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Q.Case Study – 1 On her birthday, Prema decides to donate some money to children of an orphanage home.If there are 8 children less, everyone gets ₹ 10 more. However, if there are 16 children more, everyone gets ₹ 10 less. Let the number of children in the orphanage home be x and the amount to be donated to each child be ₹ y. Based on the above information, answer the following questions :

(i) Write the system of linear equations in x and y formed of the given situation. [1]
(ii) Write the system of linear equations, obtained in
(i) above, in matrix form AX=BAX = B. [1]
(iii)
(a) Find the inverse of matrix A. [2]
(OR)
(b) Determine the values of x and y. [2]
CBSECBSE Class XII Board 2024Subjective· 4mImportance★★★★★
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The conditions give 5x−4y=405x-4y=40, 5x−8y=−805x-8y=-80; in matrix form AX=BAX=B with A=[5−45−8]A=\begin{bmatrix}5&-4\\5&-8\end{bmatrix}. Then A−1=120[8−45−5]A^{-1}=\tfrac{1}{20}\begin{bmatrix}8&-4\\5&-5\end{bmatrix} and X=A−1BX=A^{-1}B gives x=32, y=30x=32,\ y=30.

For AX=BAX=B with ∣A∣≠0|A|\neq0: X=A−1BX=A^{-1}B, where A−1=1∣A∣ adj(A)A^{-1}=\dfrac{1}{|A|}\,\text{adj}(A). For a 2×22\times2 matrix A=[abcd]A=\begin{bmatrix}a&b\\c&d\end{bmatrix}, ∣A∣=ad−bc|A|=ad-bc and adj(A)=[d−b−ca]\text{adj}(A)=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}.

Let number of children =x=x and donation per child =₹y=₹y; total donated =xy=xy (fixed).

(i) Form the equations:

  1. "88 fewer children, each gets ₹10₹10 more": (x−8)(y+10)=xy⇒10x−8y−80=0⇒5x−4y=40(x-8)(y+10)=xy\Rightarrow 10x-8y-80=0\Rightarrow 5x-4y=40.
  2. "1616 more children, each gets ₹10₹10 less": (x+16)(y−10)=xy⇒−10x+16y−160=0⇒5x−8y=−80(x+16)(y-10)=xy\Rightarrow -10x+16y-160=0\Rightarrow 5x-8y=-80.

(ii) Matrix form AX=BAX=B: [5−45−8][xy]=[40−80]\begin{bmatrix}5&-4\\5&-8\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}40\\-80\end{bmatrix}.

(iii)(a) Inverse of AA:

  1. ∣A∣=(5)(−8)−(−4)(5)=−40+20=−20≠0|A|=(5)(-8)-(-4)(5)=-40+20=-20\neq0.
  2. adj(A)=[−84−55]\text{adj}(A)=\begin{bmatrix}-8&4\\-5&5\end{bmatrix}.
  3. A−1=1−20[−84−55]=120[8−45−5]A^{-1}=\dfrac{1}{-20}\begin{bmatrix}-8&4\\-5&5\end{bmatrix}=\dfrac{1}{20}\begin{bmatrix}8&-4\\5&-5\end{bmatrix}.

(iii)(b) Values of xx and yy: …

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