Question 31 of 58advanced🔧 ApplyLong Answer4 marks

If A=[2−11−12−11−12]A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}

Verify that A3−6A2+9A−4I=0A^3 - 6A^2 + 9A - 4I = 0 and hence find A−1A^{-1}.

Correct Answer

Step 1: Calculate A2A^2

We compute the square of matrix AA by multiplying AA with itself.

A2=A⋅A=[2−11−12−11−12][2−11−12−11−12]=[6−55−56−55−56]A^2 = A \cdot A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} = \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix}

Step 2: Calculate A3A^3

Next, we compute A3A^3 by multiplying A2A^2 with AA.

A3=A2⋅A=[6−55−56−55−56][2−11−12−11−12]=[22−2121−2122−2121−2122]A^3 = A^2 \cdot A = \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} = \begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix}

Step 3: Verify the equation

We substitute the matrices into the expression A3−6A2+9A−4IA^3 - 6A^2 + 9A - 4I.

A3−6A2+9A−4IA^3 - 6A^2 + 9A - 4I =[22−2121−2122−2121−2122]−6[6−55−56−55−56]+9[2−11−12−11−12]−4[100010001]= \begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix} - 6\begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} + 9\begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} - 4\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} =[22−2121−2122−2121−2122]−[36−3030−3036−3030−3036]+[18−99−918−99−918]−[400040004]= \begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix} - \begin{bmatrix} 36 & -30 & 30 \\ -30 & 36 & -30 \\ 30 & -30 & 36 \end{bmatrix} + \begin{bmatrix} 18 & -9 & 9 \\ -9 & 18 & -9 \\ 9 & -9 & 18 \end{bmatrix} - \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} =[22−36+18−4−21+30−9+021−30+9+0−21+30−9+022−36+18−4−21+30−9+021−30+9+0−21+30−9+022−36+18−4]=[000000000]=0= \begin{bmatrix} 22-36+18-4 & -21+30-9+0 & 21-30+9+0 \\ -21+30-9+0 & 22-36+18-4 & -21+30-9+0 \\ 21-30+9+0 & -21+30-9+0 & 22-36+18-4 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = 0

Hence verified.

Step 4: Find A−1A^{-1}

From the equation A3−6A2+9A−4I=0A^3 - 6A^2 + 9A - 4I = 0, we rearrange terms to isolate II on one side:

4I=A3−6A2+9A4I = A^3 - 6A^2 + 9A 4I=A(A2−6A+9I)4I = A(A^2 - 6A + 9I)

Post-multiplying by A−1A^{-1} (since ∣A∣≠0|A| \neq 0, A−1A^{-1} exists):

4IA−1=A(A2−6A+9I)A−14IA^{-1} = A(A^2 - 6A + 9I)A^{-1} 4A−1=AA−1(A2−6A+9I)4A^{-1} = A A^{-1} (A^2 - 6A + 9I) 4A−1=I(A2−6A+9I)4A^{-1} = I(A^2 - 6A + 9I) A−1=14(A2−6A+9I)A^{-1} = \frac{1}{4}(A^2 - 6A + 9I)

Substituting the matrices:

A−1=14([6−55−56−55−56]−[12−66−612−66−612]+[900090009])A^{-1} = \frac{1}{4} \left( \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} - \begin{bmatrix} 12 & -6 & 6 \\ -6 & 12 & -6 \\ 6 & -6 & 12 \end{bmatrix} + \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} \right) A−1=14[31−1131−113]A^{-1} = \frac{1}{4} \begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix}

∴ A−1=14[31−1131−113]A^{-1} = \frac{1}{4} \begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix}

Exercise: EXERCISE 4.4 | Q: 16 | (Page 18)
For More Understanding

Explanation

This solution follows the method demonstrated in Example 15 of the provided context. First, A2A^2 and A3A^3 are calculated using matrix multiplication. The given polynomial equation is verified by substituting these matrices and showing the result is a zero matrix. Finally, the inverse is found by rearranging the polynomial equation to express A−1A^{-1} in terms of AA and II, similar to how Example 15 derives A−1A^{-1} from A2−4A+I=OA^2 - 4A + I = O.

Solution Steps

  1. Step 1: Compute A2A^2 by multiplying matrix AA with itself.

  2. Step 2: Compute A3A^3 by multiplying A2A^2 with AA.

  3. Step 3: Substitute AA, A2A^2, A3A^3, and II into the expression A3−6A2+9A−4IA^3 - 6A^2 + 9A - 4I and verify the result is the zero matrix.

  4. Step 4: Rearrange the verified equation A3−6A2+9A−4I=0A^3 - 6A^2 + 9A - 4I = 0 to solve for A−1A^{-1}.

  5. Step 5: Simplify the matrix expression 14(A2−6A+9I)\frac{1}{4}(A^2 - 6A + 9I) to get the final inverse matrix.

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