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Question

Question: If the matrix $A = \begin{bmatrix} 1 & -2 & 2 \\ 4 & 1 & 1 \\ -2 & 1 & -1 \end{bmatrix}$, find $A^2...

If the matrix

A=[1−22411−21−1]A = \begin{bmatrix} 1 & -2 & 2 \\ 4 & 1 & 1 \\ -2 & 1 & -1 \end{bmatrix}, find A2−3A+5IA^2 - 3A + 5I and write the matrix B where B is the transpose of (A2−3A+5I)(A^2 - 3A + 5I)

Answer

The matrix A2−3A+5IA^2 - 3A + 5I is:

[−94−8−6−451016]\begin{bmatrix} -9 & 4 & -8 \\ -6 & -4 & 5 \\ 10 & 1 & 6 \end{bmatrix}

The matrix B, which is the transpose of (A2−3A+5I)(A^2 - 3A + 5I), is:

B=[−9−6104−41−856]B = \begin{bmatrix} -9 & -6 & 10 \\ 4 & -4 & 1 \\ -8 & 5 & 6 \end{bmatrix}
Explanation

Solution

To solve the problem, we need to perform matrix multiplication, scalar multiplication, matrix addition/subtraction, and finally, find the transpose of the resulting matrix.

Given matrix: A=[1−22411−21−1]A = \begin{bmatrix} 1 & -2 & 2 \\ 4 & 1 & 1 \\ -2 & 1 & -1 \end{bmatrix}

Step 1: Calculate A2A^2 A2=A×A=[1−22411−21−1][1−22411−21−1]A^2 = A \times A = \begin{bmatrix} 1 & -2 & 2 \\ 4 & 1 & 1 \\ -2 & 1 & -1 \end{bmatrix} \begin{bmatrix} 1 & -2 & 2 \\ 4 & 1 & 1 \\ -2 & 1 & -1 \end{bmatrix}

The elements of A2A^2 are calculated as follows: (A2)11=(1)(1)+(−2)(4)+(2)(−2)=1−8−4=−11(A^2)_{11} = (1)(1) + (-2)(4) + (2)(-2) = 1 - 8 - 4 = -11 (A2)12=(1)(−2)+(−2)(1)+(2)(1)=−2−2+2=−2(A^2)_{12} = (1)(-2) + (-2)(1) + (2)(1) = -2 - 2 + 2 = -2 (A2)13=(1)(2)+(−2)(1)+(2)(−1)=2−2−2=−2(A^2)_{13} = (1)(2) + (-2)(1) + (2)(-1) = 2 - 2 - 2 = -2

(A2)21=(4)(1)+(1)(4)+(1)(−2)=4+4−2=6(A^2)_{21} = (4)(1) + (1)(4) + (1)(-2) = 4 + 4 - 2 = 6 (A2)22=(4)(−2)+(1)(1)+(1)(1)=−8+1+1=−6(A^2)_{22} = (4)(-2) + (1)(1) + (1)(1) = -8 + 1 + 1 = -6 (A2)23=(4)(2)+(1)(1)+(1)(−1)=8+1−1=8(A^2)_{23} = (4)(2) + (1)(1) + (1)(-1) = 8 + 1 - 1 = 8

(A2)31=(−2)(1)+(1)(4)+(−1)(−2)=−2+4+2=4(A^2)_{31} = (-2)(1) + (1)(4) + (-1)(-2) = -2 + 4 + 2 = 4 (A2)32=(−2)(−2)+(1)(1)+(−1)(1)=4+1−1=4(A^2)_{32} = (-2)(-2) + (1)(1) + (-1)(1) = 4 + 1 - 1 = 4 (A2)33=(−2)(2)+(1)(1)+(−1)(−1)=−4+1+1=−2(A^2)_{33} = (-2)(2) + (1)(1) + (-1)(-1) = -4 + 1 + 1 = -2

So, A2=[−11−2−26−6844−2]A^2 = \begin{bmatrix} -11 & -2 & -2 \\ 6 & -6 & 8 \\ 4 & 4 & -2 \end{bmatrix}

Step 2: Calculate 3A3A 3A=3[1−22411−21−1]=[3×13×(−2)3×23×43×13×13×(−2)3×13×(−1)]=[3−661233−63−3]3A = 3 \begin{bmatrix} 1 & -2 & 2 \\ 4 & 1 & 1 \\ -2 & 1 & -1 \end{bmatrix} = \begin{bmatrix} 3 \times 1 & 3 \times (-2) & 3 \times 2 \\ 3 \times 4 & 3 \times 1 & 3 \times 1 \\ 3 \times (-2) & 3 \times 1 & 3 \times (-1) \end{bmatrix} = \begin{bmatrix} 3 & -6 & 6 \\ 12 & 3 & 3 \\ -6 & 3 & -3 \end{bmatrix}

Step 3: Calculate 5I5I Since A is a 3x3 matrix, II is the 3x3 identity matrix: I=[100010001]I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} 5I=5[100010001]=[500050005]5I = 5 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}

Step 4: Calculate A2−3A+5IA^2 - 3A + 5I A2−3A+5I=[−11−2−26−6844−2]−[3−661233−63−3]+[500050005]A^2 - 3A + 5I = \begin{bmatrix} -11 & -2 & -2 \\ 6 & -6 & 8 \\ 4 & 4 & -2 \end{bmatrix} - \begin{bmatrix} 3 & -6 & 6 \\ 12 & 3 & 3 \\ -6 & 3 & -3 \end{bmatrix} + \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}

Perform the element-wise operations: =[(−11−3+5)(−2−(−6)+0)(−2−6+0)(6−12+0)(−6−3+5)(8−3+0)(4−(−6)+0)(4−3+0)(−2−(−3)+5)]= \begin{bmatrix} (-11 - 3 + 5) & (-2 - (-6) + 0) & (-2 - 6 + 0) \\ (6 - 12 + 0) & (-6 - 3 + 5) & (8 - 3 + 0) \\ (4 - (-6) + 0) & (4 - 3 + 0) & (-2 - (-3) + 5) \end{bmatrix}

=[(−14+5)(−2+6)(−8)(−6)(−9+5)(5)(4+6)(1)(−2+3+5)]= \begin{bmatrix} (-14 + 5) & (-2 + 6) & (-8) \\ (-6) & (-9 + 5) & (5) \\ (4 + 6) & (1) & (-2 + 3 + 5) \end{bmatrix}

=[−94−8−6−451016]= \begin{bmatrix} -9 & 4 & -8 \\ -6 & -4 & 5 \\ 10 & 1 & 6 \end{bmatrix}

Step 5: Find the transpose of (A2−3A+5I)(A^2 - 3A + 5I) Let C=A2−3A+5I=[−94−8−6−451016]C = A^2 - 3A + 5I = \begin{bmatrix} -9 & 4 & -8 \\ -6 & -4 & 5 \\ 10 & 1 & 6 \end{bmatrix}. The transpose of CC, denoted as B=CTB = C^T, is obtained by interchanging its rows and columns.

B=CT=[−9−6104−41−856]B = C^T = \begin{bmatrix} -9 & -6 & 10 \\ 4 & -4 & 1 \\ -8 & 5 & 6 \end{bmatrix}

The final matrix A2−3A+5IA^2 - 3A + 5I is [−94−8−6−451016]\begin{bmatrix} -9 & 4 & -8 \\ -6 & -4 & 5 \\ 10 & 1 & 6 \end{bmatrix}. The matrix B, which is the transpose of (A2−3A+5I)(A^2 - 3A + 5I), is [−9−6104−41−856]\begin{bmatrix} -9 & -6 & 10 \\ 4 & -4 & 1 \\ -8 & 5 & 6 \end{bmatrix}.