A matrix P of order \(2 \times 3\) with each entry 0 or 1 and α is a scalar which is 3 or 4. If \(\mathrm{R} = \alpha\mathrm{A}\), then number of matrices R formed is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 64
Note: The given answer is as per the official NTA Answer Key. However, the question does not make clear whether α is one fixed scalar (either 3 or 4) or may take both values, and with the second reading Option (4) could also be a possible answer, as explained below.
The paper writes \(R = \alpha A\); here A stands for the given matrix P, so \(R = \alpha P\).
A \(2 \times 3\) matrix has 6 entries, and each entry has 2 choices (0 or 1):
\(\text{number of matrices P} = 2^{6} = 64\)
α one fixed scalar (3, or else 4): multiplying by a non-zero scalar never makes two different matrices equal, so each P gives its own R:
\(\text{number of matrices R} = 64\)
α allowed to be 3 or 4: each P gives two products, 3P and 4P, which is \(2 \times 64 = 128\) products. Only the zero matrix is counted twice, because \(3O = 4O = O\). Any non-zero 3P has an entry 3 and any non-zero 4P has an entry 4, so no other products coincide:
\(\text{number of matrices R} = 128 - 1 = 127\)
Option (1) 63 — Incorrect. 63 is the number of non-zero matrices P; the zero matrix P also gives a matrix R (the zero matrix), so one fixed α gives 64, not 63.
Option (2) 64 — Correct. With α one fixed value, there is one matrix R for each of the \(2^{6} = 64\) matrices P.
Option (3) 128 — Incorrect. 128 counts 3O and 4O as two matrices, but both are the same zero matrix.
Option (4) 127 — Should also be correct, though not the answer given by NTA. If α can be either 3 or 4, the 63 non-zero matrices P give \(63 \times 2 = 126\) different matrices R, and the zero matrix adds one more: \(126 + 1 = 127\).