Question

1. Solve the matrix equation 4𝒳−ℬ=𝒳ℬ+2𝒜 if 𝒜=(−2141), ℬ=(71−72).

 

2. Calculate the inverse matrix to the matrix 𝒜=(001011111). Check whether the obtained inverse matrix is correct.

 

3. Solve the system of linear equations: 𝑥−2𝑦+𝑧=5,−2𝑥+3𝑦−𝑧=−8,−𝑥−𝑦+2𝑧=2.

 

 

4. Calculate the area of the flat shape bounded by the curves: 𝑦=√𝑥−1,𝑦=3−𝑥,𝑦=0.

 

5. Find all the extrema of the function 𝑓(𝑥)=√16−𝑥2.

 

6. Find the maximal intervals of convexity (concavity) of the function 𝑓(𝑥)=2𝑥+arctg(3𝑥). Find the respective inflection points.

 

Answer 1

1.

$$4X-B=XB+2A$$ $$A=\begin{pmatrix} -2 & 1 \\ 4 & 1 \end{pmatrix}, B=\begin{pmatrix} 7 & 1 \\ -7 & 2 \end{pmatrix}$$

Let

$$X=\begin{pmatrix} x_{11}& x_{12} \\ x_{21} & x_{22} \end{pmatrix}$$

Then

$$4X=\begin{pmatrix} 4x_{11}& 4x_{12} \\ 4x_{21} & 4x_{22} \end{pmatrix}$$ $$XB=\begin{pmatrix} x_{11}& x_{12} \\ x_{21} & x_{22} \end{pmatrix}\begin{pmatrix} 7 & 1 \\ -7 & 2 \end{pmatrix}$$ $$=\begin{pmatrix} 7x_{11}-7x_{12}& x_{11}+2x_{12} \\ 7x_{21}-7x_{22} & x_{21}+2x_{22} \end{pmatrix}$$ $$4X-BX=\begin{pmatrix} -3x_{11}+7x_{12}& -x_{11}+2x_{12} \\ -3x_{21}+7x_{22} & -x_{21}+2x_{22} \end{pmatrix}$$ $$B+2A=\begin{pmatrix} 3 & 3 \\ 1 & 4 \end{pmatrix}$$ $$\begin{pmatrix} -3x_{11}+7x_{12}& -x_{11}+2x_{12} \\ -3x_{21}+7x_{22} & -x_{21}+2x_{22} \end{pmatrix}=\begin{pmatrix} 3 & 3 \\ 1 & 4 \end{pmatrix}$$

$$-3x_{11}+7x_{12}=3$$ $$-x_{11}+2x_{12}=3$$

$$-3x_{21}+7x_{22}=1$$ $$-x_{21}+2x_{22}=4$$

$$x_{12}=-6$$ $$-x_{11}+2x_{12}=3$$

$$x_{22}=-11$$ $$-x_{21}+2x_{22}=4$$

$$x_{11}=-21, x_{12}=-6, x_{22}=-11, x_{21}=-26$$

$$X=\begin{pmatrix} -21& -6 \\ -26 & -11 \end{pmatrix}$$

2.

$$A=\begin{pmatrix} 0 & 0 & 1\\ 0 & 1 & 1\\ 1 & 1 & 1\\ \end{pmatrix}$$

$$\begin{pmatrix} 0 & 0 & 1 & | & 1 & 0 & 0\\ 0 & 1 & 1 & | & 0 & 1 & 0\\ 1 & 1 & 1 & | & 0 & 0 & 1\\ \end{pmatrix}$$

Swap the rows 1and 3

$$\begin{pmatrix} 1 & 1 & 1 & | & 0 & 0 & 1\\ 0 & 1 & 1 & | & 0 & 1 & 0\\ 0 & 0 & 1 & | & 1 & 0 & 0\\ \end{pmatrix}$$

$R_1=R_1-R_2$

$$\begin{pmatrix} 1 & 0 & 0 & | & 0 & -1 & 1\\ 0 & 1 & 1 & | & 0 & 1 & 0\\ 0 & 0 & 1 & | & 1 & 0 & 0\\ \end{pmatrix}$$

$R_2=R_2-R_3$

$$\begin{pmatrix} 1 & 0 & 0 & | & 0 & -1 & 1\\ 0 & 1 & 0 & | & -1 & 1 & 0\\ 0 & 0 & 1 & | & 1 & 0 & 0\\ \end{pmatrix}$$

On the left is the identity matrix. On the right is the inverse matrix.

$$A^{-1}=\begin{pmatrix} 0 & -1 & 1\\ -1 & 1 & 0\\ 1 & 0 & 0\\ \end{pmatrix}$$

Check

$$AA^{-1}=\begin{pmatrix} 0 & 0 & 1\\ 0 & 1 & 1\\ 1 & 1 & 1\\ \end{pmatrix}\begin{pmatrix} 0 & -1 & 1\\ -1 & 1 & 0\\ 1 & 0 & 0\\ \end{pmatrix}$$

$$=\begin{pmatrix} 0+0+1 & 0+0+0 & 0+0+0\\ 0-1+1 & 0+1+0 & 0+0+0\\ 0-1+1 & -1+1+0 & 1+0+0\\ \end{pmatrix}$$

$$=\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{pmatrix}=I_3$$

$$A^{-1}=\begin{pmatrix} 0 & -1 & 1\\ -1 & 1 & 0\\ 1 & 0 & 0\\ \end{pmatrix}$$

3.

$$x-2y+z=5\\ -2x+3y-z=-8\\ -x-y+2z=2$$

$$-x+y=-3\\ -3y+3z=7\\ -x-y+2z=2$$

$$x=y+3\\ z=y+\dfrac{7}{3}\\ -y-3-y+2y+\dfrac{14}{3}=2$$

$$x=y+13\\ z=y+1\\ \dfrac{5}{3}=2$$

No solution

4.

$$\sqrt{x-1}=0=>x=1$$

$$3-x=0=>x=3$$

$$\sqrt{x-1}=3-x=>x-1=9-6x+x^2, 1\le x\le3$$

$$x^2-7x+10=0$$

$$(x-2)(x-5)=0$$

Since $1\le x\le3,$ we take $x=2.$

$$A=\displaystyle\int_{1}^{2}\sqrt{x-1}dx+\displaystyle\int_{2}^{3}(3-x)dx$$

$$=[\dfrac{2(x-1)^{3/2}}{3}]\begin{matrix} 2 \\ 1 \end{matrix}+[3x-\dfrac{x^2}{2}]\begin{matrix} 3 \\ 2 \end{matrix}$$

$$=\dfrac{2}{3}-0+9-\dfrac{9}{2}-6+2=\dfrac{7}{6}({units}^2)$$

5.

$$16-x^2\ge0=>-4\le x\le 4$$

Domain: $[-4, 4]$

$$f'(x)=\dfrac{-2x}{2\sqrt{16-x^2}}=-\dfrac{x}{\sqrt{16-x^2}}$$

$$f'(x)=0=>-\dfrac{x}{\sqrt{16-x^2}}=0$$

$$x=0$$

Critical numbers:$-4, 0, 4.$

$$f(-4)=0=f(4)$$

$$f(0)=4$$

The function has a local minimum at $(-4,0)$ and at $(4,0).$

The function has a local maximum at $(0,4).$

 

6.

Domain: $(-\infin, \infin)$

$$f'(x)=2+\dfrac{3}{1+9x^2}$$

$$f''(x)=-\dfrac{54x}{(1+9x^2)^2}$$

$$f''(x)=0=>-\dfrac{54x}{(1+9x^2)^2}=0$$

$$x=0$$

$$f(0)=0$$

If $x<0, f''(x)>0, f(x)$ is concave upward.

If $x>0, f''(x)<0, f(x)$ is concave downward.

The function $f(x)$ is concave upward on $(-\infin, 0).$

The function $f(x)$ is concave downward on $(0,\infin).$

Point $(0,0)$ is the inflection point.