Question 22

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Jambmaths question: 

Find y, if\[left( \begin{matrix}   5 & -6  \\   2 & 7  \\\end{matrix} \right)\left( \begin{align}  & x \\ & y \\\end{align} \right)=\left( \begin{align}  & 7 \\ & -11 \\\end{align} \right)\]

Option A: 

3

Option B: 

2

Option C: 

8

Option D: 

5

Jamb Maths Solution: 

\[\begin{align}  & \left( \begin{matrix}   5 & -6  \\   2 & 7  \\\end{matrix} \right)\left( \begin{align}  & x \\ & y \\\end{align} \right)=\left( \begin{align}  & 7 \\ & -11 \\\end{align} \right) \\ & \text{Expanding the matrix} \\ & 5x-6y=7----(i) \\ & 2x-7y=-11---(ii) \\ & \text{Mulitply (i) by }2 \\ & \text{Multiply (ii) by 5} \\ & 10x-12y=14----(iii) \\ & 10x-35x=-55----(iv) \\ & \text{Subtract (iv) from (iii)} \\ & 23y=69 \\ & y=3 \\ & Method2 \\ & \text{Using inverse method} \\ & \left( \begin{matrix}   5 & -6  \\   2 & -7  \\\end{matrix} \right)\left( \begin{align}  & x \\ & y \\\end{align} \right)=\left( \begin{align}  & 7 \\ & -11 \\\end{align} \right) \\ & \text{     }A\text{        }X\text{         }B \\ & {{A}^{-1}}=\frac{1}{-35+12}\left( \begin{matrix}   -7 & 6  \\   -2 & 5  \\\end{matrix} \right)=\frac{1}{-23}\left( \begin{matrix}   -7 & 6  \\   -2 & 5  \\\end{matrix} \right) \\ & X={{A}^{-1}}B \\ & \left( \begin{align}  & x \\ & y \\\end{align} \right)=\frac{1}{-23}\left( \begin{matrix}   -7 & 6  \\   -2 & 5  \\\end{matrix} \right)\left( \begin{align}  & 7 \\ & -11 \\\end{align} \right) \\ & \left( \begin{align}  & x \\ & y \\\end{align} \right)=\frac{1}{-23}\left( \begin{align}  & -49-66 \\ & -14-55 \\\end{align} \right)=-\frac{1}{23}\left( \begin{align}  & -115 \\ & -69 \\\end{align} \right) \\ & \left( \begin{align}  & x \\ & y \\\end{align} \right)=\left( \begin{align}  & \frac{-115}{-23} \\ & \frac{-69}{-23} \\\end{align} \right)=\left( \begin{align}  & 5 \\ & 3 \\\end{align} \right) \\ & x=5,\text{ }y=3 \\\end{align}\]

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