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]