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If A=[(1,0,0),(0,1,0),(0,0,1)] then A^(2...

If `A=[(1,0,0),(0,1,0),(0,0,1)]` then `A^(2)+2A` equals

A

A

B

2A

C

3A

D

4A

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The correct Answer is:
To solve the problem, we need to find \( A^2 + 2A \) where \( A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \). ### Step 1: Calculate \( A^2 \) To find \( A^2 \), we multiply matrix \( A \) by itself: \[ A^2 = A \cdot A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \] Calculating the elements of \( A^2 \): - First row: - First column: \( 1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 = 1 \) - Second column: \( 1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0 = 0 \) - Third column: \( 1 \cdot 0 + 0 \cdot 0 + 0 \cdot 1 = 0 \) - Second row: - First column: \( 0 \cdot 1 + 1 \cdot 0 + 0 \cdot 0 = 0 \) - Second column: \( 0 \cdot 0 + 1 \cdot 1 + 0 \cdot 0 = 1 \) - Third column: \( 0 \cdot 0 + 1 \cdot 0 + 0 \cdot 1 = 0 \) - Third row: - First column: \( 0 \cdot 1 + 0 \cdot 0 + 1 \cdot 0 = 0 \) - Second column: \( 0 \cdot 0 + 0 \cdot 1 + 1 \cdot 0 = 0 \) - Third column: \( 0 \cdot 0 + 0 \cdot 0 + 1 \cdot 1 = 1 \) Thus, we have: \[ A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \] ### Step 2: Calculate \( 2A \) Next, we calculate \( 2A \): \[ 2A = 2 \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} \] ### Step 3: Add \( A^2 \) and \( 2A \) Now, we add \( A^2 \) and \( 2A \): \[ A^2 + 2A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} + \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} \] Calculating the elements: - First row: \( 1 + 2 = 3 \), \( 0 + 0 = 0 \), \( 0 + 0 = 0 \) - Second row: \( 0 + 0 = 0 \), \( 1 + 2 = 3 \), \( 0 + 0 = 0 \) - Third row: \( 0 + 0 = 0 \), \( 0 + 0 = 0 \), \( 1 + 2 = 3 \) Thus, we have: \[ A^2 + 2A = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix} \] ### Final Result The final result is: \[ A^2 + 2A = 3I \] where \( I \) is the identity matrix.
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ML KHANNA-MATRICES-PROBLEM SET(1) (MULTIPLE CHOICE QUESTIONS)
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  9. If A^(2)+A=I then A^(-1) is

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  18. Inverse of the matrix [(3,-2,-1),(-4,1,-1),(2,0,1)] is

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  19. If A=[(1,-1,1),(2,1,-3),(1,1,1)] and 10 B=[(4,2,2),(-5,0,alpha),(1,-2,...

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