Show that A² = A Question If \[ A = \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix} \] show that \( A^2 = A \). Solution We compute \( A^2 = A \times A \) \[ A^2 = \begin{bmatrix} 2 & -3 & […]
If A=[[2, -3, -5], [-1, 4, 5], [1, -3, -4]], show that A^2 = A. Read More »
Cube Root of Unity Matrix Proof Question If \( \omega \) is a complex cube root of unity, show that: \[ \left( \begin{bmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{bmatrix} + \begin{bmatrix} \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \\
Prove A^3 = pI + qA + rA^2 Question Given \[ A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{bmatrix} \] and \(I\) is the identity matrix of order 3. Prove that: \[ A^3 = pI + qA + rA^2 \] Step 1:
Find a43 and a22 of Matrix Question Find the elements \( a_{43} \) and \( a_{22} \) of the matrix: \[ A = \begin{bmatrix} 0 & 1 & 0 \\ 2 & 0 & 2 \\ 0 & 3 & 2 \\ 4 & 0 & 4 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -3 &
Verify A(B – C) = AB – AC Question Verify that \( A(B – C) = AB – AC \) for: \( A = \begin{bmatrix} 1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1 \end{bmatrix}, \; B = \begin{bmatrix} 0 & 5 & -4 \\ -2 &
Verify A(B + C) = AB + AC Question Verify that \( A(B + C) = AB + AC \) for: \( A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}, \quad C = \begin{bmatrix} 1 &
Distributive Property of Matrix Multiplication Question: Verify the distributive property of matrix multiplication: \[ A(B + C) = AB + AC \] where \[ A = \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} -1 & 0 \\ 2 & 1 \end{bmatrix}, \quad C = \begin{bmatrix} 0 & 1 \\
Verify Associativity of Matrix Multiplication Question: Verify the associativity of matrix multiplication, i.e. \((AB)C = A(BC)\), where \[ A = \begin{bmatrix} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & -1 & 1 \\ 0 & 1 & 2 \\
Verify Associativity (AB)C = A(BC) Verify \((AB)C = A(BC)\) Question: Given \[ A=\begin{bmatrix}1 & 2 & 0 \\ -1 & 0 & 1\end{bmatrix}, \quad B=\begin{bmatrix}1 & 0 \\ -1 & 2 \\ 0 & 3\end{bmatrix}, \quad C=\begin{bmatrix}1 \\ -1\end{bmatrix} \] verify that: \[ (AB)C = A(BC) \] Solution: Step 1: Compute \(AB\) \[ AB =
Evaluate A² − B² Evaluate \(A^2 – B^2\) Question: If \[ A=\begin{bmatrix} -1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & 5 & 5 \end{bmatrix}, \quad B=\begin{bmatrix} 0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4 \end{bmatrix} \] find \(A^2 – B^2\).