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Worked Examples · Example 12

Q.For A=[2312]A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}, prove that A2−4A+I=OA^2-4A+I = O, where OO is a zero matrix.

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Compute A2A^2, then 4A4A, and combine with II to show every entry of A2−4A+IA^2-4A+I is zero.

A2=A⋅AA^2=A\cdot A (row-by-column), 4A4A scales each entry by 4, and I=[1001]I=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.

  1. Compute A2A^2 with A=[2312]A=\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}:
    • (1,1): 2(2)+3(1)=7(1,1):\ 2(2)+3(1)=7, (1,2): 2(3)+3(2)=12(1,2):\ 2(3)+3(2)=12,
    • (2,1): 1(2)+2(1)=4(2,1):\ 1(2)+2(1)=4, (2,2): 1(3)+2(2)=7(2,2):\ 1(3)+2(2)=7.

A2=[71247].A^2=\begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix}.

  1. Compute 4A4A: 4A=[81248].4A=\begin{bmatrix} 8 & 12 \\ 4 & 8 \end{bmatrix}. …

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