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Matrices — JEE MCQs with answers

NCERT Class 12 · Mathematics · Ch 310 free questions503 in the full bank

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Question 1 · 3.2 Matrix · Easy
According to Definition 1 of the book, a matrix is
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Answer: B — Definition 1 reads: a matrix is an ordered rectangular array of numbers or functions, and these are called its elements or entries. So option B is correct. Option A drops the word *ordered*: moving an entry changes the matrix, which is why rows and columns are counted. Option C makes two slips: it asks for a square array (the book's $A$ is $3 \times 2$) and it drops the functions. Option D drops the functions; the book's own matrix $C$ on p.36 has entries such as $\cos x$ and $\tan x$.

A matrix is an ordered rectangular array of numbers or functions.

Question 2 · 3.3 Types of Matrices · Hard
How many different orders can a zero matrix with exactly 6 elements have?
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Answer: C — A zero matrix need not be square, so any order $m \times n$ with $mn = 6$ works: $1 \times 6$, $2 \times 3$, $3 \times 2$ and $6 \times 1$. That is 4 orders: option C. Option A assumes a zero matrix must be square, and 6 is not a perfect square. Option D treats $2 \times 3$ and $3 \times 2$ (and $1 \times 6$, $6 \times 1$) as the same order. Option B thinks there is only one zero matrix $O$, but $O$ takes its order from the context.

A matrix is said to be zero matrix or null matrix if all its elements are zero.

Question 3 · 3.5 Transpose of a Matrix · Medium
$A$ is a $2 \times 3$ matrix and $B$ is a $3 \times 4$ matrix. What is the order of $(AB)'$?
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Answer: A — $AB$ is $2 \times 4$, so $(AB)'$ is $4 \times 2$, option A. The same order comes from $B'A'$, a $4 \times 3$ times $3 \times 2$ product. Option C is the order of $AB$, before transposing. Option D assumes $(AB)' = A'B'$, then finds $A'$ ($3 \times 2$) times $B'$ ($4 \times 3$) undefined. That is the product in the wrong order; $(AB)'$ always exists when $AB$ does. Option B is the order of $A'$ alone, transposing only the first factor.

For any matrices A and B of suitable orders, we have

Question 4 · 3.4 Operations on Matrices · Medium
Let $A = \begin{bmatrix} 1 & 2 & -1 \\ 3 & 0 & 4 \end{bmatrix}$, and let $B$ be a $3 \times 3$ matrix whose second column is $\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}$; its other columns are not given. The second column of $AB$ is
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Answer: C — The $(i, 2)$th element of $AB$ uses only row $i$ of $A$ and column 2 of $B$, so the second column of $AB$ is $A$ times the second column of $B$: $1(2) + 2(-1) + (-1)(3) = -3$ and $3(2) + 0(-1) + 4(3) = 18$. Option C is correct. Option A runs down the column of $B$ from the bottom, pairing row $[a_{i1}\ a_{i2}\ a_{i3}]$ with $3, -1, 2$. Option B makes both slips, the sign of $-1$ and the reversed column. Option D reads the element $-1$ of $A$ as $+1$, so the first element becomes $2 - 2 + 3 = 3$.

we take the ith row of A and kth column of B, multiply them elementwise and take the sum of all these products

Question 5 · 3.7 Invertible Matrices · Medium
Which statement about $M = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}$ is correct?
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Answer: B — Each entry in the second row of $MB$ is (second row of $M$) times a column of $B$, which is $0 \cdot b_{1k} + 0 \cdot b_{2k} = 0$. So $MB$ always has a zero second row, while $I$ has $1$ in position $(2, 2)$. No $B$ gives $MB = I$, and $M$ has no inverse: option B. Option A: $M^2 = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}$, which is $M$ again, not $I$. Option C: $M(-M) = -M^2 = -M$. Option D: $MM' = \begin{bmatrix} 5 & 0 \\ 0 & 0 \end{bmatrix}$. None of them is $I$, as the zero-row argument predicts.

such that AB = BA = I, then B is called the inverse

Question 6 · 3.6 Symmetric and Skew Symmetric Matrices · Medium
A $4 \times 4$ skew symmetric matrix is completely fixed once you know how many of its entries?
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Answer: D — The $4$ diagonal entries are forced to be $0$, and each entry below the diagonal is the negative of its mirror above. So only the $\frac{16 - 4}{2} = 6$ entries above the diagonal are free, option D. Option A is the count for a symmetric matrix, whose diagonal is free ($6 + 4$). Option B counts every entry. Option C counts every off-diagonal entry, forgetting that they come in pairs $a_{ji} = -a_{ij}$.

all the diagonal elements of a skew symmetric matrix are zero

Question 7 · 3.2 Matrix · Medium
EXERCISE 3.1, Q3 asks for the possible orders of a matrix with 18 elements. Which list gives **all** of them, and nothing else?
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Answer: D — The number of elements is $mn = 18$, so we need every ordered pair of natural numbers with product 18, as in Example 2. They are $(1, 18), (2, 9), (3, 6), (6, 3), (9, 2), (18, 1)$, which is 6 orders, the same as the printed answer. Option D. Option A makes both slips: it drops $1 \times 18$ and $18 \times 1$, and it adds $4 \times 5$ and $5 \times 4$ (20 elements). Option B adds $4 \times 5$ and $5 \times 4$, which have 20 elements, not 18. Option C leaves out $1 \times 18$ and $18 \times 1$, but a single row or a single column of 18 elements is a matrix too.

we will find all ordered pairs of natural numbers, whose product is 8.

Question 8 · 3.3 Types of Matrices · Medium
If $\begin{bmatrix} x^2 & x \\ y & y^2 \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ 3 & 9 \end{bmatrix}$, what is $x + y$?
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Answer: D — The $(1, 2)$ elements give $x = -2$ (and then $x^2 = 4$ holds); the $(2, 1)$ elements give $y = 3$ (and then $y^2 = 9$ holds). So $x + y = 1$: option D. Option C takes the positive root $x = 2$ of $x^2 = 4$, which fails the element $x = -2$. Option B takes $x = -2$ but the negative root $y = -3$ of $y^2 = 9$, which fails $y = 3$. Option A keeps both roots of $x^2 = 4$ without checking the $(1, 2)$ element.

By equality of two matrices, equating the corresponding elements, we get

Question 9 · 3.4 Operations on Matrices · Medium
A factory's production (rows = categories 1, 2, 3; columns = boys, girls) is $\begin{bmatrix} 40 & 35 \\ 50 & 25 \\ 30 & 45 \end{bmatrix}$. If production is tripled in every category, the $(3, 1)$th element of the new production matrix is
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Answer: D — Tripling every quantity is the scalar multiple $3A$, each element multiplied by 3 (the book's doubling example, p.45); the $(3, 1)$ element is $3 \times 30 = 90$ — option D. Option A adds 3 instead of multiplying. Option B triples the whole of row 3, $3 \times (30 + 45)$. Option C is the $(3, 2)$ element, $3 \times 45 = 135$.

the new matrix is obtained by multiplying each element of the previous matrix by 2

Question 10 · 3.7 Invertible Matrices · Easy
A square matrix $A$ with $A^2 = I$ is its own inverse, since $AA = I$ in both orders. Which of these matrices is its own inverse?
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Answer: C — Square each option. Option C: $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, so it is its own inverse. This is the matrix $B$ of Example 14. Option A: its square is $\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I$, the sign trap. Option B: its square is $\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$. Option D: its square is itself, $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, which is not $I$.

such that AB = BA = I, then B is called the inverse

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