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] $

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

Solution:
$A$ must be a $2\times2$ dimensional matrix.
$A=\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] $

$\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] +\left[ \begin{array}{cc} 2 & 3 \\ -4 & 1 \end{array} \right] = \left[ \begin{array}{cc} 5 & -1 \\ 1 & 3 \end{array} \right] $

$\left[ \begin{array}{cc} a+2 & b+3 \\ c-4 & d+1 \end{array} \right] =\left[ \begin{array}{cc} 5 & -1 \\ 1 & 3 \end{array} \right] $

$a+2=5\Longrightarrow a=3$
$b+3=-1\Longrightarrow b=-4$
$c-4=1\Longrightarrow c=5$
$d+1=5\Longrightarrow d=4$
So, $A=\left[ \begin{array}{cc} 3 & -4 \\ 5 & 4 \end{array} \right]$

Solution:
We must multiply number for each element of the $3\times 1$ matrix.
$5\times \left[ \begin{array}{c} -2 \\ 3 \\ -4 \end{array} \right] =\left[ \begin{array}{c} 5\times (-2) \\ 5\times 3 \\ 5\times (-4) \end{array} \right] =\left[ \begin{array}{c} -10 \\ 15 \\ -20 \end{array} \right]$

Solution:
$X$ must be a matrix of $2\times 2$ dimension.

$X=\left[ \begin{array}{cc} a & b c & d \end{array} \right] $
So
$\frac{3}{2}\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] +\left[ \begin{array}{cc} -1 & 3 \\ 2 & -2 \end{array} \right] =\left[ \begin{array}{cc} 3 & -4 \\ 5 & 4 \end{array} \right] $

$\left[ \begin{array}{cc} \frac{3}{2}a & \frac{3}{2}b \\ \frac{3}{2}c & \frac{3}{2}d \end{array} \right] +\left[ \begin{array}{cc} -1 & 3 \\ 2 & -2 \end{array} \right] =\left[ \begin{array}{cc} 3 & -4 \\ 5 & 4 \end{array} \right]$
$\left[ \begin{array}{cc} \frac{3}{2}a-1 & \frac{3}{2}b+3 \\ \frac{3}{2}c+2 & \frac{3}{2}d-2% \end{array} \right] =\left[ \begin{array}{cc} 3 & -4 \\ 5 & 4 \end{array} \right] $

$\frac{3}{2}a-1=3\Longrightarrow a=\frac{8}{3}$

$\frac{3}{2}b+3=-4\Longrightarrow b=-\frac{14}{3}$

$\frac{3}{2}c+2=5\Longrightarrow c=\frac{6}{3}=2$

$\frac{3}{2}d-2=4\Longrightarrow d=\frac{12}{3}=4$

$X=\left[ \begin{array}{cc} \frac{8}{3} & -\frac{14}{3} \\ 2 & 4 \end{array} \right]$

Solution:
First, we need to find the dimension of $A$.
$A_{m \times n}=B_{2\times3} \cdot C_{3\times2}$ So
$m \times n=(2\times 3)\cdot (3 \times 2)$
$m\times n=2\times 2$

$A=\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] $

Then, we need to do the multiplication of $B\times C$ matrices.
$B\times C=\left[ \begin{array}{ccc} 1 & -3 & -2 \\ 2 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{cc} 2 & 1 \\ -2 & -1 \\ 3 & 0 \end{array} \right] =\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] $ where

$a=1\times 2+(-3)\times (-2)+(-2)\times 3=2+6-6=2$
$b=1\times 1+(-3)\times (-1)+(-2)\times 0=1+3+0=4$
$c=2\times 2+0\times (-2)+1\times 3=4+0+3=7$
$d=2\times 1+0 \times (-1)+1 \times 0=2+0+0=2$

So

$A=\left[ \begin{array}{cc} 2 & 4 \\ 7 & 2 \end{array} \right]$

Solution:
The determinant of the matrix $A=\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] $ is $\det A=ad-bc$

$\det A=2\times 5-(-3)\times 4=10+12=22$

Solution:
The determinant of the matrix $A=\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]$ is $\det A=ad-bc$
$\det A=3\times 0-(4)\times 0=0-0=0$

Solution:
We use the formula for the inverse of a $2\times 2$ matrix.

$A=\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] $ $\Longrightarrow A^{-1}=\frac{1}{\det A}\left[ \begin{array}{cc} d & -b \\ -c & a \end{array} \right] $ where $\det A\neq 0$

If $\det A=0$, we say that the inverse of A does not exist.
But $\det A=ad-bc=2x5-(-3)x4=10+12=22\neq 0$

So $A^{-1}=\frac{1}{\det A}\left[ \begin{array}{cc} d & -b \\ -c & a \end{array} \right] $ $=\frac{1}{22}\left[ \begin{array}{cc} 5 & 3 \\ -4 & 2 \end{array} \right] = \left[ \begin{array}{cc} \frac{5}{22} & \frac{3}{22} \\ \frac{-2}{11} & \frac{1}{11} \end{array} \right]$

$A^{-1}=\left[ \begin{array}{cc} \frac{5}{22} & \frac{3}{22} \\ \frac{-2}{11} & \frac{1}{11} \end{array} \right]$

Solution:
We use the formula for the inverse of a $2\times 2$ matrix.
$A=\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \Longrightarrow A^{-1}=\frac{1}{\det A}\left[ \begin{array}{cc} d & -b \\ -c & a \end{array} \right]$ where $\det A\neq 0$
If $\det A=0$, we say that the inverse of A does not exist.
But $\det A=ad-bc=0\times 0-(\frac{-3}{4}) \times \frac{7}{3}=0+\frac{7}{4}=\frac{7}{4}\neq 0$
So $A^{-1}=\frac{1}{\det A}\left[ \begin{array}{cc} d & -b \\ -c & a \end{array} \right] =\frac{1}{\frac{7}{4}}\left[ \begin{array}{cc} 0 & \frac{3}{4} \\ -\frac{7}{3} & 0 \end{array} \right]=\frac{4}{7}\left[ \begin{array}{cc} 0 & \frac{3}{4} \\ -\frac{7}{3} & 0 \end{array} \right] =\left[ \begin{array}{cc} 0 & \frac{3}{7} \\ -\frac{4}{3} & 0 \end{array} \right] $
$A^{-1}=\left[ \begin{array}{cc} 0 & \frac{3}{7} \\ -\frac{4}{3} & 0 \end{array} \right] $

Solution:
We use the formula for the inverse of a $2 \times 2$ matrix.
$A=\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \Longrightarrow A^{-1}=\frac{1}{\det A}\left[ \begin{array}{cc} d & -b \\ -c & a \end{array} \right]$ where $\det A \neq 0$
If $\det A=0$, we say that the inverse of A does not exist.
But $\det A=ad-bc=3 \times 8-(-4) \times (-6)=24-24=0$

So, the inverse of A does not exist.

Solution:
$A \cdot B=\left[ \begin{array}{cc} 7 & -4 \\ 4 & -3 \end{array} \right] \cdot \left[ \begin{array}{cc} \frac{3}{5} & -\frac{4}{5} \\ \frac{4}{5} & -\frac{7}{5} \end{array} \right] =\left[ \begin{array}{cc} 7x\frac{3}{5}-4x\frac{4}{5} & 7x(-\frac{4}{5})-4x(-\frac{7}{5}) \\ 4x\frac{3}{5}-3x\frac{4}{5} & 4x(-\frac{4}{5})-3x(-\frac{7}{5}) \end{array} \right]$

$A \cdot B=\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$ similarly we calculate:
$B \cdot A=\left[ \begin{array}{cc} \frac{3}{5} & -\frac{4}{5} \\ \frac{4}{5} & -\frac{7}{5} \end{array} \right] \left[ \begin{array}{cc} 7 & -4 \\ 4 & -3 \end{array} \right] =\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$

So, $A$ and $B$ are multiplicative inverse.

Solution:
$A \cdot B=\left[ \begin{array}{cc} 2 & -3 \\ 1 & -2 \end{array} \right] \left[ \begin{array}{cc} -2 & 1 \\ -3 & 2 \end{array} \right] = \left[ \begin{array}{cc} 5 & -4 \\ 4 & -3 \end{array} \right] \neq \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$

Answer: $A \cdot B \ne B \cdot A$

Solution:
$\left[ \begin{array}{cc} 8 & 9 \\ -1 & 2 \end{array} \right] \left[ \begin{array}{cc} \frac{2}{25} & -\frac{1}{5} \\ \frac{3}{25} & \frac{9}{25} \end{array} \right] = \left[ \begin{array}{cc} \frac{43}{25} & \frac{41}{25} \\ \frac{4}{25} & \frac{23}{25} \end{array} \right] \neq \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$

Solution:
$A \cdot B=\left[ \begin{array}{cc} 8 & 9 \\ -1 & 2 \end{array} \right] \left[ \begin{array}{cc} \frac{2}{25} & -\frac{9}{25} \\ \frac{1}{25} & \frac{8}{25} \end{array} \right] = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] $

$B \cdot A=\left[ \begin{array}{cc} \frac{2}{25} & -\frac{9}{25} \\ \frac{1}{25} & \frac{8}{25} \end{array} \right] \left[ \begin{array}{cc} 8 & 9 \\ -1 & 2 \end{array} \right] = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$

Solution:
If $B=A^{-1}$, then $A\cdot B= B \cdot A=I$

$\left[ \begin{array}{cc} 1 & 3 \\ -1 & 2 \end{array} \right] \left[ \begin{array}{cc} \frac{2}{5} & x \\ \frac{1}{5} & \frac{1}{5} \end{array} \right] =\left[ \begin{array}{cc} \frac{2}{5}+\frac{3}{5} & x+\frac{3}{5} \\ -\frac{2}{5}+\frac{2}{5} & -x+\frac{2}{5} \end{array} \right] =\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$

$\left\{ \begin{array}{c} x+\frac{3}{5}=0 \\ -x+\frac{2}{5}=1 \end{array} \right\} \Longrightarrow x=-\frac{3}{5}$

Solution:
In order a matrix to have no inverse its determinant must be 0. So we have to solve the equation $\det A=0$.

$\det A=2x(-2)-3x=0$ $\Longrightarrow 3x=-4\Longrightarrow x=-\frac{4}{3}$

Solution:
If we want the matrix $A$ to have no inverse, we must solve $\det A=0$. So,
$\det A=-1-(2+x)x=-1-2x-x^{2}=0$
$x^{2}+2x+1=0\Longrightarrow \left( x+1\right)^{2}=0\Longrightarrow x=-1$