KnowledgeBoat Logo
|

Mathematics

Given A = [2017] and I=[1001]\begin{bmatrix}[r] 2 & 0 \ -1 & 7 \end{bmatrix} \text{ and } I = \begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} and A2 = 9A + mI. Find m.

Matrices

16 Likes

Answer

Given,

A2=9A+mI[2017][2017]=9[2017]+m[1001][2×2+0×(1)2×0+0×71×2+7×(1)1×0+7×7]=[180963]+[m00m][40949]=[180963]+[m00m][m00m]=[40949][180963][m00m]=[418009(9)4963][m00m]=[140014]\Rightarrow A^2 = 9A + mI \\[1em] \Rightarrow \begin{bmatrix}[r] 2 & 0 \ -1 & 7 \end{bmatrix}\begin{bmatrix}[r] 2 & 0 \ -1 & 7 \end{bmatrix} = 9\begin{bmatrix}[r] 2 & 0 \ -1 & 7 \end{bmatrix} + m\begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2 \times 2 + 0 \times (-1) & 2 \times 0 + 0 \times 7 \ -1 \times 2 + 7 \times (-1) & -1 \times 0 + 7 \times 7 \end{bmatrix} = \begin{bmatrix}[r] 18 & 0 \ -9 & 63 \end{bmatrix} + \begin{bmatrix}[r] m & 0 \ 0 & m \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4 & 0 \ -9 & 49 \end{bmatrix} = \begin{bmatrix}[r] 18 & 0 \ -9 & 63 \end{bmatrix} + \begin{bmatrix}[r] m & 0 \ 0 & m \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] m & 0 \ 0 & m \end{bmatrix} = \begin{bmatrix}[r] 4 & 0 \ -9 & 49 \end{bmatrix} - \begin{bmatrix}[r] 18 & 0 \ -9 & 63 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] m & 0 \ 0 & m \end{bmatrix} = \begin{bmatrix}[r] 4 - 18 & 0 - 0 \ -9 - (-9) & 49 - 63 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] m & 0 \ 0 & m \end{bmatrix} = \begin{bmatrix}[r] -14 & 0 \ 0 & -14 \end{bmatrix}

By definition of equality of matrices we get,

⇒ m = -14.

Hence, m = -14.

Answered By

9 Likes


Related Questions