NCERT Exemplar Solutions Class 12 Maths Matrices — True or False Q82–101

NCERT Exemplar Problems, Class 12 Mathematics — Matrices, True or False questions 82 to 101. All 20 questions solved.

Twenty statements, each to be judged True or False. They test whether you remember what matrix algebra keeps from number algebra (addition is commutative and associative; $(AB)’ = B’A’$) and what it loses (multiplication is not commutative, $AB = O$ does not force $A = O$ or $B = O$, and $AB = AC$ does not force $B = C$). A single counterexample is enough to make a statement False, and the same small pair of matrices serves for several of them.

Key insight. Work out what is being claimed for every matrix, then look for one case that breaks it. The standard breakers are $A = \begin{pmatrix}0&1\\0&0\end{pmatrix}$, $B = \begin{pmatrix}1&0\\0&0\end{pmatrix}$, and matrices whose orders do not match.

Question 82

A matrix denotes a number.

Solution. A matrix is a rectangular array of numbers, not a single number. Only a $1\times1$ matrix is sometimes identified with a number, and even then the two are different objects.

Question 83

Matrices of any order can be added.

Solution. Addition is defined entry by entry, so it needs both matrices to have the same order. A $2\times3$ matrix cannot be added to a $3\times2$ matrix.

Question 84

Two matrices are equal if they have same number of rows and same number of columns.

Solution. Equal order is necessary but not sufficient. Equality also needs every corresponding entry to be equal: $\begin{pmatrix}1&2\\3&4\end{pmatrix}$ and $\begin{pmatrix}0&0\\0&0\end{pmatrix}$ have the same order but are not equal.

Question 85

Matrices of different order can not be subtracted.

Solution. Subtraction, like addition, works entry by entry, so it needs the same order. Matrices of different orders cannot be subtracted.

Question 86

Matrix addition is associative as well as commutative.

Solution. Entries are added one at a time, and addition of numbers is associative and commutative: $(A+B)+C = A+(B+C)$ and $A+B = B+A$, for matrices of the same order.

Question 87

Matrix multiplication is commutative.

Solution. In general $AB \ne BA$. For $A = \begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $B = \begin{pmatrix}1&0\\0&0\end{pmatrix}$, $AB = O$ but $BA = \begin{pmatrix}0&1\\0&0\end{pmatrix}$. Often one of the products is not even defined.

Question 88

A square matrix where every element is unity is called an identity matrix.

Solution. An identity matrix has $1$ on the main diagonal and $0$ everywhere else. A matrix of all ones, such as $\begin{pmatrix}1&1\\1&1\end{pmatrix}$, is not one: it does not leave other matrices unchanged under multiplication.

Question 89

If $A$ and $B$ are two square matrices of the same order, then $A+B = B+A$.

Solution. Addition is commutative because the entries, being numbers, add commutatively: $(A+B)_{ij} = a_{ij}+b_{ij} = b_{ij}+a_{ij} = (B+A)_{ij}$.

Question 90

If $A$ and $B$ are two matrices of the same order, then $A-B = B-A$.

Solution. $A-B = -(B-A)$, so the two are equal only when $A-B = O$, that is when $A = B$. Take $A = I$, $B = O$: $A-B = I$ but $B-A = -I$.

Question 91

If matrix $AB = O$, then $A = O$ or $B = O$ or both $A$ and $B$ are null matrices.

Solution. Matrices do not follow the zero-product rule of numbers. $A = \begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $B = \begin{pmatrix}1&0\\0&0\end{pmatrix}$ are both non-zero, yet $AB = O$.

Question 92

Transpose of a column matrix is a column matrix.

Solution. A column matrix has order $m\times1$, so its transpose has order $1\times m$: a row matrix.

Question 93

If $A$ and $B$ are two square matrices of the same order, then $AB = BA$.

Solution. Both products are defined, but they need not be equal. With $A = \begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $B = \begin{pmatrix}1&0\\0&0\end{pmatrix}$, $AB = O \ne BA$.

Question 94

If each of the three matrices of the same order are symmetric, then their sum is a symmetric matrix.

Solution. $(A+B+C)’ = A’+B’+C’ = A+B+C$, since each of $A, B, C$ equals its own transpose.

Question 95

If $A$ and $B$ are any two matrices of the same order, then $(AB)’ = A’B’$.

Solution. The correct rule is the reversal law, $(AB)’ = B’A’$. Also, two matrices of the same order need not be multipliable at all: a $2\times3$ matrix times a $2\times3$ matrix is undefined. Even for square matrices, $A’B’$ and $B’A’$ differ in general.

Question 96

If $(AB)’ = B’A’$, where $A$ and $B$ are not square matrices, then number of rows in $A$ is equal to number of columns in $B$ and number of columns in $A$ is equal to number of rows in $B$.

Solution. The identity $(AB)’ = B’A’$ is true for every pair for which $AB$ is defined; it asks only that the columns of $A$ equal the rows of $B$. It does not force the rows of $A$ to equal the columns of $B$. Take $A$ of order $2\times3$ and $B$ of order $3\times4$: $AB$ is $2\times4$, so $(AB)’$ and $B’A’$ are both $4\times2$ and they are equal, yet $A$ has $2$ rows while $B$ has $4$ columns.

The answer key prints True; that cannot be right because the identity holds in this example while the stated conclusion fails, so the statement is False.

Question 97

If $A$, $B$ and $C$ are square matrices of same order, then $AB = AC$ always implies that $B = C$.

Solution. Cancellation fails when $A$ is not invertible. Take $A = \begin{pmatrix}1&0\\0&0\end{pmatrix}$, $B = \begin{pmatrix}0&0\\0&1\end{pmatrix}$, $C = \begin{pmatrix}0&0\\1&0\end{pmatrix}$. Then $AB = AC = O$ but $B \ne C$. It does hold if $A$ is invertible, but the statement says always.

Question 98

$AA’$ is always a symmetric matrix for any matrix $A$.

Solution. If $A$ has order $m\times n$, then $A’$ is $n\times m$, so $AA’$ is always defined, of order $m\times m$. And $(AA’)’ = (A’)’A’ = AA’$, so it is symmetric.

Question 99

If $A = \begin{pmatrix}2&3&-1\\1&4&2\end{pmatrix}$ and $B = \begin{pmatrix}2&3\\4&5\\2&1\end{pmatrix}$, then $AB$ and $BA$ are defined and equal.

Solution. $A$ is $2\times3$ and $B$ is $3\times2$, so both products are defined. But $AB$ is of order $2\times2$ and $BA$ is of order $3\times3$, so they cannot be equal. Indeed $AB = \begin{pmatrix}14&20\\22&25\end{pmatrix}$.

Question 100

If $A$ is skew symmetric matrix, then $A^2$ is a symmetric matrix.

Solution. $A’ = -A$, so $(A^2)’ = (A’)^2 = (-A)^2 = A^2$. Hence $A^2$ is symmetric.

Question 101

$(AB)^{-1} = A^{-1}\cdot B^{-1}$, where $A$ and $B$ are invertible matrices satisfying commutative property with respect to multiplication.

Solution. In general $(AB)^{-1} = B^{-1}A^{-1}$. When $AB = BA$, then $(AB)^{-1} = (BA)^{-1} = A^{-1}B^{-1}$, using the same reversal law on $BA$. So the statement holds for commuting matrices.

Common mistakes

• Treating matrices like numbers. Commutativity of multiplication, the zero-product rule and cancellation all fail for matrices (Questions 87, 91, 93, 97).
• Forgetting the order condition. Addition and subtraction need equal orders, and equal orders alone do not make matrices equal (Questions 83, 84, 85).
• Writing $(AB)’ = A’B’$. The transpose of a product reverses the order of the factors (Questions 95, 96).
• Reading a rule backwards. $(AB)’ = B’A’$ needs only that $AB$ be defined, and does not say that $A$ and $B$ have matching rows and columns on both sides (Question 96).

Practise next

• Exemplar Objective and Fill in the Blanks Q53–81, the questions just before these.
• Exemplar Short Answer Q25–52, including the symmetric and skew-symmetric proofs behind Questions 94, 98 and 100.
• NCERT textbook Exercise 3.2 for matrix operations and multiplication.
• NCERT textbook Exercise 3.3 for transpose, symmetric and skew-symmetric matrices.