Question 10(ii)Given matrix, X = [1183] and I=[1001],\begin{bmatrix*}[r] 1 & 1 \ 8 & 3 \end{bmatrix*} ext{ and } I = \begin{bmatrix*}[r] 1 & 0 \ 0 & 1 \end{bmatrix*},[1813] and I=[1001], prove that X2 = 4X + 5I.

Question

Question 10(ii)Given matrix, X = [1183] and I=[1001],\begin{bmatrix*}[r] 1 & 1 \ 8 & 3 \end{bmatrix*} 	ext{ and } I = \begin{bmatrix*}[r] 1 & 0 \ 0 & 1 \end{bmatrix*},[18​13​] and I=[10​01​], prove that X2 = 4X + 5I.

Accepted Answer

Given,X2 = 4X + 5ISolving for L.H.S.,X2=[1183][1183]=[1×1+1×81×1+1×38×1+3×88×1+3×3]=[1+81+38+248+9]=[943217].X^2 = \begin{bmatrix*}[r] 1 & 1 \ 8 & 3 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 1 \ 8 & 3 \end{bmatrix*} \[1em] = \begin{bmatrix*}[r] 1 \times 1 + 1 \times 8 & 1 \times 1 + 1\times 3 \ 8 \times 1 + 3 \times 8 & 8 \times 1 + 3 \times 3 \end{bmatrix*} \[1em] = \begin{bmatrix*}[r] 1 + 8 & 1 + 3 \ 8 + 24 & 8 + 9 \end{bmatrix*} \[1em] = \begin{bmatrix*}[r] 9 & 4 \ 32 & 17 \end{bmatrix*}.X2=[18​13​][18​13​]=[1×1+1×88×1+3×8​1×1+1×38×1+3×3​]=[1+88+24​1+38+9​]=[932​417​].Solving for R.H.S.,4X+5I=4[1183]+5[1001]=[443212]+[5005]=[4+54+032+012+5]=[943217].4X + 5I = 4\begin{bmatrix*}[r] 1 & 1 \ 8 & 3 \end{bmatrix*} + 5\begin{bmatrix*}[r] 1 & 0 \ 0 & 1 \end{bmatrix*} \[1em] = \begin{bmatrix*}[r] 4 & 4 \ 32 & 12 \end{bmatrix*} + \begin{bmatrix*}[r] 5 & 0 \ 0 & 5 \end{bmatrix*} \[1em] = \begin{bmatrix*}[r] 4 + 5 & 4 + 0 \ 32 + 0 & 12 + 5 \end{bmatrix*} \[1em] = \begin{bmatrix*}[r] 9 & 4 \ 32 & 17 \end{bmatrix*}.4X+5I=4[18​13​]+5[10​01​]=[432​412​]+[50​05​]=[4+532+0​4+012+5​]=[932​417​].Since, L.H.S. = R.H.S.Hence, proved that X2 = 4X + 5I.

Solved 2025 Specimen Paper ICSE Class 10 Mathematics — Question 19

Question 10(ii)