Model Exam

Question 1

Given:

$$A=\left[\begin{array}{cc}6& -2\\ 7& 5\end{array}\right],B=\left[\begin{array}{cc}3& 0\\ 7& 5\end{array}\right],C=\left[\begin{array}{cc}2& 1\\ -1& 3\end{array}\right]$$

Find:

1) $A+B$

$$\begin{array}{cc}A+B& =\left[\begin{array}{cc}6& -2\\ 7& 5\end{array}\right]+\left[\begin{array}{cc}3& 0\\ 7& 5\end{array}\right]\\ \\ & =\left[\begin{array}{cc}6+3& -2+0\\ 7+7& 5+5\end{array}\right]\\ \\ & =\left[\begin{array}{cc}9& -2\\ 14& 10\end{array}\right]\end{array}$$

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2) $5A-B$

$$\begin{array}{c}5A=\left[\begin{array}{cc}5(3)& 5(0)\\ 7(7)& 5(5)\end{array}\right]=\left[\begin{array}{cc}30& -10\\ 35& 25\end{array}\right]\\ \\ 5A-B=\left[\begin{array}{cc}30-3& -10-0\\ 35-7& 25-5\end{array}\right]\\ \\ 5A-B=\left[\begin{array}{cc}27& -10\\ 28& 20\end{array}\right]\end{array}$$

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3) $B^T$

$$B=\left[\begin{array}{cc}3& 0\\ 7& 5\end{array}\right]\Rightarrow B^T=\left[\begin{array}{cc}3& 7\\ 0& 5\end{array}\right]$$

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4) $AB$

$$\begin{array}{c}AB=\left[\begin{array}{cc}6& -2\\ 7& 5\end{array}\right]\left[\begin{array}{cc}3& 0\\ 7& 5\end{array}\right]\\ \\ =\left[\begin{array}{cc}6(3)-2(7)& 6(0)-2(5)\\ 7(3)+5(7)& 7(0)+5(5)\end{array}\right]\\ \\ =\left[\begin{array}{cc}18-14& 0-10\\ 21+35& 0+25\end{array}\right]\\ \\ AB=\left[\begin{array}{cc}4& -10\\ 56& 25\end{array}\right]\end{array}$$

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5) $(BC{)}^T$

$$\begin{array}{c}BC=\left[\begin{array}{cc}3& 0\\ 7& 5\end{array}\right]\left[\begin{array}{cc}2& 1\\ -1& 3\end{array}\right]\\ \\ BC=\left[\begin{array}{cc}3(2)+0(-1)& 3(1)+0(3)\\ 7(2)+5(-1)& 7(1)+5(3)\end{array}\right]\\ \\ BC=\left[\begin{array}{cc}6+0& 3+0\\ 14-5& 7+15\end{array}\right]\\ \\ BC=\left[\begin{array}{cc}6& 9\\ 3& 22\end{array}\right]\end{array}$$

Question 2

Given:

$$A=\left[\begin{array}{cc}4& 2\\ 8& 5\end{array}\right],C=\left[\begin{array}{ccc}1& 2& -1\\ 3& 0& 4\\ 1& 2& -1\end{array}\right]$$

Find:

1) $|A|$

$$|A|=4(5)-2(8)=20-16=4$$

2) $A^{-1}$

$$\begin{array}{c}{A}^{-1}=\frac{1}{|A|}\ast adj(A)\\ \\ adj(A)=\left[\begin{array}{cc}5& -2\\ -8& 4\end{array}\right]\\ \\ A^{-1}=\frac{1}{4}\ast \left[\begin{array}{cc}5& -2\\ -8& 4\end{array}\right]\end{array}$$

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3) $|C|$

$$\begin{array}{c}|C|=1\left|\begin{array}{cc}0& 4\\ 2& 1\end{array}\right|-2\left|\begin{array}{cc}3& 4\\ 1& 1\end{array}\right|-1\left|\begin{array}{cc}3& 0\\ 1& 2\end{array}\right|\\ \\ |C|=1(0-8)-2(3-4)-1(6-0)\\ \\ |C|=-8+2-6=-12\end{array}$$

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Bonus questions: find $C^{-1}$

$$\begin{array}{c}{C}^{-1}=\frac{1}{|C|}\ast adj(C),|C|=-12\\ \\ adj(C)={\left[\begin{array}{ccc}0(1)-2(4)& 3(1)-4(1)& 3(2)-0(1)\\ 2(1)+1(2)& 1(1)+1(1)& 1(2)-2(1)\\ 2(4)+1(0)& 1(4)+1(3)& 1(0)-2(3)\end{array}\right]}^T\\ \\ adj(C)={\left[\begin{array}{ccc}-8& -1& 6\\ 4& 2& 0\\ 8& 7& -6\end{array}\right]}^T\end{array}$$

Applying the Sign rule:

$$\begin{array}{c}adj(C)={\left[\begin{array}{ccc}-8& 1& 6\\ -4& 2& 0\\ 8& -7& -6\end{array}\right]}^T\\ \\ adj(C)=\left[\begin{array}{ccc}-8& -4& 8\\ 1& 2& -7\\ 6& 0& -6\end{array}\right]\\ \\ C^{-1}=\frac{1}{-12}\left[\begin{array}{ccc}-8& -4& 8\\ 1& 2& -7\\ 6& 0& -6\end{array}\right]\\ \\ C^{-1}=\frac{1}{12}\left[\begin{array}{ccc}8& 4& -8\\ -1& -2& 7\\ -6& 0& 6\end{array}\right]\end{array}$$

Question 3

If:

$$\begin{array}{c}4x+2y=8\\ 8x+5y=18\end{array}$$

Solve the previous system using:

1. Cramer's Method

$$\begin{array}{c}\mathrm{\Delta }=\left|\begin{array}{cc}4& 2\\ 8& 5\end{array}\right|=4(5)-2(8)\\ \\ \mathrm{\Delta }=20-16=4\\ \\ {\mathrm{\Delta }}_x=\left|\begin{array}{cc}8& 2\\ 18& 5\end{array}\right|=8(5)-2(18)\\ \\ {\mathrm{\Delta }}_x=40-36=4\\ \\ {\mathrm{\Delta }}_y=\left|\begin{array}{cc}4& 8\\ 8& 18\end{array}\right|=4(18)-8(8)\\ \\ {\mathrm{\Delta }}_y=72-64=8\\ \\ \mathrm{\Delta }=4,{\mathrm{\Delta }}_x=4,{\mathrm{\Delta }}_y=8\\ \\ x=\frac{{\mathrm{\Delta }}_x}{\mathrm{\Delta }}=\frac{4}{4}=1\\ \\ y=\frac{{\mathrm{\Delta }}_y}{\mathrm{\Delta }}=\frac{8}{4}=2\\ \\ x=1,y=2\end{array}$$

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2. Inverse Matrix Method

$$\begin{array}{c}AX=B,\therefore X=A^{-1}B\\ \\ A=\left[\begin{array}{cc}4& 2\\ 8& 5\end{array}\right],B=\left[\begin{array}{c}8\\ 18\end{array}\right]\\ \\ A^{-1}=\frac{1}{|A|}\cdot adj(A)\\ \\ |A|=4(5)-2(8)=20-16=4\\ \\ A^{-1}=\frac{1}{4}\left[\begin{array}{cc}5& -2\\ -8& 4\end{array}\right]\\ \\ A^{-1}\cdot B=\frac{1}{4}\left[\begin{array}{c}5(8)-2(18)\\ -8(8)+4(18)\end{array}\right]\\ \\ X=\frac{1}{4}\left[\begin{array}{c}4\\ 8\end{array}\right]=\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]\\ \\ x=1,y=2\end{array}$$

Question 4

1. Find the area of $\mathrm{△}abc$ where $a=(1,4),b=(2,5),c=(1,2)$

$$\begin{array}{c}\text{area of}\mathrm{△}abc=\frac{1}{2}\left|\begin{array}{ccc}1& 4& 1\\ 2& 5& 1\\ 1& 2& 1\end{array}\right|\\ \\ =1\left|\begin{array}{cc}5& 1\\ 2& 1\end{array}\right|-4\left|\begin{array}{cc}2& 1\\ 1& 1\end{array}\right|+1\left|\begin{array}{cc}2& 5\\ 1& 2\end{array}\right|\\ \\ =\frac{1}{2}\cdot 1(5-2)-4(2-1)+1(4-5)\\ \\ =\frac{1}{2}\cdot |-2|=1c{m}^2\end{array}$$

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2. Find the Value of $K$ That Makes the Volume of the Parallelepiped Equal to Zero

Given:

$$\begin{array}{c}\overrightarrow{u}=\left(\begin{array}{c}1\\ 2\\ 5\end{array}\right),\overrightarrow{v}=\left(\begin{array}{c}0\\ 1\\ k\end{array}\right),\overrightarrow{w}=\left(\begin{array}{c}2\\ 1\\ 3\end{array}\right)\\ \\ \\ Volume=\left|\begin{array}{ccc}1& 0& 2\\ 2& 1& 1\\ 5& k& 3\end{array}\right|=0\\ \\ =1(3-k)-0(6-5)+2(2k-5)\\ \\ =3-k+4k-10\\ \\ =3k-7=0,3k=7\\ \\ k=\frac{7}{3}\approx 2.33c{m}^3\end{array}$$