# NCERT Class 12 Maths Matrices

## Miscellaneous Exercise

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## Chapter 3 Ex.3.1 Question 1

In the matrix $$A = \left( {\begin{array}{*{20}{c}}2&5&{19}&{ - 7}\\{35}&{ - 2}&{\frac{5}{2}}&{12}\\{\sqrt 3 }&1&{ - 5}&{17}\end{array}} \right)$$, write:

(i) The order of the matrix

(ii) The number of elements

(iii) Write the elements $${a_{13}},{a_{21}},{a_{33}},{a_{24}},{a_{23}}$$

### Solution

(i) Since, in the given matrix, the number of rows is $$3$$and the number of columns is $$4$$, the order of the matrix is $$3 \times 4$$.

(ii) Since the order of the matrix is $$3 \times 4$$, there are $$3 \times 4 = 12$$elements.

(iii) Here,

\begin{align}{a_{13}} &= 19\\{a_{21}} &= 35\\{a_{33}}& = - 5\\{a_{24}} &= 12\\{a_{23}} &= \frac{5}{2}\end{align}

## Chapter 3 Ex.3.1 Question 2

If a matrix has $$24$$ elements, what are the possible order it can have? What, if it has $$13$$ elements?

### Solution

We know that if a matrix is of the order $$m \times n$$, it has $$mn$$ elements. Thus, to find all the possible orders of a matrix having $${\rm{24}}$$ elements, we have to find all the ordered pairs of natural numbers whose product is $${\rm{24}}$$.

The ordered pairs are: $$\left( {{\rm{1}},{\rm{24}}} \right),\left( {{\rm{24}},{\rm{1}}} \right),\left( {{\rm{2}},{\rm{12}}} \right),\left( {{\rm{12}},{\rm{2}}} \right),\left( {{\rm{3,8}}} \right),\left( {{\rm{8}},{\rm{3}}} \right),\left( {{\rm{4}},{\rm{6}}} \right)$$ and $$\left( {{\rm{6,4}}} \right)$$.

Hence, the possible orders of a matrix having $${\rm{24}}$$elements are:

$$\left( {1 \times 24} \right),\;\left( {24 \times 1} \right),\;\left( {2 \times 12} \right),\;\left( {12 \times 2} \right),\;\left( {3 \times 8} \right),\;\left( {8 \times 3} \right),\;\left( {4 \times 6} \right)$$ and $$\left( {{\rm{6}} \times {\rm{4}}} \right)$$.

$$\left( {{\rm{1}},{\rm{13}}} \right)$$ and $$~\left( \text{13},\text{1} \right)$$ are the ordered pairs of natural numbers whose product is$$~\text{13}$$.

Hence, the possible orders of a matrix having $${\rm{13}}$$ elements are $$\left( \text{1}~\times ~\text{13} \right)$$ and $$\left( \text{13}~\times ~\text{1} \right)$$.

## Chapter 3 Ex.3.1 Question 3

If a matrix has $$18$$ elements, what are the possible order it can have? What, if it has $$5$$ elements?

### Solution

We know that if a matrix is of the order $$m \times n$$, it has $$mn$$ elements. Thus, to find all the possible orders of a matrix having $$18$$ elements, we have to find all the ordered pairs of natural numbers whose product is $$18$$.

The ordered pairs are: $$\left( {{\rm{1}},18} \right),\left( {18,{\rm{1}}} \right),\left( {{\rm{2}},9} \right),\left( {9,{\rm{2}}} \right),\left( {{\rm{3,6}}} \right)$$ and $$\left( {6,{\rm{3}}} \right)$$.

Hence, the possible orders of a matrix having 18 elements are:

$$\left( {1 \times 18} \right),\;\left( {18 \times 1} \right),\;\left( {2 \times 9} \right),\;\left( {9 \times 2} \right),\;\left( {3 \times 6} \right)$$ and $$\left( {6 \times 3} \right)$$.

$$\left( {1 \times 5} \right)$$ and $$~\left( 5\times 1 \right)$$ are the ordered pairs of natural numbers whose product is 5.

Hence, the possible orders of a matrix having 5 elements are $$\left( {1 \times 5} \right)$$ and $$~\left( 5\times 1 \right)$$.

## Chapter 3 Ex.3.1 Question 4

Construct a $$2 \times 2$$ matrix, $$A = \left[ {{a_{ij}}} \right]$$, whose elements are given by:

(i) $${a_{ij}} = \frac{{{{\left( {i + j} \right)}^2}}}{2}$$

(ii) $${a_{ij}} = \frac{i}{j}$$

(iii) $${a_{ij}} = \frac{{{{\left( {i + 2j} \right)}^2}}}{2}$$

### Solution

In general, a $$\text{2}~\times ~\text{2}$$ matrix is given by $$A = \left( {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{array}} \right)$$

(i) $${a_{ij}} = \frac{{{{\left( {i + j} \right)}^2}}}{2};\;\;\;i,j = 1,2$$

Therefore,

\begin{align}{a_{11}}& = \frac{{{{\left( {1 + 1} \right)}^2}}}{2} = \frac{4}{2} = 2\\{a_{12}} &= \frac{{{{\left( {1 + 2} \right)}^2}}}{2} = \frac{9}{2}\\{a_{21}} &= \frac{{{{\left( {2 + 1} \right)}^2}}}{2} = \frac{9}{2}\\{a_{22}} &= \frac{{{{\left( {2 + 2} \right)}^2}}}{2} = \frac{{16}}{2} = 8\end{align}

Thus, the required matrix is $$A = \left( {\begin{array}{*{20}{c}}2&{\frac{9}{2}}\\{\frac{9}{2}}&8\end{array}} \right)$$

(ii) $${a_{ij}} = \frac{i}{j};\;\;\;i,j = 1,2$$

Therefore,

\begin{align}{a_{11}} &= \frac{1}{1} &= 1\\{a_{12}} = \frac{1}{2}\\{a_{21}} &= \frac{2}{1} = 2\\{a_{22}} &= \frac{2}{2} = 1\end{align}

Thus, the required matrix is $$A = \left( {\begin{array}{*{20}{c}}1&{\frac{1}{2}}\\2&1\end{array}} \right)$$

(iii) $${a_{ij}} = \frac{{{{\left( {i + 2j} \right)}^2}}}{2};\;\;\;i,j = 1,2$$

Therefore,

\begin{align}{a_{11}}& = \frac{{{{\left( {1 + 2} \right)}^2}}}{2} = \frac{9}{2}\\{a_{12}} &= \frac{{{{\left( {1 + 4} \right)}^2}}}{2} = \frac{{25}}{2}\\{a_{21}} &= \frac{{{{\left( {2 + 2} \right)}^2}}}{2} = 8\\{a_{22}} &= \frac{{{{\left( {2 + 4} \right)}^2}}}{2} = 18\end{align}

Thus, the required matrix is $$A = \left( {\begin{array}{*{20}{c}}{\frac{9}{2}}&{\frac{{25}}{2}}\\8&{18}\end{array}} \right)$$

## Chapter 3 Ex.3.1 Question 5

In general, a $$\text{3}\times ~4$$ matrix whose elements are given by

(i) $${a_{ij}} = \frac{1}{2}\left| { - 3i + j} \right|$$

(ii) $${a_{ij}} = 2i - j$$

### Solution

In general, a $$3 \times 4$$ matrix is given by $$A = \left( {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}&{{a_{24}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}&{{a_{34}}}\end{array}} \right)$$

(i) Given $${a_{ij}} = \frac{1}{2}\left| { - 3i + j} \right|;\;\;i = 1,2,3\;\;j = 1,2,3,4$$

\begin{align}{a_{11}} &= \frac{1}{2}\left| { - 3\left( 1 \right) + 1} \right| = \frac{1}{2}\left| { - 3 + 1} \right| = \frac{1}{2}\left| { - 2} \right| = \frac{2}{2} = 1\\{a_{21}} &= \frac{1}{2}\left| { - 3\left( 2 \right) + 1} \right| = \frac{1}{2}\left| { - 6 + 1} \right| = \frac{1}{2}\left| { - 5} \right| = \frac{5}{2}\\{a_{31}} &= \frac{1}{2}\left| { - 3\left( 3 \right) + 1} \right| = \frac{1}{2}\left| { - 9 + 1} \right| = \frac{1}{2}\left| { - 8} \right| = \frac{8}{2} = 4\end{align}

\begin{align}{a_{12}} &= \frac{1}{2}\left| { - 3\left( 1 \right) + 2} \right| = \frac{1}{2}\left| { - 3 + 2} \right| = \frac{1}{2}\left| { - 1} \right| = \frac{1}{2}\\{a_{22}} &= \frac{1}{2}\left| { - 3\left( 2 \right) + 2} \right| = \frac{1}{2}\left| { - 6 + 2} \right| = \frac{1}{2}\left| { - 4} \right| = \frac{4}{2} = 2\\{a_{32}} &= \frac{1}{2}\left| { - 3\left( 3 \right) + 2} \right| = \frac{1}{2}\left| { - 9 + 2} \right| = \frac{1}{2}\left| { - 7} \right| = \frac{7}{2}\end{align}

\begin{align}{a_{13}} &= \frac{1}{2}\left| { - 3\left( 1 \right) + 3} \right| = \frac{1}{2}\left| { - 3 + 3} \right| = 0\\{a_{23}} &= \frac{1}{2}\left| { - 3\left( 2 \right) + 3} \right| = \frac{1}{2}\left| { - 6 + 3} \right| = \frac{1}{2}\left| { - 3} \right| = \frac{3}{2}\\{a_{33}} &= \frac{1}{2}\left| { - 3\left( 3 \right) + 3} \right| = \frac{1}{2}\left| { - 9 + 3} \right| = \frac{1}{2}\left| { - 6} \right| = \frac{6}{2} = 3\end{align}

\begin{align}{a_{14}} &= \frac{1}{2}\left| { - 3\left( 1 \right) + 4} \right| = \frac{1}{2}\left| { - 3 + 4} \right| = \frac{1}{2}\left| 1 \right| = \frac{1}{2}\\{a_{24}}& = \frac{1}{2}\left| { - 3\left( 2 \right) + 4} \right| = \frac{1}{2}\left| { - 6 + 4} \right| = \frac{1}{2}\left| { - 2} \right| = \frac{2}{2} = 1\\{a_{34}} &= \frac{1}{2}\left| { - 3\left( 3 \right) + 4} \right| = \frac{1}{2}\left| { - 9 + 4} \right| = \frac{1}{2}\left| { - 5} \right| = \frac{5}{2}\end{align}

Thus, the required matrix is $$A = \left( {\begin{array}{*{20}{c}}1&{\frac{1}{2}}&0&{\frac{1}{2}}\\{\frac{5}{2}}&2&{\frac{3}{2}}&1\\4&{\frac{7}{2}}&3&{\frac{5}{2}}\end{array}} \right)$$

(ii) $${a_{ij}} = 2i - j;\;\;i = 1,2,3\;\;j = 1,2,3,4$$

\begin{align}{a_{11}} &= 2\left( 1 \right) - 1 = 2 - 1 = 1\\{a_{21}} &= 2\left( 2 \right) - 1 = 4 - 1 = 3\\{a_{31}}& = 2\left( 3 \right) - 1 = 6 - 1 = 5\end{align}

\begin{align}{a_{12}} &= 2\left( 1 \right) - 2 = 2 - 2 = 0\\{a_{22}} &= 2\left( 2 \right) - 2 = 4 - 2 = 2\\{a_{32}} &= 2\left( 3 \right) - 2 = 6 - 2 = 4\end{align}

\begin{align}{a_{13}} &= 2\left( 1 \right) - 3 = 2 - 3 = - 1\\{a_{23}}& = 2\left( 2 \right) - 3 = 4 - 3 = 1\\{a_{33}} &= 2\left( 3 \right) - 3 = 6 - 3 = 3\end{align}

\begin{align}{a_{14}}& = 2\left( 1 \right) - 4 = 2 - 4 = - 2\\{a_{24}} &= 2\left( 2 \right) - 4 = 4 - 4 = 0\\{a_{34}} &= 2\left( 3 \right) - 4 = 6 - 4 = 2\end{align}

Thus, the required matrix is $$A = \left( {\begin{array}{*{20}{c}}1&0&{ - 1}&{ - 2}\\3&2&1&0\\5&4&3&2\end{array}} \right)$$

## Chapter 3 Ex.3.1 Question 6

Find the value of $$x, y$$ and $$z$$ from the following equation:

(i) $$\left( {\begin{array}{*{20}{c}}4&3\\x&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}y&z\\1&5\end{array}} \right)$$

(ii) $$\left( {\begin{array}{*{20}{c}}{x + y}&2\\{5 + z}&{xy}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6&2\\5&8\end{array}} \right)$$

(iii) $$\left( {\begin{array}{*{20}{c}}{x + y + z}\\{x + z}\\{y + z}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}9\\5\\7\end{array}} \right)$$

### Solution

(i) $$\left( {\begin{array}{*{20}{c}}4&3\\x&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}y&z\\1&5\end{array}} \right)$$

As the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

$$x = 1,\;y = 4$$ and $$z = 3$$

(ii) $$\left( {\begin{array}{*{20}{c}}{x + y}&2\\{5 + z}&{xy}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6&2\\5&8\end{array}} \right)$$

As the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

\begin{align}x + y &= 6\\xy &= 8\\5 + z &= 5\end{align}

Hence,

\begin{align}&\Rightarrow\; 5 + z = 5\\& \Rightarrow\; z = 0\end{align}

We know that $${\left( {a - b} \right)^2} = {\left( {a + b} \right)^2} - 4ab$$

\begin{align}&\Rightarrow \;{\left( {x - y} \right)^2} = {\left( 6 \right)^2} - 8 \times 4\\ &\Rightarrow\; {\left( {x - y} \right)^2} = 36 - 32\\& \Rightarrow\; {\left( {x - y} \right)^2} = 4\\& \Rightarrow \;\left( {x - y} \right) = \pm 2\end{align}

Equating $$x - y = 2$$ and $$x + y = 6$$, we get $$x = 4,\;y = 2$$

Similarly, Equating $$x - y = - 2$$ and $$x + y = 6$$, we get $$x = 2,\;y = 4$$

Thus, $$x = 4,\;y = 2,\;z = 0$$ or $$x = 2,\;y = 4,\;z = 0$$

(iii)$$\left( {\begin{array}{*{20}{c}}{x + y + z}\\{x + z}\\{y + z}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}9\\5\\7\end{array}} \right)$$

As the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

\begin{align}x + y + z &= 9\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\x + z &= 5\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\y + z &= 7\;\;\;\;\;\;\;\;\;\;\;\ldots \left( 3 \right)\end{align}

From $$\left( 1 \right)$$ and$$\left( 2 \right)$$, we have

\begin{align} &\Rightarrow y + 5 = 9\\& \Rightarrow y = 4\end{align}

From $$\left( 3 \right)$$, we have

\begin{align} &\Rightarrow 4 + z = 7\\& \Rightarrow z = 3\end{align}

Therefore,

\begin{align} &\Rightarrow x + z = 5\\& \Rightarrow x + 3 = 5\\ &\Rightarrow x = 2\end{align}

Thus, $$x = 2,\;y = 4,\;z = 3$$

## Chapter 3 Ex.3.1 Question 7

Find the value of $$a,~b,c$$ and $$d$$ from the equation:

$$\left( {\begin{array}{*{20}{c}}{a - b}&{2a + c}\\{2a - b}&{3c + d}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 1}&5\\0&{13}\end{array}} \right)$$

### Solution

$$\left( {\begin{array}{*{20}{c}}{a - b}&{2a + c}\\{2a - b}&{3c + d}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 1}&5\\0&{13}\end{array}} \right)$$

As the two matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

\begin{align}a - b = - 1\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\2a - b = 0\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\2a + c = 5\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\\3c + d = 13\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\end{align}

From $$\left( 2 \right)$$,

$$b = 2a$$

Putting this value in $$\left( 1 \right)$$,

\begin{align} &\Rightarrow a - 2a = - 1\\ &\Rightarrow a = 1\end{align}

Hence,

$$a = 1$$

Putting in $$\left( 3 \right)$$,

\begin{align} &\Rightarrow 2\left( 1 \right) + c = 5\\& \Rightarrow c = 3\end{align}

Putting $$c = 3$$ in $$\left( 4 \right)$$,

\begin{align}& \Rightarrow 3\left( 3 \right) + d = 13\\& \Rightarrow d = 4\end{align}

Thus, $$a = 1,\;b = 2,\;c = 3$$ and $$d = 4$$.

## Chapter 3 Ex.3.1 Question 8

$$A = {\left[ {{a_{ij}}} \right]_{m \times n}}$$ is a square matrix, if

(A) $$m < n$$

(B) $$m > n$$

(C) $$m = n$$

(D) None of these

### Solution

It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns.

Therefore, $$A = {\left[ {{a_{ij}}} \right]_{m \times n}}$$ is a square matrix, if $$m = n$$.

Thus, the correct option is C.

## Chapter 3 Ex.3.1 Question 9

Which of the given values of $$x$$ and $$y$$ make the following pair of matrices equal

$\left[ {\begin{array}{*{20}{c}}{3x + 7}&5\\{y + 1}&{2 - 3x}\end{array}} \right],\;\left[ {\begin{array}{*{20}{c}}0&{y - 2}\\8&4\end{array}} \right]$

(A) $$x = \frac{{ - 1}}{3},y = 7$$

(B) Not possible to find

(C) $$y = 7,x = \frac{{ - 2}}{3}$$

(D) $$x = \frac{{ - 1}}{3},y = \frac{{ - 2}}{3}$$

### Solution

The given matrices are $$\left[ {\begin{array}{*{20}{c}}{3x + 7}&5\\{y + 1}&{2 - 3x}\end{array}} \right]$$ and $$\left[ {\begin{array}{*{20}{c}}0&{y - 2}\\8&4\end{array}} \right]$$

Equating the corresponding elements, we get:

\begin{align}&3x + 7 = 0 \Rightarrow x = \frac{{ - 7}}{3}\\&y - 2 = 5 \Rightarrow y = 7\\&y + 1 = 8 \Rightarrow y = 7\\&2 - 3x = 4 \Rightarrow x = \frac{{ - 2}}{3}\end{align}

We find that on comparing the corresponding elements of the two matrices, we get two different values of x, which is not possible.

Hence, it is not possible to find the values of $$x$$ and $$y$$ for which the given matrices are equal.

Thus, the correct option is $$B$$.

## Chapter 3 Ex.3.1 Question 10

The number of all possible matrices of order $$3 \times 3$$ with each entry $$0$$ or $$1$$ is:

(A) $$27$$

(B) $$18$$

(C) $$81$$

(D) $$512$$

### Solution

The given matrix of the order $$3 \times 3$$has $$9$$elements and each of these elements can be either 0 or 1.

Now, each of the $$9$$ elements can be filled in two possible ways.

Hence, by the multiplication principle, the required number of possible matrices is $${{2}^{9}}~=512$$.

Thus, the correct option is D.

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