Q.Toshit, Kuldeep and Kanishk bought vegetables from a shop. All three bought potatoes, onion and tomatoes in different quantities. Toshit bought 2 kg potatoes, 3 kg onion and 1 kg tomatoes, Kuldeep bought 3 kg potatoes, 2 kg onion and 2 kg tomatoes and Kanishk bought 4 kg potatoes, 2 kg onion and 1 kg tomatoes. Toshit, Kuldeep and Kanishk paid bills of Rs. 180, Rs. 220 and Rs. 190 respectively. Find the price of potatoes, onion and tomatoes per kg using matrix method.

Potato Rs $20$/kg, onion Rs $30$/kg, tomato Rs $50$/kg (solved by the matrix method $X=A^{-1}B$).

Concept. Write the system as $AX=B$; if $\det A eq0$ then $X=A^{-1}B=\dfrac{1}{\det A}(\text{adj }A)B$.

Set up. Let $x,y,z$ be prices per kg of potato, onion, tomato.

• Toshit: $2x+3y+z=180$.
• Kuldeep: $3x+2y+2z=220$.
• Kanishk: $4x+2y+z=190$.

So $A=\begin{bmatrix}2&3&1\\3&2&2\\4&2&1\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix},\ B=\begin{bmatrix}180\\220\\190\end{bmatrix}.$

Determinant.

$\det A=2(2\cdot1-2\cdot2)-3(3\cdot1-2\cdot4)+1(3\cdot2-2\cdot4)=2(-2)-3(-5)+1(-2)=-4+15-2=9 eq0.$

Adjoint (transpose of the cofactor matrix).

$\text{adj }A=\begin{bmatrix}-2&-1&4\\5&-2&-1\\-2&8&-5\end{bmatrix}.$

Solve $X=\dfrac{1}{9}(\text{adj }A)B$.

• $x=\dfrac19\big[(-2)(180)+(-1)(220)+(4)(190)\big]=\dfrac19(-360-220+760)=\dfrac{180}{9}=20.$
• $y=\dfrac19\big[(5)(180)+(-2)(220)+(-1)(190)\big]=\dfrac19(900-440-190)=\dfrac{270}{9}=30.$
• $z=\dfrac19\big[(-2)(180)+(8)(220)+(-5)(190)\big]=\dfrac19(-360+1760-950)=\dfrac{450}{9}=50.$

Check. $3(20)+2(30)+2(50)=60+60+100=220$ ✓ and $4(20)+2(30)+50=190$ ✓.

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