PYQs / GATE CE / 2021 / Set 1 / Q12 GATE CE 2021 Set 1 — Question 12 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 & Operations Linear Algebra Engineering Mathematics
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Last updated 5 September 2026
Question If
P = [ 1 2 3 4 ] P = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} P = [ 1 3 2 4 ] and
Q = [ 0 1 1 0 ] Q = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} Q = [ 0 1 1 0 ] then
Q T P T Q^T P^T Q T P T is
Correct answer (D) bmatrix 2 & 4 \ 1 & 3 bmatrix
Solution We need to find
Q T P T Q^T P^T Q T P T .
First, find the transposes of
P P P and
Q Q Q :
P T = [ 1 3 2 4 ] P^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} P T = [ 1 2 3 4 ] Q T = [ 0 1 1 0 ] Q^T = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} Q T = [ 0 1 1 0 ] Now, multiply
Q T Q^T Q T and
P T P^T P T :
Q T P T = [ 0 1 1 0 ] [ 1 3 2 4 ] = [ ( 0 × 1 + 1 × 2 ) ( 0 × 3 + 1 × 4 ) ( 1 × 1 + 0 × 2 ) ( 1 × 3 + 0 × 4 ) ] = [ 2 4 1 3 ] Q^T P^T = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} (0 \times 1 + 1 \times 2) & (0 \times 3 + 1 \times 4) \\ (1 \times 1 + 0 \times 2) & (1 \times 3 + 0 \times 4) \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix} Q T P T = [ 0 1 1 0 ] [ 1 2 3 4 ] = [ ( 0 × 1 + 1 × 2 ) ( 1 × 1 + 0 × 2 ) ( 0 × 3 + 1 × 4 ) ( 1 × 3 + 0 × 4 ) ] = [ 2 1 4 3 ] Alternatively, using the property
( A B ) T = B T A T (AB)^T = B^T A^T ( A B ) T = B T A T , we have
Q T P T = ( P Q ) T Q^T P^T = (PQ)^T Q T P T = ( P Q ) T .
P Q = [ 1 2 3 4 ] [ 0 1 1 0 ] = [ 2 1 4 3 ] PQ = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix} P Q = [ 1 3 2 4 ] [ 0 1 1 0 ] = [ 2 4 1 3 ] ( P Q ) T = [ 2 4 1 3 ] (PQ)^T = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix} ( P Q ) T = [ 2 1 4 3 ] Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker. Review answer and solution without JavaScript Interactive answer checking needs JavaScript. The published solution is available below.
Correct answer (D) bmatrix 2 & 4 \ 1 & 3 bmatrix
Solution We need to find
Q T P T Q^T P^T Q T P T .
First, find the transposes of
P P P and
Q Q Q :
P T = [ 1 3 2 4 ] P^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} P T = [ 1 2 3 4 ] Q T = [ 0 1 1 0 ] Q^T = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} Q T = [ 0 1 1 0 ] Now, multiply
Q T Q^T Q T and
P T P^T P T :
Q T P T = [ 0 1 1 0 ] [ 1 3 2 4 ] = [ ( 0 × 1 + 1 × 2 ) ( 0 × 3 + 1 × 4 ) ( 1 × 1 + 0 × 2 ) ( 1 × 3 + 0 × 4 ) ] = [ 2 4 1 3 ] Q^T P^T = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} (0 \times 1 + 1 \times 2) & (0 \times 3 + 1 \times 4) \\ (1 \times 1 + 0 \times 2) & (1 \times 3 + 0 \times 4) \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix} Q T P T = [ 0 1 1 0 ] [ 1 2 3 4 ] = [ ( 0 × 1 + 1 × 2 ) ( 1 × 1 + 0 × 2 ) ( 0 × 3 + 1 × 4 ) ( 1 × 3 + 0 × 4 ) ] = [ 2 1 4 3 ] Alternatively, using the property
( A B ) T = B T A T (AB)^T = B^T A^T ( A B ) T = B T A T , we have
Q T P T = ( P Q ) T Q^T P^T = (PQ)^T Q T P T = ( P Q ) T .
P Q = [ 1 2 3 4 ] [ 0 1 1 0 ] = [ 2 1 4 3 ] PQ = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix} P Q = [ 1 3 2 4 ] [ 0 1 1 0 ] = [ 2 4 1 3 ] ( P Q ) T = [ 2 4 1 3 ] (PQ)^T = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix} ( P Q ) T = [ 2 1 4 3 ] Understand the concept, then try another question Revisit Engineering Mathematics with concept notes, common mistakes and an original worked example before your next attempt.
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