Matrices and Determinants: Problems with Solutions

Solution:
The dimensions of the matrices are $n\times m$, where
$n$ is the number of rows and
$m$ is the number of columns.

In this case, the matrix of the example is $4 \times 5$ because it has $4$ rows and $5$ columns.

Dimension of $A: 4\times 5$

Solution:
$A_{2,4}$ is the element that is in the second row and the fourth column of $A$

$A_{2,4}=0$.

Solution:
$A_{3,2}$ is the element that is in the third row and the second column of $A$

$A_{3,2}=-1$.

Solution:
Identity matrix (unit matrix) is a square matrix with ones on the main diagonal and zeros elsewhere.
Example:
$\left[ \begin{array}{ccccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] $

Solution:
$\left[ \begin{array}{ccc} 3 & -2 & 3 \\ 5 & 1 & 0 \end{array} \right] $

This matrix represents the coefficients of the variables and the constant terms of the system of equations.

Solution:
$\left[ \begin{array}{cc} 2 & -1 \\ 1 & 3 \end{array} \right] +\left[ \begin{array}{cc} 1 & 0 \\ 2 & -1 \end{array} \right] =\left[ \begin{array}{cc} 3 & -1 \\ 3 & 2 \end{array} \right] $

$2+1=3$
$-1+0=-1$
$1+2=3$
$3+(-1)=2$

Solution:
We subtract the known matrix from both sides; matrices are subtracted element by element:
[tex]A=\begin{bmatrix}5 & -1 \\ 1 & 5\end{bmatrix}-\begin{bmatrix}2 & 3 \\ -4 & 1\end{bmatrix}=\begin{bmatrix}5-2 & -1-3 \\ 1-(-4) & 5-1\end{bmatrix}[/tex]
Answer: [tex]A=\begin{bmatrix}3 & -4 \\ 5 & 4\end{bmatrix}[/tex]

Solution:
We multiply every element of the matrix by [tex]5[/tex]:
[tex]5\begin{bmatrix}-2 \\ 3 \\ -4\end{bmatrix}=\begin{bmatrix}5\times(-2) \\ 5\times3 \\ 5\times(-4)\end{bmatrix}=\begin{bmatrix}-10 \\ 15 \\ -20\end{bmatrix}[/tex]
Answer: [tex]5\begin{bmatrix}-2 \\ 3 \\ -4\end{bmatrix}=\begin{bmatrix}-10 \\ 15 \\ -20\end{bmatrix}[/tex]

Solution:
[tex]\frac{3}{2}X=\begin{bmatrix}3 & -4 \\ 5 & 4\end{bmatrix}-\begin{bmatrix}-1 & 3 \\ 2 & -2\end{bmatrix}=\begin{bmatrix}3-(-1) & -4-3 \\ 5-2 & 4-(-2)\end{bmatrix}=\begin{bmatrix}4 & -7 \\ 3 & 6\end{bmatrix}[/tex]
We multiply both sides by [tex]\frac{2}{3}[/tex]:
[tex]X=\frac{2}{3}\begin{bmatrix}4 & -7 \\ 3 & 6\end{bmatrix}=\begin{bmatrix}\frac{8}{3} & -\frac{14}{3} \\ 2 & 4\end{bmatrix}[/tex]
Answer: [tex]X=\begin{bmatrix}\frac{8}{3} & -\frac{14}{3} \\ 2 & 4\end{bmatrix}[/tex]

Solution:
[tex]B[/tex] is a [tex]2\times 3[/tex] matrix and [tex]C[/tex] is a [tex]3\times 2[/tex] matrix, so the product [tex]A=BC[/tex] is a [tex]2\times 2[/tex] matrix.
Each element is a row of [tex]B[/tex] times a column of [tex]C[/tex]:
[tex]a_{11}=1\times2+(-3)\times(-2)+(-2)\times3=2+6-6=2[/tex]
[tex]a_{12}=1\times1+(-3)\times(-1)+(-2)\times0=1+3+0=4[/tex]
[tex]a_{21}=2\times2+0\times(-2)+1\times3=4+0+3=7[/tex]
[tex]a_{22}=2\times1+0\times(-1)+1\times0=2+0+0=2[/tex]
Answer: [tex]A=\begin{bmatrix}2 & 4 \\ 7 & 2\end{bmatrix}[/tex]

Solution:
For [tex]A=\begin{bmatrix}a & b \\ c & d\end{bmatrix}[/tex] we have [tex]\det A=ad-bc[/tex].
[tex]\det A=2\times5-(-3)\times4=10+12=22[/tex]
Answer: [tex]\det A=22[/tex]

Solution:
For [tex]A=\begin{bmatrix}a & b \\ c & d\end{bmatrix}[/tex] we have [tex]\det A=ad-bc[/tex].
[tex]\det A=3\times0-4\times0=0[/tex]
(A matrix with a row of zeros always has determinant [tex]0[/tex].)
Answer: [tex]\det A=0[/tex]

Solution:
For [tex]A=\begin{bmatrix}a & b \\ c & d\end{bmatrix}[/tex] with [tex]\det A\neq 0[/tex]: [tex]A^{-1}=\frac{1}{\det A}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}[/tex]. If [tex]\det A=0[/tex], the inverse does not exist.
[tex]\det A=2\times5-(-3)\times4=10+12=22\neq 0[/tex]
[tex]A^{-1}=\frac{1}{22}\begin{bmatrix}5 & 3 \\ -4 & 2\end{bmatrix}=\begin{bmatrix}\frac{5}{22} & \frac{3}{22} \\ -\frac{2}{11} & \frac{1}{11}\end{bmatrix}[/tex]
Answer: [tex]A^{-1}=\begin{bmatrix}\frac{5}{22} & \frac{3}{22} \\ -\frac{2}{11} & \frac{1}{11}\end{bmatrix}[/tex]

Solution:
For [tex]A=\begin{bmatrix}a & b \\ c & d\end{bmatrix}[/tex] with [tex]\det A\neq 0[/tex]: [tex]A^{-1}=\frac{1}{\det A}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}[/tex]. If [tex]\det A=0[/tex], the inverse does not exist.
[tex]\det A=0\times0-\left(-\frac{3}{4}\right)\times\frac{7}{3}=\frac{7}{4}\neq 0[/tex]
[tex]A^{-1}=\frac{4}{7}\begin{bmatrix}0 & \frac{3}{4} \\ -\frac{7}{3} & 0\end{bmatrix}=\begin{bmatrix}0 & \frac{3}{7} \\ -\frac{4}{3} & 0\end{bmatrix}[/tex]
Answer: [tex]A^{-1}=\begin{bmatrix}0 & \frac{3}{7} \\ -\frac{4}{3} & 0\end{bmatrix}[/tex]

Solution:
For [tex]A=\begin{bmatrix}a & b \\ c & d\end{bmatrix}[/tex] with [tex]\det A\neq 0[/tex]: [tex]A^{-1}=\frac{1}{\det A}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}[/tex]. If [tex]\det A=0[/tex], the inverse does not exist.
[tex]\det A=3\times8-(-4)\times(-6)=24-24=0[/tex]
The determinant is [tex]0[/tex], so [tex]A[/tex] has no inverse.

Solution:
[tex]AB=\begin{bmatrix}\frac{21}{5}-\frac{16}{5} & -\frac{28}{5}+\frac{28}{5} \\ \frac{12}{5}-\frac{12}{5} & -\frac{16}{5}+\frac{21}{5}\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}[/tex]
[tex]BA=\begin{bmatrix}\frac{21}{5}-\frac{16}{5} & -\frac{12}{5}+\frac{12}{5} \\ \frac{28}{5}-\frac{28}{5} & -\frac{16}{5}+\frac{21}{5}\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}[/tex]
Both products equal the identity matrix, so [tex]B=A^{-1}[/tex].

Solution:
[tex]AB=\begin{bmatrix}-4+9 & 2-6 \\ -2+6 & 1-4\end{bmatrix}=\begin{bmatrix}5 & -4 \\ 4 & -3\end{bmatrix}\neq \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}[/tex]
[tex]AB[/tex] is not the identity matrix, so [tex]B[/tex] is not the inverse of [tex]A[/tex].

Solution:
[tex]AB=\begin{bmatrix}\frac{16}{25}+\frac{27}{25} & -\frac{40}{25}+\frac{81}{25} \\ -\frac{2}{25}+\frac{6}{25} & \frac{5}{25}+\frac{18}{25}\end{bmatrix}=\begin{bmatrix}\frac{43}{25} & \frac{41}{25} \\ \frac{4}{25} & \frac{23}{25}\end{bmatrix}\neq \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}[/tex]
[tex]AB[/tex] is not the identity matrix, so [tex]B[/tex] is not the inverse of [tex]A[/tex].

Solution:
[tex]AB=\begin{bmatrix}\frac{16}{25}+\frac{9}{25} & -\frac{72}{25}+\frac{72}{25} \\ -\frac{2}{25}+\frac{2}{25} & \frac{9}{25}+\frac{16}{25}\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}[/tex]
[tex]BA=\begin{bmatrix}\frac{16}{25}+\frac{9}{25} & \frac{18}{25}-\frac{18}{25} \\ \frac{8}{25}-\frac{8}{25} & \frac{9}{25}+\frac{16}{25}\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}[/tex]
Both products equal the identity matrix, so [tex]B=A^{-1}[/tex].

Solution:
If [tex]B=A^{-1}[/tex], then [tex]AB=I[/tex]:
[tex]AB=\begin{bmatrix}\frac{2}{5}+\frac{3}{5} & x+\frac{3}{5} \\ -\frac{2}{5}+\frac{2}{5} & -x+\frac{2}{5}\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}[/tex]
[tex]x+\frac{3}{5}=0\;\Rightarrow\; x=-\frac{3}{5}[/tex]
Check: [tex]-x+\frac{2}{5}=\frac{3}{5}+\frac{2}{5}=1[/tex].
Answer: [tex]x=-\frac{3}{5}[/tex]

Solution:
A matrix has no inverse exactly when its determinant is [tex]0[/tex].
[tex]\det A=2\times(-2)-3x=-4-3x[/tex]
[tex]-4-3x=0\;\Rightarrow\; x=-\frac{4}{3}[/tex]
Answer: [tex]x=-\frac{4}{3}[/tex]

Solution:
A matrix has no inverse exactly when its determinant is [tex]0[/tex].
[tex]\det A=1\times(-1)-(2+x)x=-1-2x-x^{2}[/tex]
[tex]-x^{2}-2x-1=0\;\Rightarrow\; x^{2}+2x+1=0\;\Rightarrow\;(x+1)^{2}=0\;\Rightarrow\; x=-1[/tex]
Answer: [tex]x=-1[/tex]

Solution:
Transposing turns the rows of [tex]B[/tex] into columns: [tex]B^{T}=\begin{bmatrix}1 & 2\\ 0 & 1\\ -2 & 1\end{bmatrix}[/tex]. Like [tex]F[/tex], it has [tex]3[/tex] rows and [tex]2[/tex] columns, so the difference [tex]2F-B^{T}[/tex] is defined:
[tex]2F-B^{T}=\begin{bmatrix}2 & -4\\ 6 & 2\\ -2 & 2\end{bmatrix}-\begin{bmatrix}1 & 2\\ 0 & 1\\ -2 & 1\end{bmatrix}=\begin{bmatrix}1 & -6\\ 6 & 1\\ 0 & 1\end{bmatrix}[/tex]
Multiply this matrix by [tex]C[/tex] (row by column); the product again has [tex]3[/tex] rows and [tex]2[/tex] columns:
[tex](2F-B^{T})C=\begin{bmatrix}1\times 1+(-6)\times 0 & 1\times (-2)+(-6)\times (-1)\\ 6\times 1+1\times 0 & 6\times (-2)+1\times (-1)\\ 0\times 1+1\times 0 & 0\times (-2)+1\times (-1)\end{bmatrix}=\begin{bmatrix}1 & 4\\ 6 & -13\\ 0 & -1\end{bmatrix}[/tex]
Finally add [tex]3F[/tex]:
[tex]A=\begin{bmatrix}1 & 4\\ 6 & -13\\ 0 & -1\end{bmatrix}+\begin{bmatrix}3 & -6\\ 9 & 3\\ -3 & 3\end{bmatrix}=\begin{bmatrix}4 & -2\\ 15 & -10\\ -3 & 2\end{bmatrix}[/tex]

Solution:
First compute [tex]A^2=AA[/tex], multiplying row by column (not squaring the entries):
[tex]A^2=\begin{bmatrix}3\times 3+2\times 1 & 3\times 2+2\times 4\\ 1\times 3+4\times 1 & 1\times 2+4\times 4\end{bmatrix}=\begin{bmatrix}11 & 14\\ 7 & 18\end{bmatrix}[/tex]
The constant term [tex]5[/tex] of the polynomial is replaced by [tex]5E=\begin{bmatrix}5 & 0\\ 0 & 5\end{bmatrix}[/tex], not by a matrix of fives. Hence
[tex]f(A)=\begin{bmatrix}22 & 28\\ 14 & 36\end{bmatrix}+\begin{bmatrix}9 & 6\\ 3 & 12\end{bmatrix}+\begin{bmatrix}5 & 0\\ 0 & 5\end{bmatrix}=\begin{bmatrix}36 & 34\\ 17 & 53\end{bmatrix}[/tex]

Solution:
First write the system in standard form, with the unknowns on the left-hand side:
[tex]\begin{cases} x-y=2 \\ 2x-3y=1 \end{cases}[/tex]
The determinant of the system is non-zero, so the system has a unique solution:
[tex]\Delta=\begin{vmatrix} 1 & -1 \\ 2 & -3 \end{vmatrix}=-3+2=-1\ne 0[/tex]
Replace the first, respectively the second, column by the column of constant terms:
[tex]\Delta_x=\begin{vmatrix} 2 & -1 \\ 1 & -3 \end{vmatrix}=-6+1=-5 , \quad \Delta_y=\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}=1-4=-3[/tex]
By Cramer's rule
[tex]x=\frac{\Delta_x}{\Delta}=\frac{-5}{-1}=5 , \quad y=\frac{\Delta_y}{\Delta}=\frac{-3}{-1}=3[/tex]
Check: [tex]5=2+3[/tex] and [tex]2\times 5-3\times 3=1[/tex]. The solution is [tex]x=5[/tex], [tex]y=3[/tex].

Solution:
Subtracting the first row from the second and from the third row does not change the value of the determinant:
[tex]D=\begin{vmatrix} -3 & 6 & 1 \\ 8 & -10 & 0 \\ -4 & 2 & 0 \end{vmatrix}[/tex]
Expand along the third column, where only the entry [tex]a_{13}=1[/tex] is non-zero:
[tex]D=1\times (-1)^{1+3}\begin{vmatrix} 8 & -10 \\ -4 & 2 \end{vmatrix}=8\times 2-(-10)\times (-4)=16-40=-24[/tex]
Check by the rule of Sarrus:
[tex]D=(-3)\times (-4)\times 1+6\times 1\times (-7)+1\times 5\times 8-1\times (-4)\times (-7)-(-3)\times 1\times 8-6\times 5\times 1[/tex]
[tex]D=12-42+40-28+24-30=-24[/tex]

Solution:
A matrix [tex]S[/tex] is symmetric when [tex]S^{T}=S[/tex]. Since [tex]A[/tex] is square, [tex]A[/tex] and [tex]A^{T}[/tex] have the same size, so the sum [tex]A+A^{T}[/tex] is defined. Transposition has the properties
[tex](X+Y)^{T}=X^{T}+Y^{T} , \quad (X^{T})^{T}=X[/tex]
Using them and the commutativity of matrix addition, we get
[tex](A+A^{T})^{T}=A^{T}+(A^{T})^{T}=A^{T}+A=A+A^{T}[/tex]
Hence [tex]A+A^{T}[/tex] is symmetric. (Entry by entry: the entry in row [tex]i[/tex] and column [tex]j[/tex] of [tex]A+A^{T}[/tex] is [tex]a_{ij}+a_{ji}[/tex], which does not change when [tex]i[/tex] and [tex]j[/tex] are swapped.)

Solution:
Denote the determinant by [tex]D[/tex]. By the rule of Sarrus, the three products taken with a plus sign are
[tex]3x\times 2\times 5=30x , \quad 5\times (-x)\times x=-5x^2 , \quad (-6)\times 4\times (-3)=72[/tex]
and the three products taken with a minus sign are
[tex](-6)\times 2\times x=-12x , \quad 3x\times (-x)\times (-3)=9x^2 , \quad 5\times 4\times 5=100[/tex]
Hence
[tex]D=30x-5x^2+72-(-12x+9x^2+100)=-14x^2+42x-28=-14(x^2-3x+2)[/tex]
The equation [tex]D=0[/tex] is equivalent to
[tex]x^2-3x+2=0 \iff (x-1)(x-2)=0 \iff x_1=1 , \ x_2=2[/tex]

Solution:
Let [tex]E[/tex] be the identity matrix of the same order. We use the associativity of matrix multiplication and the equalities [tex]XX^{-1}=X^{-1}X=E[/tex] and [tex]XE=EX=X[/tex]:
[tex](ABC)(C^{-1}B^{-1}A^{-1})=AB(CC^{-1})B^{-1}A^{-1}=A(BB^{-1})A^{-1}=AA^{-1}=E[/tex]
In the same way
[tex](C^{-1}B^{-1}A^{-1})(ABC)=C^{-1}B^{-1}(A^{-1}A)BC=C^{-1}(B^{-1}B)C=C^{-1}C=E[/tex]
So [tex]C^{-1}B^{-1}A^{-1}[/tex] is a two-sided inverse of [tex]ABC[/tex]. Therefore [tex]ABC[/tex] is invertible, and since the inverse matrix is unique, [tex](ABC)^{-1}=C^{-1}B^{-1}A^{-1}[/tex].
Note that the order of the factors is reversed: in general [tex]A^{-1}B^{-1}C^{-1}[/tex] is not the inverse, because matrix multiplication is not commutative.

Solution:
Adding the other three columns to the first column does not change the determinant. Every row sums to [tex]5[/tex], so all entries of the first column become [tex]5[/tex], and the common factor [tex]5[/tex] can be taken out:
[tex]D=\begin{vmatrix}5 & 1 & 1 & 1\\ 5 & 2 & 1 & 1\\ 5 & 1 & 2 & 1\\ 5 & 1 & 1 & 2\end{vmatrix}=5\begin{vmatrix}1 & 1 & 1 & 1\\ 1 & 2 & 1 & 1\\ 1 & 1 & 2 & 1\\ 1 & 1 & 1 & 2\end{vmatrix}[/tex]
Subtract the first row from each of the other rows (this again does not change the determinant). The result is an upper triangular determinant, which equals the product of its diagonal entries:
[tex]D=5\begin{vmatrix}1 & 1 & 1 & 1\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{vmatrix}=5\times 1\times 1\times 1\times 1=5[/tex]