5 Marks Questions

Question 15 Marks

If $A=\left[\begin{array}{ccc}3 & 1 & 2 \\ 3 & 2 & -3 \\ 2 & 0 & -1\end{array}\right]$, then find $A^{-1}$, also find the solution of system of equations as follows :
$ 3 x+3 y+2 z=1$
$x+2 y=4$
$2 x-3 y-z=5 $

Answer

$\begin{array}{rlrl} & A =\left[\begin{array}{ccc}3 & 1 & 2 \\ 3 & 2 & -3 \\ 2 & 0 & -1\end{array}\right] \\ & \therefore | A | =3(-2+0)-1(-1+6)+2(0-4) \\ & =-6-3-8=-17 \neq 0\end{array}$
Hence $A^{-1}$ exists.
Matrix of elements of cofactors of A .
$=\left[\begin{array}{ccc}\left|\begin{array}{cc}2 & -3 \\ 0 & -1\end{array}\right| & -\left|\begin{array}{ll}3 & -3 \\ 2 & -1\end{array}\right| & \left|\begin{array}{ll}3 & 2 \\ 2 & 0\end{array}\right| \\ -\left|\begin{array}{cc}1 & 2 \\ 0 & -1\end{array}\right| & \left|\begin{array}{cc}3 & 2 \\ 2 & -1\end{array}\right| & -\left\lvert\, \begin{array}{ll}3 & 1 \\ 2 & 0\end{array}\right. \\ \left|\begin{array}{cc}1 & 2 \\ 2 & -3\end{array}\right| & -\left|\begin{array}{cc}3 & 2 \\ 3 & -3\end{array}\right| & \left|\begin{array}{ll}3 & 1 \\ 3 & 2\end{array}\right|\end{array}\right]$
$\begin{array}{c}=\left[\begin{array}{ccc}-2 & -3 & -4 \\ 1 & -7 & 2 \\ -7 & 15 & 3\end{array}\right] \\ \therefore A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{1}{-17}\left[\begin{array}{ccc}-2 & 1 & -7 \\ -3 & -7 & 15 \\ -4 & 2 & 3\end{array}\right]\end{array}$
given system of equations in the form of matrix,
$\left[\begin{array}{ccc}3 & 3 & 2 \\ 1 & 2 & 0 \\ 2 & -3 & -1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 4 \\ 5\end{array}\right]$
$\begin{aligned} \Rightarrow & & A ^{ T } X & = B \\ \Rightarrow & & X & =\left( A ^{ T }\right)^{-1} B=\left( A ^{-1}\right)^{ T } B \end{aligned}$
$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-17}\left[\begin{array}{ccc}-2 & -3 & -4 \\ 1 & -7 & 2 \\ -7 & 15 & 3\end{array}\right]\left[\begin{array}{l}1 \\ 4 \\ 5\end{array}\right]$
$=-\frac{1}{17}\left[\begin{array}{c}-2-20-20 \\ 1-28+10 \\ -7+60+15\end{array}\right]=\frac{-1}{17}\left[\begin{array}{c}-34 \\ -17 \\ 68\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}2 \\ 1 \\ -4\end{array}\right]$
Hence $x=2, y=1, z=-4$

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Question 25 Marks

Find the inverse matrix of matrix $\left[\begin{array}{ccc}3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2\end{array}\right]$ and after that with the help of this, find the solution of system of equations $: \left[\begin{array}{lll} 3 & 0 & 3 \\ 2 & 1 & 0 \\ 4 & 0 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 8 \\ 1 \\ 4 \end{array}\right]+\left[\begin{array}{c} 2 y \\ z \\ 3 y \end{array}\right] $

Answer

Suppose $\begin{array}{ll} A=\left[\begin{array}{ccc}3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2\end{array}\right] \end{array} $
$ \therefore |A|=\left|\begin{array}{ccc}3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2\end{array}\right|$
$\therefore |\mathrm{A}|=-3+16-30=-17$
$\because |\mathrm{~A}| \neq 0 $
$\therefore \mathrm{~A}^{-1} \text { is existed. }$
With the help of corresponding factors of elements of determinant $A$, matrix will be $:$
$\left[\begin{array}{ccc}-1 & -8 & -10 \\ -5 & -6 & 1 \\ -1 & 9 & 7\end{array}\right]$
$ \begin{aligned} & \therefore \operatorname{adj} \mathrm{A}=\left[\begin{array}{ccc} -1 & -8 & -10 \\ -5 & -6 & 1 \\ -1 & 9 & 7
\end{array}\right]=\left[\begin{array}{ccc} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{array}\right] \end{aligned} .....(2)$
$ \therefore \mathrm{A}^{-1}=\frac{\operatorname{adjA}}{|\mathrm{A}|}=\frac{-1}{17}\left[\begin{array}{ccc} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{array}\right]^{\prime}......(3) $
Given system of equations $:$
$\begin{aligned} & {\left[\begin{array}{lll}3 & 0 & 3 \\ 2 & 1 & 0 \\ 4 & 0 & 2\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}8 \\ 1 \\ 4\end{array}\right]+\left[\begin{array}{c}2 y \\ z \\ 3 y\end{array}\right]} \end{aligned} $
$ \Rightarrow \left[\begin{array}{c}3 x+3 z \\ 2 x+y \\ 4 x+2 z\end{array}\right]=\left[\begin{array}{c}8+2 y \\ 1+z \\ 4+3 y\end{array}\right]$
$\begin{aligned} 3 x+3 z & =8+2 y \\ 2 x+y & =1+z \\ 4 x+2 z & =4+3 y \end{aligned} \quad \Rightarrow \quad \begin{array}{r} 3 x-2 y+3 z=8 \\ 2 x+y-z=1 \\ 4 x-3 y+2 z=4 \end{array}....(4)$
On writing system of equation $(4)$ in the form of matrix $:$
$\left[\begin{array}{ccc}3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}8 \\ 1 \\ 4\end{array}\right]$
$\mathrm{AX} =\mathrm{B} $
$\Rightarrow \mathrm{X} =\mathrm{A}^{-1} \mathrm{~B} $
$\begin{array}{rlrl} \mathrm{X} & =\frac{-1}{17}\left[\begin{array}{ccc}-1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7\end{array}\right]\left[\begin{array}{l}8 \\ 1 \\ 4\end{array}\right] \text { from equation (3) }\end{array}$
$=\frac{-1}{17}\left[\begin{array}{c}-8-5-4 \\ -64-6+36 \\ -80+1+28\end{array}\right]=\frac{-1}{17}\left[\begin{array}{c}-17 \\ -34 \\ -51\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$
$\begin{aligned} & \Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right] \end{aligned} $
$ \Rightarrow x=1, y=2, z=3$

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MATHS 5 Marks Questions

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