow\\, A_q \\quad (p\\ne q)\\]
\non \\(A\\) and let \\(B\\) be the result.
\nDefine
\n\\[b_{ij} = \\begin{cases}
\nk & \\text{if} \\, (i,\\,j) = (p,\\,q) \\\\[5pt]
\n0 & \\text{otherwise,}
\n\\end{cases}\\]
\nand take
\n\\[R = (r_{ij})_{n\\times n}= I_n + (b_{ij})_{n\\times n}.\\]
\nThen we have
\n\\[AR = B.\\]
\nBut for
\n\\[\\det(\\partial_{pq} R)=0\\]
\nand
\n\\[\\begin{aligned}
\n\\det(R)
\n&= \\sum_{i=1}^n (-1)^{i+q} r_{iq} \\det(\\partial_{iq} R) \\\\[6pt]
\n&= \\sum_{i=1}^n (-1)^{i+q} \\delta_{iq} \\det(\\partial_{iq} R) \\\\[6pt]
\n&= \\det(I_n) = 1,
\n\\end{aligned}\\]
\nwe have
\n\\[\\det(AR) = \\det(A)\\det(R) = 1\\cdot \\det(A) = \\det(A).\\]
\nAs 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
\n\\[kA^2 + A^3 \\,\\rightarrow\\, A^3 ,\\]
\nthat is, the result of the operation is
\n\\[B = [A^1 ,\\, A^2 ,\\, kA^2 + A^3 ].\\]
\nObserve that
\n\\[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]
\n\\left[\\begin{array}{ccc} 1 & 0 & 0 \\\\ 0 & 1 & k \\\\ 0 & 0 & 1 \\end{array}\\right]
\n=
\n[A^1 ,\\, A^2 ,\\, kA^2 + A^3 ].\\]
\nSince
\n\\[\\begin{aligned}
\n\\det(R)
\n&= \\det\\left[\\begin{array}{ccc} 1 & 0 & 0 \\\\ 0 & 1 & k \\\\ 0 & 0 & 1 \\end{array}\\right] \\\\[6pt]
\n&= 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]
\n&= 0 + 0 + \\det\\left[\\begin{array}{cc} 1 & 0 \\\\ 0 & 1 \\end{array}\\right] = 1 = \\det(I_3 ),
\n\\end{aligned}\\]
\nwe have
\n\\[\\det(B) = \\det(AR) = \\det(A)\\det(R) = \\det(A)\\cdot 1 = \\det(A).\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.6<\/span>
\nConsider the matrix
\n\\[A = \\begin{pmatrix}
\nx_1 & a_1 & b_1 \\\\
\nx_2 & a_2 & b_2 \\\\
\nx_3 & a_3 & b_3
\n\\end{pmatrix}\\]
\nin \\(M_3 (K),\\) where the \\(a\\text{‘s}\\) and \\(b\\text{‘s}\\) are fixed and the \\(x\\text{‘s}\\) may vary. Show that the set of all \\((x_1,\\,x_2,\\,x_3)\\) such that \\(\\det(A)=0\\) is a subspace of \\(K^3\\) of dimension at least \\(2.\\)\n<\/p>\n

\n

Solution.<\/span>
\nLet \\(A^j\\) be the \\(j\\)th column of \\(A.\\)<\/p>\n

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

<\/p>\n

Problem 8.7<\/span>
\nEvaluate the determinant of the matrix
\n\\[A=\\begin{pmatrix}
\n0&1&0&0&0\\\\
\n0&0&1&0&0\\\\
\n0&0&0&0&1\\\\
\n1&0&0&0&0\\\\
\n0&0&0&1&0
\n\\end{pmatrix}.\\]
\nThis is an example of a permutation matrix<\/span> since it acts by permuting the canonical basis vectors of \\(K^5.\\)\n<\/p>\n

\n

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

<\/p>\n

Problem 8.8<\/span>
\nEvaluate the determinant of the following matrix using expansion by any row or column as appropriate.
\n\\[A = \\left(\\begin{array}{ccrc}
\n1&1&-1&2 \\\\
\n0&1&2&0 \\\\
\n4&0&3&1 \\\\
\n0&2&0&0
\n\\end{array} \\right).
\n\\]\n<\/p>\n

\n

Solution.<\/span>
\nAlong the \\(4\\)th row, we have
\n\\[\\begin{aligned}
\n\\det(A)
\n&= -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]
\n&= 2\\cdot \\det\\left[\\begin{array}{crc} 1 & -1 & 2 \\\\ 0 & 2 & 0 \\\\ 4 & 3 & 1 \\end{array}\\right] \\\\[6pt]
\n&= 2 \\times 2 \\times \\det\\left[\\begin{array}{cc} 1 & 2 \\\\ 4 & 1 \\end{array}\\right] \\\\[6pt]
\n&= 2 \\times 2 \\times (-7) = -28.
\n\\end{aligned}\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.9<\/span>
\nEvaluate the determinant of the following matrix using expansion by any row or column as appropriate.
\n\\[A = \\left(\\begin{array}{ccrc}
\n0&1&-1&2 \\\\
\n1&1&2&0 \\\\
\n4&0&2&-1 \\\\
\n3&2&0&4
\n\\end{array} \\right).
\n\\]\n<\/p>\n

\n

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

<\/p>\n

Problem 8.10<\/span>
\nEvaluate the determinant of the following matrix without expanding by either rows or columns.
\n\\[A = \\begin{pmatrix}
\n1&1&5 \\\\ 2&0&2 \\\\ 4&0&0
\n\\end{pmatrix}.
\n\\]\n<\/p>\n

\n

Solution.<\/span>
\nUsing the rule of Sarrus, we have
\n\\[\\begin{aligned}
\n\\det(A)
\n&= 1 \\times 0 \\times 0 + 1 \\times 2 \\times 4 + 5 \\times 2 \\times 0 \\\\[6pt]
\n&\\quad\\quad -1 \\times 2 \\times 0 – 1 \\times 2 \\times 0 – 5 \\times 0 \\times 4 \\\\[6pt]
\n&=8.
\n\\end{aligned}\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.11<\/span>
\nEvaluate the determinant of the matrix
\n\\[A= \\left( \\begin{array}{crcc}
\n2 & -1 & 3 & 1 \\\\
\n0 & 2 & 2 & 0 \\\\
\n3 & 0 & 1 & 0 \\\\
\n0 & 0 & 2 & 0
\n\\end{array}\\right).\\]\n<\/p>\n

\n

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

\\end{aligned}\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.12<\/span>
\nEvaluate the determinant of the matrix
\n\\[A= \\begin{pmatrix}
\n2 & 6 & 1 & 8 & 0 \\\\
\n0 & 4 & 5 & 7 & 1 \\\\
\n0 & 0 & 4 & 9 & 7 \\\\
\n0 & 0 & 0 & 1 & 5 \\\\
\n0 & 0 & 0 & 1 & 0
\n\\end{pmatrix}.\\]\n<\/p>\n

\n

Solution.<\/span>
\n\\[\\begin{aligned}
\n\\det(A)
\n&=-\\det \\left[\\begin{array}{ccccc}
\n2 & 6 & 1 & 0 & 8 \\\\
\n0 & 4 & 5 & 1 & 7 \\\\
\n0 & 0 & 4 & 7 & 9 \\\\
\n0 & 0 & 0 & 5 & 1 \\\\
\n0 & 0 & 0 & 0 & 1
\n\\end{array}\\right] \\\\[6pt]
\n&= -2 \\times 4 \\times 4 \\times 5 \\times 1 = -160.
\n\\end{aligned}\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.13<\/span>
\nConsider a matrix \\(A = \\left( a_{ij} \\right) \\in M_n (K)\\) whose elements are all zero above the minor diagonal<\/span> (from bottom left to top right). Give a succinct formula for the determinant of \\(A.\\)\n<\/p>\n

\n

Solution.<\/span>
\nWe 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
\n\\[A^1 \\,\\leftrightarrow \\, A^n \\quad\\text{and}\\quad A^2 \\,\\leftrightarrow\\, A^{n-1}\\]
\nare needed.<\/p>\n

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
\n\\[\\det(A) = (\\text{product of minor diagonal entries of } A),\\]
\nand if \\(n=4k+2\\) or \\(n=4k+3,\\) then
\n\\[\\det(A) = -(\\text{product of minor diagonal entries of } A).\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.14<\/span>
\nLet \\(A\\in M_n (K)\\) have factorization \\(A=LU\\) into the product of a lower triangular matrix \\(L\\) and an upper triangular matrix \\(U.\\) Show that the determinant of \\(A\\) is equal to the product of the diagonal terms in both \\(L\\) and \\(U.\\)\n<\/p>\n

\n

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

<\/p>\n

Problem 8.15<\/span>
\nLet \\(A,\\,B\\in M_n(K)\\) and suppose that \\(A\\) is singular, i.e., not invertible. Show that the product \\(AB\\) is also singular.\n<\/p>\n

\n

Solution.<\/span>
\nSince \\(\\det(A) =0,\\) we have
\n\\[\\det(AB) = \\det(A)\\det(B) = 0\\times \\det(B) =0.\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.16<\/span>
\nSuppose that \\(A\\) is an invertible matrix such that both \\(A\\) and \\(A^{-1}\\) consist entirely of integers. Show that the determinant of \\(A\\) is either \\(+1\\) or \\(-1.\\)\n<\/p>\n

\n

Solution.<\/span>
\nLet \\(\\det(A) =p,\\) \\(\\det(A^{-1})=q.\\) Since \\(A\\) and \\(A^{-1}\\) are integer matrices, both \\(p\\) and \\(q\\) are integers. Besides, since
\n\\[pq = \\det(A)\\det(A^{-1}) = \\det(AA^{-1}) = \\det(I_n) = 1,\\]
\nwe have \\(p=\\pm 1.\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.17<\/span>
\nConsider each of the elementary row operations(defined on matrices in Section 5.3). Analyze how each affects the determinant of a square matrix.\n<\/p>\n

\n

Solution.<\/span>
\n(i) Interchanging two rows changes the sign of the determinant.<\/p>\n

(ii) Multiplication of a row by a scalar \\(k\\) multiplies the determinant by \\(k.\\)<\/p>\n

(iii) Addition of a scalar multiple of one row to another changes nothing of the determinant. (See Problem 5.)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.18<\/span>
\nFor all \\(A\\in M_n(\\mathbb{R}),\\) prove that \\(\\det (AA^T ) \\ge 0.\\)\n<\/p>\n

\n

Solution.<\/span>
\n\\[\\begin{aligned}
\n\\det(AA^T)
\n&= \\det(A) \\det(A^T) \\\\[6pt]
\n&= \\det(A) \\det(A) \\\\[6pt]
\n&= (\\det(A))^2 \\ge 0.
\n\\end{aligned}\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.19<\/span>
\nProve that if \\(A\\) is similar to \\(B,\\) then \\(\\det (A) = \\det (B).\\)\n<\/p>\n

\n

Solution.<\/span>
\nBy the definition of similarity, there is an invertible matrix \\(P\\) such that
\n\\[A = P^{-1}BP.\\]
\nHence
\n\\[\\begin{aligned}
\n\\det(A)
\n&= \\det(P^{-1}BP) \\\\[6pt]
\n&= \\det(P^{-1})\\det(B)\\det(P) \\\\[6pt]
\n&= (\\det(P))^{-1} \\det(B) \\det(P) \\\\[6pt]
\n&= \\det(B).
\n\\end{aligned}\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.20<\/span>
\nProve that the following matrices are not similar.
\n\\[A = \\left( \\begin{array}{crc} 1 & -1 & 0 \\\\ 0 & 2 & 5 \\\\ 0 & 0 & 3 \\end{array}\\right) ,\\quad
\nB = \\left( \\begin{array}{rcc} 2 & 0 & 0 \\\\ -1 & 4 & 0 \\\\ 0 & 3 & 7 \\end{array}\\right).\\]\n<\/p>\n

\n

Solution.<\/span>
\nSince
\n\\[\\det(A) = 6 \\ne 56 = \\det(B),\\]
\ntwo matrices are not similar. (By the preceding problem.)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.21<\/span>
\nDetermine whether the following set of vectors is linearly independent.
\n\\[(1,\\,2,\\,-1),\\,\\, (6,\\,0,\\,2),\\,\\, (4,\\,-4,\\,2).\\]\n<\/p>\n

\n

Solution.<\/span>
\nLet
\n\\[A=\\left[\\begin{array}{rcr} 1 & 6 & 4 \\\\ 2 & 0 & -4 \\\\ -1 & 2 & 2 \\end{array}\\right].\\]
\nThen \\(\\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.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.22<\/span>
\nShow that
\n\\[\\det \\begin{pmatrix} 1 & x_1 & x_1 ^2 \\\\ 1 & x_2 & x_2 ^2 \\\\ 1 & x_3 & x_3 ^2 \\end{pmatrix}
\n= \\left( x_2 – x_1 \\right) \\left( x_3 – x_1 \\right) \\left( x_3 – x_2 \\right).\\]
\nConclude that the vectors
\n\\[ (1,\\,x_1 ,\\,x_1^2 ),\\,\\, (1,\\,x_2 ,\\,x_2^2 ) ,\\,\\, (1 ,\\,x_3 ,\\,x_3 ^2 )\\]
\nare linearly independent if and only if \\(x_1 ,\\) \\(x_2 \\) and \\(x_3\\) are distinct. Generalize this to higher dimensions.\n<\/p>\n

\n

Solution.<\/span>
\nWe immediately give the formula for the general case. Let
\n\\[A = \\left[\\begin{array}{ccccc}
\n1 & x_1 & x_1^2 & \\cdots & x_1^{n-1} \\\\
\n1 & x_2 & x_2^2 & \\cdots & x_2^{n-1} \\\\
\n1 & x_3 & x_3^2 & \\cdots & x_3^{n-1} \\\\
\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\
\n1 & x_n & x_n^2 & \\cdots & x_n^{n-1}
\n\\end{array}\\right].\\]
\nAs a polynomial of \\(x_1,\\) \\(x_2,\\) \\(\\cdots,\\) \\(x_n,\\) the degree of \\(\\det(A)\\) is
\n\\[1+2+3+\\cdots + (n-1) = \\frac{n(n-1)}{2}.\\]
\nIF \\(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
\n\\[p(x_1 ,\\,x_2 ,\\, \\cdots ,\\,x_n ) = \\prod_{i > j}(x_i – x_j).\\]
\nThen 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
\n\\[\\det(A) = kp(x_1 ,\\,x_2 ,\\, \\cdots ,\\,x_n )\\]
\nfor some \\(k.\\)<\/p>\n

If we take the first term of each factor of \\(p,\\) then their product becomes
\n\\[x_2 \\, x_3 ^2 \\, x_4^3 \\cdots x_n^{n-1}\\]
\nwhich 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.\\)<\/p>\n

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.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.23<\/span>
\nLet \\(f_1,\\) \\(\\cdots ,\\) \\(f_n\\) be family of functions in \\(C^{\\infty} (\\mathbb{R}).\\) Note that if
\n\\[\\sum_{j=1}^n \\lambda_j f_j (x) =0\\]
\nfor some family of coefficients \\(\\lambda_j \\in \\mathbb{R},\\) then
\n\\[\\sum_{j=1}^n \\lambda_j f_j ^{(k)} (x) =0\\]
\nfor all \\(k\\ge 0.\\) With this background, show that if \\(f_1,\\) \\(\\cdots,\\) \\(f_n\\) are linearly dependent, then
\n\\[\\det \\begin{pmatrix}
\nf_1 & f_2 & \\cdots & f_n \\\\
\nf_1^{(1)} & f_2^{(1)} & \\cdots & f_n^{(1)} \\\\
\n\\vdots & \\vdots & \\ddots & \\vdots \\\\
\nf_1^{(n-1)} & f_2^{(n-1)} & \\cdots & f_n^{(n-1)}
\n\\end{pmatrix} =0.\\]
\n(This is called the Wronskian determinant<\/span> of the family \\(f_1,\\) \\(\\cdots,\\) \\(f_n\\) and is critical in the theory of differential equation.)\n<\/p>\n

\n

Solution.<\/span>
\nSince the family \\(f_1 ,\\) \\(f_2 ,\\) \\(\\cdots ,\\) \\(f_n\\) is linearly dependent, there exist scalars \\(\\lambda_j ‘ \\text{s}\\) such that
\n\\[\\lambda_1 f_1 + \\lambda_2 f_2 + \\cdots + \\lambda_n f_n = 0\\]
\nand not all \\(\\lambda_j ‘ \\text{s}\\) are zeros. Let
\n\\[A=\\left[\\begin{array}{ccccc}
\nf_1 & f_2 & f_3 & \\cdots & f_n \\\\
\nf_1 ^{(1)} & f_2 ^{(1)} & f_3 ^{(1)} & \\cdots & f_n ^{(1)} \\\\
\nf_1 ^{(2)} & f_2 ^{(2)} & f_3 ^{(2)} & \\cdots & f_n ^{(2)} \\\\
\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\
\nf_1 ^{(n-1)} & f_2 ^{(n-1)} & f_3 ^{(n-1)} & \\cdots & f_n ^{(n-1)}
\n\\end{array}\\right].\\]
\nLet \\(A^j\\) be the \\(j\\)th column of \\(A.\\) Then
\n\\[\\lambda_1 A^1 + \\lambda_2 A^2 + \\cdots + \\lambda_n A^n = 0.\\]
\nHence the family of column vectors of \\(A\\) is linearly dependent. Consequently we have \\(\\det(A) =0.\\)<\/p>\n<\/div>\n

<\/p>\n

Problem 8.24<\/span>
\nUse the previous exercise to show that the functions \\(e^x,\\) \\(e^{2x}\\) and \\(e^{3x}\\) are linearly independent; that is, show that
\n\\[\\det \\begin{pmatrix} e^x & e^{2x} & e^{3x} \\\\ e^x & 2e^{2x} & 3e^{3x} \\\\ e^x & 4e^{2x} & 9e^{3x} \\end{pmatrix} \\ne 0.\\]\n<\/p>\n

\n

Solution.<\/span><\/p>\n

\\[\\begin{aligned}
\n\\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]
\n&=
\n\\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]
\n&=
\ne^x e^{2x} e^{3x}
\n\\det \\left[ \\begin{array}{ccc} 1 & 1 & 1 \\\\ 1 & 2 & 3 \\\\ 1 & 2^2 & 3^2 \\end{array}\\right]
\n\\ne 0.<\/p>\n

\\end{aligned}\\]<\/p>\n<\/div>\n

<\/p>\n

Problem 8.25<\/span>
\nUse Problem 22 and 23 to show that the collection of functions
\n\\[e^{\\lambda_1 x}, \\,\\, e^{\\lambda_2 x} ,\\,\\, \\cdots ,\\,\\, e^{\\lambda_n x} \\,\\,\\, ( \\lambda_j \\in \\mathbb{R} )\\]
\nis linearly independent if and only if the numbers \\(\\lambda_j\\) are distinct. This generalizes the previous problem.\n<\/p>\n

\n

Solution.<\/span>
\nIf \\(\\lambda_i = \\lambda_j\\) for some \\(i\\ne j,\\) then the family of given functions is linearly dependent.<\/p>\n

Suppose now that \\(\\lambda_j ‘ \\text{s}\\) are all distinct. Let
\n\\[A = \\left[\\begin{array}{ccccc}
\ne^{\\lambda_1 x} & e^{\\lambda_2 x} & e^{\\lambda_3 x} & \\cdots & e^{\\lambda_n x} \\\\
\n\\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} \\\\
\n\\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} \\\\
\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\
\n\\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} \\\\
\n\\end{array}\\right].\\]
\nThen
\n\\[\\det(A) = e^{\\lambda_1 x} e^{\\lambda_2 x} \\cdots e^{\\lambda_n x} \\det\\left[\\begin{array}{ccccc}
\n1 & 1 & 1 & \\cdots & 1 \\\\
\n\\lambda_1 & \\lambda_2 & \\lambda_3 & \\cdots & \\lambda_n \\\\
\n\\lambda_1 ^2 & \\lambda_2 ^2 & \\lambda_3 ^2 & \\cdots & \\lambda_n ^2 \\\\
\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\
\n\\lambda_1 ^{n-1} & \\lambda_2 ^{n-1} & \\lambda_3 ^{n-1} & \\cdots & \\lambda_n ^{n-1} \\\\
\n\\end{array}\\right] \\ne 0\\]
\nby the Problem 22. Hence by Problem 23, the family of \\(e^{\\lambda_j x}\\)’s is linearly independent.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.26<\/span>
\nLet
\n\\[A= \\begin{pmatrix} 1 & 0 & 1 \\\\ 0 & 2 & 1 \\\\ 1 & 4 & 3 \\end{pmatrix}.\\]
\nFind \\(A^{-1}\\) or prove that it does not exist.\n<\/p>\n

\n

Solution.<\/span>
\nIf \\(A^j\\) is the \\(j\\)th column vector of \\(A,\\) then
\n\\[2A^1 + A^2 = 2A^3,\\]
\nthat is, the family of column vectors of \\(A\\) is not linearly independent. Hence \\(\\det(A) =0\\) and \\(A^{-1}\\) does not exist.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.27<\/span>
\nShow that \\(\\operatorname{SL}_n(K)\\) is a subgroup of \\(\\operatorname{GL}_n(K).\\)\n<\/p>\n

\n

Solution.<\/span>
\nIf \\(A,\\) \\(B\\in \\operatorname{SL}_n (K),\\) then
\n\\[\\det(AB) = \\det(A)\\det(B) = 1\\times 1 = 1,\\]
\nhence \\(AB\\in\\operatorname{SL}_n (K).\\)<\/p>\n

If \\(A\\in\\operatorname{SL}_n (K),\\) then \\(\\det(A) =1.\\) Hence \\(A^{-1}\\) exists and
\n\\[\\det(A^{-1}) = (\\det(A))^{-1} = 1^{-1} = 1,\\]
\nthat is, \\(A^{-1} \\in \\operatorname{SL}_n (K).\\)<\/p>\n

Therefore \\(\\operatorname{SL}_n(K)\\) is a subgroup of \\(\\operatorname{GL}_n(K).\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 8.28<\/span>
\nConsider \\(2\\times 2\\) matrix
\n\\[A = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}.\\]
\nLet \\(P = (a,\\,c)\\) and \\(Q = (b,\\,d)\\) be the points in \\(\\mathbb{R}^2\\) defined by the columns of \\(A.\\) These points span a parallelogram whose vertices are \\(\\mathbf{0},\\) \\(P,\\) \\(Q\\) and \\(P+Q.\\) Find the area of this object.\n<\/p>\n

\n

Solution.<\/span>
\nKeep calm and sketch a picture to derive
\n\\[\\left\\lvert \\det(A) \\right\\rvert.\\]\n<\/p>\n<\/div>\n

<\/p>\n

<\/p>\n

<\/p>\n","protected":false},"excerpt":{"rendered":"

This set of exercises is retrieved from the eighth chapter of Linear Algebra by Robert J. Valenza. Note that these solutions are not fully elaborated; You have to fill the descriptions by yourself. Problem 8.1 Using the recursive definition given in the proof of the existence of determinant, systematically evaluate the determinant of the following matrix: \\(A=\\begin{pmatrix}1&2&1\\\\0&1&1\\\\1&0&2\\end{pmatrix}.\\) Solution. \\(\\begin{aligned} \\det (A) &= 1 \\cdot…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_lmt_disableupdate":"","_lmt_disable":"","footnotes":""},"categories":[57],"tags":[417,402,400,412,414],"class_list":["post-5231","post","type-post","status-publish","format-standard","hentry","category-linear-algebra","tag-determinant","tag-exercises","tag-linear-algebra","tag-linear-transformation","tag-matrix"],"_links":{"self":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts\/5231","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/comments?post=5231"}],"version-history":[{"count":80,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts\/5231\/revisions"}],"predecessor-version":[{"id":6484,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts\/5231\/revisions\/6484"}],"wp:attachment":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/media?parent=5231"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/categories?post=5231"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/tags?post=5231"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}