Exercise-4.5-Part-2, Class 12th, Maths, NCERT

Question 15.
If

$$A=\left(\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right),$$

find $A^{-1}$. Using $A^{-1}$, solve the following system of equations:

$$\{\begin{array}{l}2x-3y+5z=11,\\ 3x+2y-4z=-5,\\ x+y-2z=-3.\end{array}$$

Solution:

We have

$$A=\left(\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right)\mathrm{}$$

Let $\mathbf{b}=\left(\begin{array}{c}11\\ -5\\ -3\end{array}\right)\mathrm{}$

We know that

$$A\mathbf{x}=\mathbf{b}\Rightarrow \mathbf{x}=A^{-1}\mathbf{b}\mathrm{}$$

Step 1: Find $A^{-1}$

Using the adjoint and determinant method,

$$\det (A)=-1$$

The adjugate (adjoint) of $A$ is:

$$\text{adj}(A)=\left(\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right)\mathrm{}$$

Hence,

$$A^{-1}=\frac{1}{\det (A)}\cdot \text{adj}(A)=-1\times \left(\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right)=\left(\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right)\mathrm{.}$$

Step 2: Use $A^{-1}$ to find $x,y,z$

$$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=A^{-1}\mathbf{b}=\left(\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right)\left(\begin{array}{c}11\\ -5\\ -3\end{array}\right)\mathrm{}$$

Now multiply:

$$\begin{array}{rl}{\displaystyle x}& {\displaystyle =0(11)+1(-5)+(-2)(-3)=-5+6=1,}\\ {\displaystyle y}& {\displaystyle =-2(11)+9(-5)+(-23)(-3)=-22-45+69=2,}\\ {\displaystyle z}& {\displaystyle =-1(11)+5(-5)+(-13)(-3)=-11-25+39=3.}\end{array}$$

Final Answers:

$$A^{-1}=\left(\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right),x=1,y=2,z=3$$

Question 16.
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60.
The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90.
The cost of 6 kg onion, 2 kg wheat and 3 kg rice is ₹70.
Find the cost of each item per kilogram using the matrix method.

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Solution:

Let the cost per kilogram of onion, wheat and rice be ₹$o$, ₹$w$ and ₹$r$ respectively.

Then, according to the question, we have:

$$\begin{array}{rl}{\displaystyle 4o+3w+2r}& {\displaystyle =60,}\\ {\displaystyle 2o+4w+6r}& {\displaystyle =90,}\\ {\displaystyle 6o+2w+3r}& {\displaystyle =70}\end{array}$$

Step 1: Write in matrix form

$$\left(\begin{array}{ccc}4& 3& 2\\ 2& 4& 6\\ 6& 2& 3\end{array}\right)\left(\begin{array}{c}o\\ w\\ r\end{array}\right)=\left(\begin{array}{c}60\\ 90\\ 70\end{array}\right)$$

That is,

$$A\mathbf{x}=\mathbf{b},$$

where

$$A=\left(\begin{array}{ccc}4& 3& 2\\ 2& 4& 6\\ 6& 2& 3\end{array}\right),\mathbf{x}=\left(\begin{array}{c}o\\ w\\ r\end{array}\right),\mathbf{b}=\left(\begin{array}{c}60\\ 90\\ 70\end{array}\right)\mathrm{}$$

Step 2: Find $A^{-1}$

We have

$$\det (A)=50,$$

and the adjugate of $A$ is

$$\text{adj}(A)=\left(\begin{array}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right)\mathrm{}$$

Hence,

$$A^{-1}=\frac{1}{50}\left(\begin{array}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right)=\left(\begin{array}{ccc}0& -\frac{1}{10}& \frac{1}{5}\\ \frac{3}{5}& 0& -\frac{2}{5}\\ -\frac{2}{5}& \frac{1}{5}& \frac{1}{5}\end{array}\right)\mathrm{}$$

Step 3: Find $\mathbf{x}=A^{-1}\mathbf{}$

$$\left(\begin{array}{c}o\\ w\\ r\end{array}\right)=\left(\begin{array}{ccc}0& -\frac{1}{10}& \frac{1}{5}\\ \frac{3}{5}& 0& -\frac{2}{5}\\ -\frac{2}{5}& \frac{1}{5}& \frac{1}{5}\end{array}\right)\left(\begin{array}{c}60\\ 90\\ 70\end{array}\right)\mathrm{}$$

Now multiplying:

$$\begin{array}{rl}{\displaystyle o}& {\displaystyle =0(60)+(-\frac{1}{10})(90)+(\frac{1}{5})(70)=-9+14=5,}\\ {\displaystyle w}& {\displaystyle =(\frac{3}{5})(60)+0(90)+(-\frac{2}{5})(70)=36-28=8,}\\ {\displaystyle r}& {\displaystyle =(-\frac{2}{5})(60)+(\frac{1}{5})(90)+(\frac{1}{5})(70)=-24+18+14=8.}\end{array}$$

Final Answers:

$$\boxed{{\displaystyle \begin{array}{rl}{\displaystyle \text{Cost of onion per kg}}& {\displaystyle =\text{₹}5,}\\ {\displaystyle \text{Cost of wheat per kg}}& {\displaystyle =\text{₹}8,}\\ {\displaystyle \text{Cost of rice per kg}}& {\displaystyle =\text{₹}8}\end{array}}}$$