a) Using row equivalent method or inverse matrix method, solve the following system of equations. x + y + z = 1 , x + 2y + 3z = 4 , x + 3y + 7z = 13 . b) If one root of the equation is the square of…

a) Solution of the system of equations using the inverse matrix method

We are given the system of equations: $$\{\begin{array}{l}x+y+z=1\\ x+2y+3z=4\\ x+3y+7z=13\end{array}$$

This can be written in matrix form as $AX=B$, where: $$A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 3& 7\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],B=\left[\begin{array}{c}1\\ 4\\ 13\end{array}\right]$$

Step 1: Find the determinant of $A$

$$\det (A)=\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 3& 7\end{array}\right|=1(2\cdot 7-3\cdot 3)-1(1\cdot 7-3\cdot 1)+1(1\cdot 3-2\cdot 1)$$ $$=1(14-9)-1(7-3)+1(3-2)$$ $$=1(5)-1(4)+1(1)=5-4+1=2$$

Since $\det (A)\ne 0$, the inverse of $A$ exists.

Step 2: Find the inverse of $A$

The inverse of $A$ is given by: $$A^{-1}=\frac{1}{\det (A)}\cdot \text{adj}(A)$$

First, compute the cofactor matrix: $$C_{11}=\left|\begin{array}{cc}2& 3\\ 3& 7\end{array}\right|=14-9=5$$ $$C_{12}=-\left|\begin{array}{cc}1& 3\\ 1& 7\end{array}\right|=-(7-3)=-4$$ $$C_{13}=\left|\begin{array}{cc}1& 2\\ 1& 3\end{array}\right|=3-2=1$$ $$C_{21}=-\left|\begin{array}{cc}1& 1\\ 1& 7\end{array}\right|=-(7-1)=-6$$ $$C_{22}=\left|\begin{array}{cc}1& 1\\ 1& 7\end{array}\right|=7-1=6$$ $$C_{23}=-\left|\begin{array}{cc}1& 1\\ 1& 3\end{array}\right|=-(3-1)=-2$$ $$C_{31}=\left|\begin{array}{cc}1& 1\\ 2& 3\end{array}\right|=3-2=1$$ $$C_{32}=-\left|\begin{array}{cc}1& 1\\ 1& 3\end{array}\right|=-(3-1)=-2$$ $$C_{33}=\left|\begin{array}{cc}1& 1\\ 1& 2\end{array}\right|=2-1=1$$

Thus, the cofactor matrix is: $$\text{cof}(A)=\left[\begin{array}{ccc}5& -4& 1\\ -6& 6& -2\\ 1& -2& 1\end{array}\right]$$

The adjugate matrix is the transpose of the cofactor matrix: $$\text{adj}(A)=\left[\begin{array}{ccc}5& -6& 1\\ -4& 6& -2\\ 1& -2& 1\end{array}\right]$$

Now, the inverse of $A$ is: $$A^{-1}=\frac{1}{2}\left[\begin{array}{ccc}5& -6& 1\\ -4& 6& -2\\ 1& -2& 1\end{array}\right]$$

Step 3: Solve for $X$

$$X=A^{-1}B=\frac{1}{2}\left[\begin{array}{ccc}5& -6& 1\\ -4& 6& -2\\ 1& -2& 1\end{array}\right]\left[\begin{array}{c}1\\ 4\\ 13\end{array}\right]$$

Compute the matrix multiplication: $$X=\frac{1}{2}\left[\begin{array}{c}5(1)+(-6)(4)+1(13)\\ -4(1)+6(4)+(-2)(13)\\ 1(1)+(-2)(4)+1(13)\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}5-24+13\\ -4+24-26\\ 1-8+13\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}-6\\ -6\\ 6\end{array}\right]$$

Thus: $$X=\left[\begin{array}{c}-3\\ -3\\ 3\end{array}\right]$$

So, the solution is: $$\boxed{x=-3,y=-3,z=3}$$

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b) Proof that $b^3+a^{2}c+a{c}^2=3abc$ if one root of the quadratic equation is the square of the other

Let the quadratic equation be: $$a{x}^2+bx+c=0$$

Let the roots be $\alpha$ and ${\alpha }^2$ (since one root is the square of the other).

Step 1: Use the sum and product of roots

For a quadratic equation $a{x}^2+bx+c=0$, the sum and product of the roots are given by: $$\alpha +{\alpha }^2=-\frac{b}{a}\text{(1)}$$ $$\alpha \cdot {\alpha }^2=\frac{c}{a}⟹{\alpha }^3=\frac{c}{a}\text{(2)}$$

Step 2: Express $b$ and $c$ in terms of $\alpha$ and $a$

From equation (1): $$b=-a(\alpha +{\alpha }^2)$$

From equation (2): $$c=a{\alpha }^3$$

Step 3: Substitute $b$ and $c$ into the expression $b^3+a^{2}c+a{c}^2$

$$b^3+a^{2}c+a{c}^2={\left[-a(\alpha +{\alpha }^2)\right]}^3+a^2(a{\alpha }^3)+a(a{\alpha }^3{)}^2$$

Simplify each term: $${\left[-a(\alpha +{\alpha }^2)\right]}^3=-a^3(\alpha +{\alpha }^2{)}^3$$ $$a^{2}c=a^2(a{\alpha }^3)=a^3{\alpha }^3$$ $$a{c}^2=a(a{\alpha }^3{)}^2=a^3{\alpha }^6$$

Thus: $$b^3+a^{2}c+a{c}^2=-a^3(\alpha +{\alpha }^2{)}^3+a^3{\alpha }^3+a^3{\alpha }^6$$

Factor out $a^3$: $$=a^3\left[-(\alpha +{\alpha }^2{)}^3+{\alpha }^3+{\alpha }^6\right]$$

Step 4: Expand $(\alpha +{\alpha }^2{)}^3$

$$(\alpha +{\alpha }^2{)}^3={\alpha }^3+3{\alpha }^4+3{\alpha }^5+{\alpha }^6$$

Substitute back: $$=a^3\left[-({\alpha }^3+3{\alpha }^4+3{\alpha }^5+{\alpha }^6)+{\alpha }^3+{\alpha }^6\right]$$ $$=a^3\left[-{\alpha }^3-3{\alpha }^4-3{\alpha }^5-{\alpha }^6+{\alpha }^3+{\alpha }^6\right]$$ $$=a^3\left[-3{\alpha }^4-3{\alpha }^5\right]$$ $$=a^3\left[-3{\alpha }^4(1+\alpha )\right]$$

But from equation (1), $\alpha +{\alpha }^2=-\frac{b}{a}$, so: $$\alpha (1+\alpha )=-\frac{b}{a}$$

Thus: $$-3{\alpha }^4(1+\alpha )=-3{\alpha }^4\left(-\frac{b}{a}\right)=3{\alpha }^4\frac{b}{a}$$

But from equation (2), ${\alpha }^3=\frac{c}{a}$, so ${\alpha }^4=\alpha \cdot {\alpha }^3=\alpha \cdot \frac{c}{a}$.

Substitute: $$3{\alpha }^4\frac{b}{a}=3\left(\alpha \cdot \frac{c}{a}\right)\frac{b}{a}=3\frac{bc\alpha }{a^2}$$

But from equation (2), $\alpha ={\left(\frac{c}{a}\right)}^{1/3}$, so this seems inconsistent. Instead, let's use a different approach by directly substituting ${\alpha }^3=\frac{c}{a}$ into the expression.

Step 5: Alternative approach using ${\alpha }^3=\frac{c}{a}$

From equation (1): $$\alpha +{\alpha }^2=-\frac{b}{a}$$

Multiply both sides by $\alpha$: $${\alpha }^2+{\alpha }^3=-\frac{b}{a}\alpha$$

But ${\alpha }^3=\frac{c}{a}$, so: $${\alpha }^2+\frac{c}{a}=-\frac{b}{a}\alpha$$

Multiply through by $a$: $$a{\alpha }^2+c=-b\alpha$$

Rearrange: $$a{\alpha }^2+b\alpha +c=0$$

This is the original quadratic equation, so it is consistent.

Now, let's compute $b^3+a^{2}c+a{c}^2$ using ${\alpha }^3=\frac{c}{a}$ and $\alpha +{\alpha }^2=-\frac{b}{a}$.

We know: $$(\alpha +{\alpha }^2{)}^3={\alpha }^3+3{\alpha }^4+3{\alpha }^5+{\alpha }^6$$

But ${\alpha }^3=\frac{c}{a}$, so: $${\alpha }^4=\alpha \cdot {\alpha }^3=\alpha \cdot \frac{c}{a}$$ $${\alpha }^5={\alpha }^2\cdot {\alpha }^3={\alpha }^2\cdot \frac{c}{a}$$ $${\alpha }^6=({\alpha }^3{)}^2={\left(\frac{c}{a}\right)}^2$$

Substitute these into $(\alpha +{\alpha }^2{)}^3$: $$(\alpha +{\alpha }^2{)}^3=\frac{c}{a}+3\alpha \cdot \frac{c}{a}+3{\alpha }^2\cdot \frac{c}{a}+{\left(\frac{c}{a}\right)}^2$$

But from equation (1), $\alpha +{\alpha }^2=-\frac{b}{a}$, so: $${\left(-\frac{b}{a}\right)}^3=\frac{c}{a}+\frac{3c}{a}(\alpha +{\alpha }^2)+{\left(\frac{c}{a}\right)}^2$$

Substitute $\alpha +{\alpha }^2=-\frac{b}{a}$: $$-\frac{b^3}{a^3}=\frac{c}{a}+\frac{3c}{a}\left(-\frac{b}{a}\right)+\frac{c^2}{a^2}$$ $$-\frac{b^3}{a^3}=\frac{c}{a}-\frac{3bc}{a^2}+\frac{c^2}{a^2}$$

Multiply through by $a^3$: $$-b^3=a^{2}c-3abc+a{c}^2$$

Rearrange: $$b^3+a^{2}c+a{c}^2=3abc$$

Thus, we have proven that: $$\boxed{b^3+a^{2}c+a{c}^2=3abc}$$