Matrices and Determinants: Difficult Problems with Solutions

Solution:
Let [tex]a,b,c\in\mathbb{Z}[/tex] be the diagonal entries of [tex]A[/tex]. The product of diagonal matrices is a diagonal matrix whose diagonal entries are the products of the corresponding diagonal entries, so
[tex]A^3-3A+2E=\begin{bmatrix}a^3-3a+2 & 0 & 0\\ 0 & b^3-3b+2 & 0\\ 0 & 0 & c^3-3c+2\end{bmatrix}[/tex]
Hence [tex]A^3-3A+2E=O[/tex] exactly when each of [tex]a,b,c[/tex] is a root of the equation
[tex]t^3-3t+2=(t-1)^2(t+2)=0[/tex]
Its only roots are [tex]t=1[/tex] (double) and [tex]t=-2[/tex], and both are integers. Each of the three diagonal entries is chosen independently from [tex]\{1, -2\}[/tex], so the number of matrices is
[tex]2^3=8[/tex]

Solution:
Write [tex]A=12E+N[/tex], where [tex]E[/tex] is the identity matrix and
[tex]N=\begin{bmatrix}0 & 4\\ 0 & 0\end{bmatrix},\quad N^2=\begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix}=O[/tex]
The matrices [tex]12E[/tex] and [tex]N[/tex] commute, so the binomial theorem can be used; all terms with [tex]k\ge 2[/tex] contain [tex]N^k=O[/tex] and vanish:
[tex]A^n=\sum_{k=0}^{n}\binom{n}{k}12^{n-k}N^k=12^nE+n\times 12^{n-1}N=\begin{bmatrix}12^n & 4n\times 12^{n-1}\\ 0 & 12^n\end{bmatrix}[/tex]
For [tex]n=124[/tex] we have [tex]4\times 124=496[/tex], so
[tex]A^{124}=\begin{bmatrix}12^{124} & 496\times 12^{123}\\ 0 & 12^{124}\end{bmatrix}[/tex]

Solution:
By Sarrus' rule
[tex]\det A=1\times 3\times 2+2\times (-2)\times 0+3\times (-1)\times 5-3\times 3\times 0-1\times (-2)\times 5-2\times (-1)\times 2=6+0-15-0+10+4=5\ne 0[/tex]
so the inverse matrix exists. Compute the cofactors [tex]A_{ij}=(-1)^{i+j}M_{ij}[/tex], where the minor [tex]M_{ij}[/tex] is obtained by deleting row [tex]i[/tex] and column [tex]j[/tex]; for example
[tex]A_{11}=\begin{vmatrix}3 & -2\\ 5 & 2\end{vmatrix}=16,\quad A_{21}=-\begin{vmatrix}2 & 3\\ 5 & 2\end{vmatrix}=11,\quad A_{32}=-\begin{vmatrix}1 & 3\\ -1 & -2\end{vmatrix}=-1[/tex]
In the same way we obtain all of them:
[tex]A_{11}=16, A_{12}=2, A_{13}=-5, A_{21}=11, A_{22}=2, A_{23}=-5, A_{31}=-13, A_{32}=-1, A_{33}=5[/tex]
The inverse is [tex]A^{-1}=\frac{1}{\det A}A^{*}[/tex], where the adjugate matrix [tex]A^{*}[/tex] is the transpose of the matrix of cofactors (the cofactor [tex]A_{ij}[/tex] goes to row [tex]j[/tex], column [tex]i[/tex]):
[tex]A^{-1}=\frac{1}{5}\begin{bmatrix}16 & 11 & -13\\ 2 & 2 & -1\\ -5 & -5 & 5\end{bmatrix}[/tex]
Check: multiplying the first row of [tex]A[/tex] by the columns of [tex]A^{-1}[/tex] gives the first row of [tex]E[/tex]:
[tex]\frac{16+4-15}{5}=1,\quad \frac{11+4-15}{5}=0,\quad \frac{-13-2+15}{5}=0[/tex]

Solution:
Compute the determinant of the system: subtract the second row from the third (the value does not change) and expand along the new third row:
[tex]\Delta=\begin{vmatrix} 2 & -3 & 1 \\ 1 & 5 & -4 \\ 1 & 5 & -9 \end{vmatrix}=\begin{vmatrix} 2 & -3 & 1 \\ 1 & 5 & -4 \\ 0 & 0 & -5 \end{vmatrix}=-5\times (-1)^{3+3}\begin{vmatrix} 2 & -3 \\ 1 & 5 \end{vmatrix}=-5\times 13=-65\ne 0[/tex]
Hence the system has a unique solution. Replace the third column by the column of constant terms and apply the same row operation:
[tex]\Delta_3=\begin{vmatrix} 2 & -3 & 2 \\ 1 & 5 & -5 \\ 1 & 5 & -4 \end{vmatrix}=\begin{vmatrix} 2 & -3 & 2 \\ 1 & 5 & -5 \\ 0 & 0 & 1 \end{vmatrix}=1\times \begin{vmatrix} 2 & -3 \\ 1 & 5 \end{vmatrix}=13[/tex]
By Cramer's rule
[tex]x_3=\frac{\Delta_3}{\Delta}=\frac{13}{-65}=-\frac{1}{5}[/tex]
Check: subtracting the second equation from the third gives [tex]-5x_3=1[/tex], i.e. [tex]x_3=-\frac{1}{5}[/tex].

Solution:
The equation has the form [tex]XA=B[/tex], where
[tex]A=\begin{bmatrix} 2 & 0 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{bmatrix},\quad B=\begin{bmatrix} 0 & -1 & 2 \\ 1 & -2 & 3 \end{bmatrix}[/tex]
Expanding along the first row:
[tex]\det A=2\times \big((-1)\times (-1)-1\times 2\big)-0+1\times \big(1\times 2-(-1)\times 1\big)=-2+3=1\ne 0[/tex]
so [tex]A^{-1}[/tex] exists. Multiplying the equation on the right by [tex]A^{-1}[/tex] gives [tex]X=BA^{-1}[/tex] (the product [tex]A^{-1}B[/tex] is not even defined, since [tex]A^{-1}[/tex] has [tex]3[/tex] columns and [tex]B[/tex] has [tex]2[/tex] rows). The cofactors of [tex]A[/tex] are
[tex]A_{11}=-1, A_{12}=2, A_{13}=3, A_{21}=2, A_{22}=-3, A_{23}=-4, A_{31}=1, A_{32}=-1, A_{33}=-2[/tex]
Writing them as columns (the cofactor [tex]A_{ij}[/tex] goes to row [tex]j[/tex], column [tex]i[/tex]) and dividing by [tex]\det A=1[/tex]:
[tex]A^{-1}=\begin{bmatrix} -1 & 2 & 1 \\ 2 & -3 & -1 \\ 3 & -4 & -2 \end{bmatrix}[/tex]
Therefore
[tex]X=BA^{-1}=\begin{bmatrix} 0 & -1 & 2 \\ 1 & -2 & 3 \end{bmatrix}\begin{bmatrix} -1 & 2 & 1 \\ 2 & -3 & -1 \\ 3 & -4 & -2 \end{bmatrix}=\begin{bmatrix} 4 & -5 & -3 \\ 4 & -4 & -3 \end{bmatrix}[/tex]
Check:
[tex]XA=\begin{bmatrix} 4 & -5 & -3 \\ 4 & -4 & -3 \end{bmatrix}\begin{bmatrix} 2 & 0 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{bmatrix}=\begin{bmatrix} 0 & -1 & 2 \\ 1 & -2 & 3 \end{bmatrix}=B[/tex]

Solution:
Use the properties of determinants instead of multiplying large numbers. Subtracting the first column from the second does not change the determinant; every entry of the second column becomes [tex]-10[/tex], and this common factor can be taken out:
[tex]D=\begin{vmatrix} 1947 & -10 & 1 \\ 1967 & -10 & 37 \\ 1967 & -10 & 38 \end{vmatrix}=-10\begin{vmatrix} 1947 & 1 & 1 \\ 1967 & 1 & 37 \\ 1967 & 1 & 38 \end{vmatrix}[/tex]
Now subtract the second row from the third, and then the first row from the second:
[tex]D=-10\begin{vmatrix} 1947 & 1 & 1 \\ 20 & 0 & 36 \\ 0 & 0 & 1 \end{vmatrix}[/tex]
Expanding along the third row:
[tex]D=-10\times 1\times \begin{vmatrix} 1947 & 1 \\ 20 & 0 \end{vmatrix}=-10\times (0-20)=200[/tex]

Solution:
By Laplace's theorem, if we fix two columns [tex]j_1\lt j_2[/tex], the determinant equals the sum, over all pairs of rows [tex]i_1\lt i_2[/tex], of the products [tex](-1)^{i_1+i_2+j_1+j_2}\times M\times M'[/tex], where [tex]M[/tex] is the second-order minor in these rows and columns and [tex]M'[/tex] is its complementary minor (formed by the remaining rows and columns). Fix columns [tex]2[/tex] and [tex]3[/tex]. In these columns rows [tex]2[/tex] and [tex]4[/tex] contain only zeros, so every minor that uses one of these rows is [tex]0[/tex], and only the minor in rows [tex]1[/tex] and [tex]3[/tex] remains:
[tex]M=\begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix}=4-6=-2[/tex]
Its complementary minor is formed by rows [tex]2[/tex], [tex]4[/tex] and columns [tex]1[/tex], [tex]4[/tex]:
[tex]M'=\begin{vmatrix} 3 & 2 \\ 2 & 3 \end{vmatrix}=9-4=5[/tex]
The sign is [tex](-1)^{1+3+2+3}=-1[/tex], hence
[tex]D=(-1)\times (-2)\times 5=10[/tex]

Solution:
Instead of computing nine cofactors, compute [tex]A^2[/tex]. For example, the first row of [tex]A[/tex] times its three columns gives [tex]1+4+4=9[/tex], [tex]2+2-4=0[/tex] and [tex]2-4+2=0[/tex]; the other rows are handled in the same way:
[tex]A^2=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}=\begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix}=9E[/tex]
Therefore
[tex]A\left(\frac{1}{9}A\right)=\left(\frac{1}{9}A\right)A=\frac{1}{9}A^2=E\Rightarrow A^{-1}=\frac{1}{9}A=\frac{1}{9}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}[/tex]
This agrees with the adjugate formula: [tex]\det A=-27[/tex] and [tex]A^{*}=-3A[/tex], so [tex]A^{-1}=\frac{1}{-27}\times (-3A)=\frac{1}{9}A[/tex].

Solution:
Factor the polynomial:
[tex]f(x)=x^2-6x+5=(x-1)(x-5)[/tex]
Since [tex]A[/tex] commutes with [tex]E[/tex], the same factorization holds for the matrix:
[tex](A-E)(A-5E)=A^2-5A-A+5E=A^2-6A+5E=f(A)[/tex]
The determinant of a product equals the product of the determinants:
[tex]\det f(A)=\det(A-E)\times \det(A-5E)[/tex]
We expand both determinants along the second row:
[tex]\det(A-E)=\begin{vmatrix}0 & -1 & 2\\ 0 & 2 & 0\\ -2 & 1 & -4\end{vmatrix}=2\begin{vmatrix}0 & 2\\ -2 & -4\end{vmatrix}=2\times 4=8[/tex]
[tex]\det(A-5E)=\begin{vmatrix}-4 & -1 & 2\\ 0 & -2 & 0\\ -2 & 1 & -8\end{vmatrix}=-2\begin{vmatrix}-4 & 2\\ -2 & -8\end{vmatrix}=-2\times 36=-72[/tex]
[tex]\det f(A)=8\times (-72)=-576[/tex]
Check by direct computation:
[tex]f(A)=\begin{bmatrix}-4 & 4 & -16\\ 0 & -4 & 0\\ 16 & -4 & 28\end{bmatrix},\quad \det f(A)=-4\begin{vmatrix}-4 & -16\\ 16 & 28\end{vmatrix}=-4\times 144=-576[/tex]

Solution:
Let [tex]A=\begin{bmatrix}a & b\\ c & d\end{bmatrix}[/tex] with [tex]a,b,c,d[/tex] all non-zero. Then
[tex]A^2=\begin{bmatrix}a^2+bc & b(a+d)\\ c(a+d) & d^2+bc\end{bmatrix}[/tex]
From [tex]b(a+d)=0[/tex] and [tex]b\ne 0[/tex] we get [tex]d=-a[/tex], and from [tex]a^2+bc=0[/tex] we get [tex]c=-\frac{a^2}{b}[/tex]. Then the other two entries are also zero:
[tex]c(a+d)=c\times 0=0 , \quad d^2+bc=a^2-a^2=0[/tex]
Conversely, for any [tex]a\ne 0[/tex] and [tex]b\ne 0[/tex] the matrix below has non-zero entries and its square is the zero matrix. The answer is
[tex]\begin{bmatrix}a & b\\ -\frac{a^2}{b} & -a\end{bmatrix},\ a\ne 0,\ b\ne 0[/tex]

Solution:
Write [tex]A=E+N[/tex], where [tex]E[/tex] is the identity matrix and
[tex]N=\begin{bmatrix}0 & 0 & 1\\ 0 & 0 & 0\\ 1 & 0 & 0\end{bmatrix}[/tex]
Since [tex]EX=XE=X[/tex], the condition [tex]AX=XA[/tex] is equivalent to [tex]NX=XN[/tex]. For [tex]X=\left(x_{ij}\right)[/tex], multiplying by [tex]N[/tex] on the left swaps rows [tex]1[/tex] and [tex]3[/tex] and makes row [tex]2[/tex] zero, while multiplying on the right swaps columns [tex]1[/tex] and [tex]3[/tex] and makes column [tex]2[/tex] zero:
[tex]NX=\begin{bmatrix}x_{31} & x_{32} & x_{33}\\ 0 & 0 & 0\\ x_{11} & x_{12} & x_{13}\end{bmatrix},\qquad XN=\begin{bmatrix}x_{13} & 0 & x_{11}\\ x_{23} & 0 & x_{21}\\ x_{33} & 0 & x_{31}\end{bmatrix}[/tex]
Comparing the corresponding entries, we get
[tex]x_{12}=x_{32}=x_{21}=x_{23}=0 , \quad x_{11}=x_{33} , \quad x_{13}=x_{31}[/tex]
while [tex]x_{22}[/tex] is arbitrary. These conditions are equivalent to [tex]NX=XN[/tex], so with [tex]x_{11}=a[/tex], [tex]x_{13}=b[/tex], [tex]x_{22}=c[/tex] the answer is
[tex]X=\begin{bmatrix}a & 0 & b\\ 0 & c & 0\\ b & 0 & a\end{bmatrix},\ a,b,c\in\mathbb{R}[/tex]

Solution:
Let [tex]A[/tex] be such a matrix of odd order [tex]n[/tex]. Transposition does not change the determinant:
[tex]\det A^{T}=\det A[/tex]
The matrix [tex]-A[/tex] is obtained by multiplying each of the [tex]n[/tex] rows of [tex]A[/tex] by [tex]-1[/tex], and each such multiplication multiplies the determinant by [tex]-1[/tex]. Since [tex]n[/tex] is odd,
[tex]\det(-A)=(-1)^n\det A=-\det A[/tex]
From [tex]A^{T}=-A[/tex] we get
[tex]\det A=\det A^{T}=\det(-A)=-\det A\ \Rightarrow\ 2\det A=0\ \Rightarrow\ \det A=0[/tex]

Solution:
Subtract the first row from each of the other rows; this does not change the determinant. In the [tex]k[/tex]-th row [tex]\left(k\ge 2\right)[/tex] all entries become [tex]0[/tex] except the diagonal one, which becomes
[tex](k-1-x)-1=k-2-x[/tex]
The determinant becomes upper triangular, so it equals the product of its diagonal entries:
[tex]\begin{vmatrix}1 & 1 & 1 & \ldots & 1\\ 0 & -x & 0 & \ldots & 0\\ 0 & 0 & 1-x & \ldots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \ldots & n-1-x\end{vmatrix}=(-x)(1-x)(2-x)\ldots(n-1-x)[/tex]
This product equals zero exactly when [tex]x[/tex] is one of the numbers [tex]0,1,2,\ldots,n-1[/tex]. Hence
[tex]x\in\{0, 1, 2, \ldots, n-1\}[/tex]

Solution:
Denote the given matrix by [tex]M[/tex]. By the addition formula every entry is a sum of two products, for example
[tex]\sin (x+p)=\sin x\cos p+\cos x\sin p[/tex]
Hence [tex]M=ST[/tex], where
[tex]S=\begin{bmatrix}\sin x & \cos x & 0\\ \sin y & \cos y & 0\\ \sin z & \cos z & 0\end{bmatrix},\qquad T=\begin{bmatrix}\cos p & \cos q & \cos r\\ \sin p & \sin q & \sin r\\ 0 & 0 & 0\end{bmatrix}[/tex]
Indeed, the entry of [tex]ST[/tex] in the first row and first column is
[tex]\sin x\cos p+\cos x\sin p+0\times 0=\sin (x+p)[/tex]
and the other entries are obtained in the same way. The determinant of a product equals the product of the determinants, and [tex]\det S=0[/tex] because [tex]S[/tex] has a zero column. Therefore
[tex]\det M=\det S\times \det T=0[/tex]

Solution:
Adding to a column a linear combination of the other columns does not change the determinant. Denote the columns by [tex]C_1,\ldots,C_6[/tex] and replace the last column by
[tex]10^5C_1+10^4C_2+10^3C_3+10^2C_4+10C_5+C_6[/tex]
In each row the new last entry is the six-digit number written with the digits of that row, for example in the first row
[tex]5\times 10^5+1\times 10^4+0\times 10^3+8\times 10^2+0\times 10+7=510807[/tex]
So the new last column is
[tex]\begin{bmatrix}510807\\ 448218\\ 831828\\ 426009\\ 668289\\ 224109\end{bmatrix}=2019\times \begin{bmatrix}253\\ 222\\ 412\\ 211\\ 331\\ 111\end{bmatrix}[/tex]
Expanding the new determinant along its last column, we get
[tex]D=2019\left(253A_{16}+222A_{26}+412A_{36}+211A_{46}+331A_{56}+111A_{66}\right)[/tex]
where the cofactors [tex]A_{i6}[/tex] are integers, because up to sign they are determinants of matrices with integer entries. Hence [tex]D[/tex] is divisible by [tex]2019[/tex]. (In fact, [tex]D=8076=4\times 2019[/tex].)