Question 21

loading...

Prev/Next links

Jambmaths question: 

A matrix P has an inverse ${{P}^{-1}}=\left( \begin{matrix}   1 & -3  \\   0 & 1  \\\end{matrix} \right)$find P

Option A: 

$\left( \begin{matrix}   -1 & 3  \\   0 & -1  \\\end{matrix} \right)$

Option B: 

\[\left( \begin{matrix}   1 & 3  \\   0 & -1  \\\end{matrix} \right)\]

Option C: 

\[\left( \begin{matrix}   1 & -3  \\   0 & -1  \\\end{matrix} \right)\]

Option D: 

\[\left( \begin{matrix}   1 & 3  \\   0 & 1  \\\end{matrix} \right)\]

Jamb Maths Solution: 

$\begin{align}  & \mathbf{NOTE}:\text{ The product of a matrix and its inverse will only given an identity matrix i}.\text{e }I. \\ & \text{Also if}~\text{Matrix }A=\left( \begin{matrix}   a & b  \\   c & d  \\\end{matrix} \right)\text{ }~\text{its inverse will be given as }{{A}^{-1}}=\frac{1}{ad-bc}\left( \begin{matrix}   d & -b  \\   -c & a  \\\end{matrix} \right) \\ & \text{Given }{{P}^{-1}}=\left( \begin{matrix}   1 & -3  \\   0 & 1  \\\end{matrix} \right) \\ & P=\frac{1}{(1\times 1)-[0\times (-3)]}\left( \begin{matrix}   1 & 3  \\   0 & 1  \\\end{matrix} \right)=\frac{1}{1}\left( \begin{matrix}   1 & 3  \\   0 & 1  \\\end{matrix} \right) \\ & P=\left( \begin{matrix}   1 & 3  \\   0 & 1  \\\end{matrix} \right) \\\end{align}$

Jamb Maths Topic: 
Year of Exam: