Matrices and Determinants: Very Difficult Problems with Solutions

Solution:
From [tex]M^{T}M=E[/tex] it follows that [tex]M^{T}=M^{-1}[/tex], hence also
[tex]MM^{T}=E[/tex]
Therefore
[tex]M-E=M-MM^{T}=M\left(E-M^{T}\right)=M(E-M)^{T}[/tex]
Using [tex]\det(XY)=\det X\times \det Y[/tex], [tex]\det X^{T}=\det X[/tex] and [tex]\det M=1[/tex], we get
[tex]\det(M-E)=\det M\times \det\left((E-M)^{T}\right)=\det(E-M)[/tex]
On the other hand, [tex]E-M=-(M-E)[/tex] and the order [tex]3[/tex] is odd, so
[tex]\det(E-M)=(-1)^3\det(M-E)=-\det(M-E)[/tex]
Thus [tex]\det(M-E)=-\det(M-E)[/tex], i.e. [tex]2\det(M-E)=0[/tex], hence
[tex]\det(M-E)=0[/tex]

Solution:
Multiply out, keeping the order of the factors, and use the hypotheses [tex]A^3=B^3[/tex] and [tex]A^2B=B^2A[/tex] ([tex]O[/tex] denotes the zero matrix):
[tex](A^2+B^2)(A-B)=A^3-A^2B+B^2A-B^3=(A^3-B^3)-(A^2B-B^2A)=O[/tex]
Suppose that [tex]A^2+B^2[/tex] is invertible. Multiplying the last equality from the left by [tex](A^2+B^2)^{-1}[/tex], we get
[tex]A-B=(A^2+B^2)^{-1}O=O\Rightarrow A=B[/tex]
which contradicts the assumption that the matrices are different. Hence [tex]A^2+B^2[/tex] is not invertible, i.e. [tex]\det(A^2+B^2)=0[/tex].

Solution:
Expand [tex]D_n[/tex] ([tex]n\ge 3[/tex]) along the first row. The minor of the entry [tex]a_{11}=2[/tex] is a determinant of the same form of order [tex]n-1[/tex], i.e. [tex]D_{n-1}[/tex]. The minor [tex]M_{12}[/tex] of the entry [tex]a_{12}=1[/tex] has first column [tex](1, 0, \ldots, 0)^{T}[/tex]; expanding it along this column gives [tex]M_{12}=D_{n-2}[/tex]. Hence
[tex]D_n=2\times D_{n-1}-1\times M_{12}=2D_{n-1}-D_{n-2}\Rightarrow D_n-D_{n-1}=D_{n-1}-D_{n-2}[/tex]
Thus [tex]D_1, D_2, D_3, \ldots[/tex] is an arithmetic progression. Its first two terms are
[tex]D_1=2, \quad D_2=\begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix}=4-1=3[/tex]
so its common difference is [tex]d=D_2-D_1=1[/tex] and
[tex]D_n=D_1+(n-1)d=2+(n-1)=n+1[/tex]

Solution:
Add the first column to the second, then the new second column to the third, and so on, and finally the new column [tex]n[/tex] to column [tex]n+1[/tex]. These operations do not change the determinant, and afterwards column [tex]k[/tex] is the sum of the original columns [tex]1, 2, \ldots, k[/tex]. In the first row the last entry becomes
[tex]S=a_0+a_1+\ldots+a_n[/tex]
Row [tex]i[/tex] ([tex]2\le i\le n+1[/tex]) originally has [tex]-1[/tex] in column [tex]i-1[/tex], [tex]1[/tex] in column [tex]i[/tex] and zeros elsewhere. Its partial sums are therefore [tex]0[/tex] before column [tex]i-1[/tex], [tex]-1[/tex] in column [tex]i-1[/tex] and [tex]-1+1=0[/tex] from column [tex]i[/tex] on.
So the last column becomes [tex](S, 0, \ldots, 0)^{T}[/tex], and after deleting the first row and the last column we are left with the diagonal matrix of order [tex]n[/tex] with [tex]-1[/tex] on the diagonal; its determinant is [tex](-1)^n[/tex]. Expanding along the last column, we get
[tex]D=S\times (-1)^{1+(n+1)}\times (-1)^n=S\times (-1)^{2n+2}=S=a_0+a_1+\ldots+a_n[/tex]

Solution:
For any number [tex]\beta[/tex], using [tex]P^2=P[/tex] we get
[tex](E-\alpha P)(E+\beta P)=E+\beta P-\alpha P-\alpha\beta P^2=E+(\beta-\alpha-\alpha\beta)P[/tex]
The coefficient of [tex]P[/tex] vanishes when [tex]\beta(1-\alpha)=\alpha[/tex], which, since [tex]\alpha\ne 1[/tex], gives
[tex]\beta=\frac{\alpha}{1-\alpha}[/tex]
With this [tex]\beta[/tex] we have [tex](E-\alpha P)(E+\beta P)=E[/tex]. The matrices [tex]E[/tex] and [tex]P[/tex] commute, so the product in the reverse order gives the same expression: [tex](E+\beta P)(E-\alpha P)=E+(\beta-\alpha-\alpha\beta)P=E[/tex]. Hence [tex]E-\alpha P[/tex] is invertible and
[tex](E-\alpha P)^{-1}=E+\frac{\alpha}{1-\alpha}P[/tex]
Remark: the condition [tex]\alpha\ne 1[/tex] is essential. For [tex]\alpha=1[/tex] we have [tex](E-P)P=P-P^2=O[/tex] (the zero matrix), so for [tex]P\ne O[/tex] the matrix [tex]E-P[/tex] is not invertible (otherwise multiplying by its inverse would give [tex]P=O[/tex]).

Solution:
Let [tex]J[/tex] be the matrix of order [tex]n[/tex] whose entries are all equal to [tex]1[/tex], and [tex]E[/tex] the identity matrix. Every entry of [tex]J^2[/tex] is a sum of [tex]n[/tex] ones, so
[tex]J^2=nJ, \qquad A=9J-E[/tex]
We look for the inverse in the form [tex]\alpha E+\beta J[/tex]. Such matrices commute with [tex]A[/tex] (all of them are linear combinations of [tex]E[/tex] and [tex]J[/tex]), so it suffices to check the product in one order:
[tex](9J-E)(\alpha E+\beta J)=9\alpha J+9n\beta J-\alpha E-\beta J=-\alpha E+(9\alpha+9n\beta-\beta)J[/tex]
This equals [tex]E[/tex] if the coefficient of [tex]E[/tex] is [tex]1[/tex] and the coefficient of [tex]J[/tex] is [tex]0[/tex]:
[tex]\alpha=-1, \qquad -9+(9n-1)\beta=0\Rightarrow \beta=\frac{9}{9n-1}[/tex]
Here [tex]9n-1\ne 0[/tex] for every natural [tex]n[/tex]. Hence [tex]A^{-1}=-E+\frac{9}{9n-1}J[/tex]: its entries off the main diagonal are equal to [tex]\frac{9}{9n-1}[/tex], and its diagonal entries are
[tex]-1+\frac{9}{9n-1}=\frac{10-9n}{9n-1}[/tex]
that is,
[tex]A^{-1}=\frac{1}{9n-1}\begin{bmatrix}10-9n & 9 & \ldots & 9\\ 9 & 10-9n & \ldots & 9\\ \vdots & \vdots & \ddots & \vdots\\ 9 & 9 & \ldots & 10-9n\end{bmatrix}[/tex]

Solution:
Subtract from each row the previous one, starting from the bottom: subtract row [tex]n-1[/tex] from row [tex]n[/tex], then row [tex]n-2[/tex] from row [tex]n-1[/tex], and so on, finally the first row from the second (in this way the original rows are always used). The determinant does not change. For [tex]i\ge 2[/tex] the new row [tex]i[/tex] has the entries
[tex]|i-j|-|i-1-j|=\begin{cases} 1 & (j\le i-1)\\ -1 & (j\ge i)\end{cases}[/tex]
So the first row is [tex](0, 1, 2, \ldots, n-1)[/tex], and row [tex]i[/tex] ([tex]i\ge 2[/tex]) consists of [tex]i-1[/tex] ones followed by [tex]n-i+1[/tex] entries equal to [tex]-1[/tex].
Now add the last column to each of the columns [tex]1, 2, \ldots, n-1[/tex]. The first row becomes [tex](n-1, n, \ldots, 2n-3, n-1)[/tex]; in row [tex]i[/tex] ([tex]i\ge 2[/tex]) the entries in the columns [tex]j\le i-1[/tex] become [tex]1+(-1)=0[/tex], those in the columns [tex]i\le j\le n-1[/tex] become [tex]-1+(-1)=-2[/tex], and the last entry stays [tex]-1[/tex].
The resulting matrix is upper triangular with diagonal entries [tex]n-1, -2, \ldots, -2, -1[/tex] (the number [tex]-2[/tex] appears [tex]n-2[/tex] times). Hence
[tex]\det A=(n-1)\times (-2)^{n-2}\times (-1)=(-1)^{n-1}(n-1)\times 2^{n-2}[/tex]

Solution:
Let [tex]n\ge 2[/tex]. Write the first column as a sum of two columns:
[tex](x_1, a, \ldots, a)^{T}=(x_1-a, 0, \ldots, 0)^{T}+(a, a, \ldots, a)^{T}[/tex]
A determinant is linear in each column, so [tex]D_n=D'+D''[/tex], where [tex]D'[/tex] and [tex]D''[/tex] have first columns [tex](x_1-a, 0, \ldots, 0)^{T}[/tex] and [tex](a, a, \ldots, a)^{T}[/tex], respectively, and the other columns as in [tex]D_n[/tex]. Expanding [tex]D'[/tex] along its first column, we get [tex]D'=(x_1-a)D_{n-1}[/tex], where [tex]D_{n-1}[/tex] is the determinant of the same form built from [tex]x_2,\ldots,x_n[/tex].
In [tex]D''[/tex] subtract the first column (all its entries are [tex]a[/tex]) from each of the other columns. Then column [tex]j[/tex] ([tex]j\ge 2[/tex]) has [tex]x_j-a[/tex] in row [tex]j[/tex] and zeros elsewhere, i.e. [tex]D''[/tex] becomes lower triangular with diagonal [tex]a, x_2-a, \ldots, x_n-a[/tex]. Hence
[tex]D_n=(x_1-a)D_{n-1}+a\prod_{j=2}^{n}(x_j-a)[/tex]
Now we prove by induction on [tex]n[/tex] that
[tex]D_n=\prod_{i=1}^{n}(x_i-a)+a\sum_{i=1}^{n}\prod_{j\ne i}(x_j-a)[/tex]
Base case: [tex]D_1=x_1=(x_1-a)+a[/tex] (an empty product equals [tex]1[/tex]).
Inductive step: if the formula holds for [tex]x_2,\ldots,x_n[/tex], then multiplying [tex]D_{n-1}[/tex] by [tex]x_1-a[/tex] gives [tex]\prod_{i=1}^{n}(x_i-a)[/tex] and all the terms [tex]a\prod_{j\ne i}(x_j-a)[/tex] with [tex]i\ge 2[/tex], while the added term [tex]a\prod_{j\ne 1}(x_j-a)[/tex] is exactly the term with [tex]i=1[/tex]. Hence the formula holds for every [tex]n[/tex]. When all [tex]x_i\ne a[/tex], it can also be written as
[tex]D_n=\prod_{i=1}^{n}(x_i-a)\left(1+\sum_{i=1}^{n}\frac{a}{x_i-a}\right)[/tex]

Solution:
Lemma: a square matrix [tex]M[/tex] with integer entries has an inverse with integer entries if and only if [tex]\det M=\pm 1[/tex].
Indeed, if [tex]M[/tex] and [tex]M^{-1}[/tex] have integer entries, then [tex]\det M[/tex] and [tex]\det M^{-1}[/tex] are integers and
[tex]\det M\times \det M^{-1}=\det(MM^{-1})=\det E=1\Rightarrow \det M=\pm 1[/tex]
Conversely, if [tex]\det M=\pm 1[/tex], then [tex]M^{-1}=\frac{1}{\det M}M^{*}=\pm M^{*}[/tex], where the adjugate matrix [tex]M^{*}[/tex] consists of the cofactors of the entries of [tex]M[/tex], which are integers.
Now let [tex]A=(a_{ij})[/tex], [tex]B=(b_{ij})[/tex] and
[tex]f(t)=\det(A+tB)=(a_{11}+tb_{11})(a_{22}+tb_{22})-(a_{12}+tb_{12})(a_{21}+tb_{21})[/tex]
This is a polynomial in [tex]t[/tex] of degree at most [tex]2[/tex]. By the lemma, [tex]f(0), f(1), f(2), f(3), f(4)\in\{1, -1\}[/tex]. These are five values taken from two possible ones, so one of them, call it [tex]c[/tex], is taken at least three times. Then the polynomial [tex]f(t)-c[/tex] has degree at most [tex]2[/tex] and at least three distinct roots, so it is the zero polynomial, i.e. [tex]f(t)=c[/tex] for every [tex]t[/tex]. In particular,
[tex]\det(A+5B)=f(5)=c=\pm 1[/tex]
and by the lemma [tex]A+5B[/tex] is invertible and its inverse has integer entries.