PYQs / GATE EE / 2025 / Set 1 / Q14 GATE EE 2025 Set 1 — Question 14 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +1 / -0.33 Easy Matrix Algebra Linear Algebra Engineering Mathematics Cayley-Hamilton Theorem
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Last updated 5 September 2026
Question Let
P = [ 2 1 0 − 1 0 0 0 0 1 ] P = \begin{bmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} P = 2 − 1 0 1 0 0 0 0 1 and let
I I I be the identity matrix. Then
P 2 P^2 P 2 is equal to
Solution To find
P 2 P^2 P 2 , we multiply matrix
P P P by itself:
P = [ 2 1 0 − 1 0 0 0 0 1 ] P = \begin{bmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} P = 2 − 1 0 1 0 0 0 0 1 P 2 = [ 2 1 0 − 1 0 0 0 0 1 ] [ 2 1 0 − 1 0 0 0 0 1 ] = [ ( 2 ) ( 2 ) + ( 1 ) ( − 1 ) + ( 0 ) ( 0 ) ( 2 ) ( 1 ) + ( 1 ) ( 0 ) + ( 0 ) ( 0 ) ( 2 ) ( 0 ) + ( 1 ) ( 0 ) + ( 0 ) ( 1 ) ( − 1 ) ( 2 ) + ( 0 ) ( − 1 ) + ( 0 ) ( 0 ) ( − 1 ) ( 1 ) + ( 0 ) ( 0 ) + ( 0 ) ( 0 ) ( − 1 ) ( 0 ) + ( 0 ) ( 0 ) + ( 0 ) ( 1 ) ( 0 ) ( 2 ) + ( 0 ) ( − 1 ) + ( 1 ) ( 0 ) ( 0 ) ( 1 ) + ( 0 ) ( 0 ) + ( 1 ) ( 0 ) ( 0 ) ( 0 ) + ( 0 ) ( 0 ) + ( 1 ) ( 1 ) ] P^2 = \begin{bmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} (2)(2) + (1)(-1) + (0)(0) & (2)(1) + (1)(0) + (0)(0) & (2)(0) + (1)(0) + (0)(1) \\ (-1)(2) + (0)(-1) + (0)(0) & (-1)(1) + (0)(0) + (0)(0) & (-1)(0) + (0)(0) + (0)(1) \\ (0)(2) + (0)(-1) + (1)(0) & (0)(1) + (0)(0) + (1)(0) & (0)(0) + (0)(0) + (1)(1) \end{bmatrix} P 2 = 2 − 1 0 1 0 0 0 0 1 2 − 1 0 1 0 0 0 0 1 = ( 2 ) ( 2 ) + ( 1 ) ( − 1 ) + ( 0 ) ( 0 ) ( − 1 ) ( 2 ) + ( 0 ) ( − 1 ) + ( 0 ) ( 0 ) ( 0 ) ( 2 ) + ( 0 ) ( − 1 ) + ( 1 ) ( 0 ) ( 2 ) ( 1 ) + ( 1 ) ( 0 ) + ( 0 ) ( 0 ) ( − 1 ) ( 1 ) + ( 0 ) ( 0 ) + ( 0 ) ( 0 ) ( 0 ) ( 1 ) + ( 0 ) ( 0 ) + ( 1 ) ( 0 ) ( 2 ) ( 0 ) + ( 1 ) ( 0 ) + ( 0 ) ( 1 ) ( − 1 ) ( 0 ) + ( 0 ) ( 0 ) + ( 0 ) ( 1 ) ( 0 ) ( 0 ) + ( 0 ) ( 0 ) + ( 1 ) ( 1 ) P 2 = [ 3 2 0 − 2 − 1 0 0 0 1 ] P^2 = \begin{bmatrix} 3 & 2 & 0 \\ -2 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} P 2 = 3 − 2 0 2 − 1 0 0 0 1 Now, let's evaluate option (A)
2 P − I 2P - I 2 P − I :
2 P − I = 2 [ 2 1 0 − 1 0 0 0 0 1 ] − [ 1 0 0 0 1 0 0 0 1 ] = [ 4 2 0 − 2 0 0 0 0 2 ] − [ 1 0 0 0 1 0 0 0 1 ] = [ 3 2 0 − 2 − 1 0 0 0 1 ] 2P - I = 2 \begin{bmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 2 & 0 \\ -2 & 0 & 0 \\ 0 & 0 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2 & 0 \\ -2 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} 2 P − I = 2 2 − 1 0 1 0 0 0 0 1 − 1 0 0 0 1 0 0 0 1 = 4 − 2 0 2 0 0 0 0 2 − 1 0 0 0 1 0 0 0 1 = 3 − 2 0 2 − 1 0 0 0 1 Since
P 2 = 2 P − I P^2 = 2P - I P 2 = 2 P − I , option (A) is correct.
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Solution To find
P 2 P^2 P 2 , we multiply matrix
P P P by itself:
P = [ 2 1 0 − 1 0 0 0 0 1 ] P = \begin{bmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} P = 2 − 1 0 1 0 0 0 0 1 P 2 = [ 2 1 0 − 1 0 0 0 0 1 ] [ 2 1 0 − 1 0 0 0 0 1 ] = [ ( 2 ) ( 2 ) + ( 1 ) ( − 1 ) + ( 0 ) ( 0 ) ( 2 ) ( 1 ) + ( 1 ) ( 0 ) + ( 0 ) ( 0 ) ( 2 ) ( 0 ) + ( 1 ) ( 0 ) + ( 0 ) ( 1 ) ( − 1 ) ( 2 ) + ( 0 ) ( − 1 ) + ( 0 ) ( 0 ) ( − 1 ) ( 1 ) + ( 0 ) ( 0 ) + ( 0 ) ( 0 ) ( − 1 ) ( 0 ) + ( 0 ) ( 0 ) + ( 0 ) ( 1 ) ( 0 ) ( 2 ) + ( 0 ) ( − 1 ) + ( 1 ) ( 0 ) ( 0 ) ( 1 ) + ( 0 ) ( 0 ) + ( 1 ) ( 0 ) ( 0 ) ( 0 ) + ( 0 ) ( 0 ) + ( 1 ) ( 1 ) ] P^2 = \begin{bmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} (2)(2) + (1)(-1) + (0)(0) & (2)(1) + (1)(0) + (0)(0) & (2)(0) + (1)(0) + (0)(1) \\ (-1)(2) + (0)(-1) + (0)(0) & (-1)(1) + (0)(0) + (0)(0) & (-1)(0) + (0)(0) + (0)(1) \\ (0)(2) + (0)(-1) + (1)(0) & (0)(1) + (0)(0) + (1)(0) & (0)(0) + (0)(0) + (1)(1) \end{bmatrix} P 2 = 2 − 1 0 1 0 0 0 0 1 2 − 1 0 1 0 0 0 0 1 = ( 2 ) ( 2 ) + ( 1 ) ( − 1 ) + ( 0 ) ( 0 ) ( − 1 ) ( 2 ) + ( 0 ) ( − 1 ) + ( 0 ) ( 0 ) ( 0 ) ( 2 ) + ( 0 ) ( − 1 ) + ( 1 ) ( 0 ) ( 2 ) ( 1 ) + ( 1 ) ( 0 ) + ( 0 ) ( 0 ) ( − 1 ) ( 1 ) + ( 0 ) ( 0 ) + ( 0 ) ( 0 ) ( 0 ) ( 1 ) + ( 0 ) ( 0 ) + ( 1 ) ( 0 ) ( 2 ) ( 0 ) + ( 1 ) ( 0 ) + ( 0 ) ( 1 ) ( − 1 ) ( 0 ) + ( 0 ) ( 0 ) + ( 0 ) ( 1 ) ( 0 ) ( 0 ) + ( 0 ) ( 0 ) + ( 1 ) ( 1 ) P 2 = [ 3 2 0 − 2 − 1 0 0 0 1 ] P^2 = \begin{bmatrix} 3 & 2 & 0 \\ -2 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} P 2 = 3 − 2 0 2 − 1 0 0 0 1 Now, let's evaluate option (A)
2 P − I 2P - I 2 P − I :
2 P − I = 2 [ 2 1 0 − 1 0 0 0 0 1 ] − [ 1 0 0 0 1 0 0 0 1 ] = [ 4 2 0 − 2 0 0 0 0 2 ] − [ 1 0 0 0 1 0 0 0 1 ] = [ 3 2 0 − 2 − 1 0 0 0 1 ] 2P - I = 2 \begin{bmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 2 & 0 \\ -2 & 0 & 0 \\ 0 & 0 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2 & 0 \\ -2 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} 2 P − I = 2 2 − 1 0 1 0 0 0 0 1 − 1 0 0 0 1 0 0 0 1 = 4 − 2 0 2 0 0 0 0 2 − 1 0 0 0 1 0 0 0 1 = 3 − 2 0 2 − 1 0 0 0 1 Since
P 2 = 2 P − I P^2 = 2P - I P 2 = 2 P − I , option (A) is correct.
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