Alternative English Year - 2017 Full Marks : 100
Chapter 3 Matrices Class 12 Ncert Solutions
Exercise 3.1
1. In the matrix A=\( \begin{bmatrix} 2&5&19&-7\\ 35&-2&\frac{5}{2}&12\\\sqrt3&1&-5&17\\ \end{bmatrix} \), write : (i) The order of the matrix, (ii)The number of elements,
(iii) Write the elements \(a_{13},a_{21},a_{33},a_{24},a_{23}.\)
Sol. (i) The given matrix have 3rows and 4 columns, so the order of the matrix is \(3\times4\)
(ii)The number of elements is 12 .
(iii)\(a_{13}=19,a_{21}=35,a_{33}=-5,a_{24}=12,a_{23}=\frac{5}{2}.\)
2. If a matrix has 24 elements, what are the possible orders it can have?What if it has 13 elemnts?
Sol. If a matrix have 24 elements then we can have matrix with:
(i)1 row and 24 columns i.e. order=\(1\times24\)
(ii)2 rows and 12 columns i.e. order= \(2\times12\)
(iii) 3 rows and 8 columns i.e. order =\(3\times8\)
(iv) 4 rows and 6 columns i.e. order =\(4\times6\)
(v) 6 rows and 4 columns i.e. order=\(6\times4\)
(vi) 8 rows and 3 columns i.e. order =\(8\times4\)
(vii) 12 rows and 2 columns i.e. order =\(12\times2\)
(viii) 24 rows ans 1 columns i.e. order=\(24\times1\)
Similarly, if the matrix has 13 elements it can only have two possible orders \(1\times 13\) and \(13\times1 \)
3. If a matrix has 18 elements, what are the possible orders it can have? What if it has 5 elements?
Sol. If a matrix have 18 elements then we can have matrix with:
(i)1 row and 18 columns i.e. order=\(1\times18\)
(ii)2 rows and 9 columns i.e. order= \(2\times9\)
(iii) 3 rows and 6 columns i.e. order =\(3\times6\)
(iv) 6 rows and 3 columns i.e. order =\(6\times3\)
(v) 9 rows and 2 columns i.e. order=\(9\times2\)
(vi) 18 rows and 1 columns i.e. order =\(18\times1\)
Similarly, if it has 5 elements it can only have two possible orders \(1\times 5\) and \(5\times1 \). Since 5 is a prime number and it has only 2 factors, 1 and 5.
4. Construct a \(2\times2\) matrix, A=[\(a_{ij}\)], whose elemnts are given by:
(i) \((a_{ij})={(i+j)^2\over2}\) (ii) \((a_{ij})=\frac{i}{j}\) (iii) \((a_{ij})={(i+2j)^2\over2}\)
Sol. To construct a \(2\times2\) matrix we need the elements \(a_{11},a_{12},a_{21},a_{22}\).Since a \(2\times2\) matrix A is given as \[A= \begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\\end{bmatrix}\]
(i) \((a_{ij})={(i+j)^2\over2}\)
Therefore, \((a_{11})={(1+1)^2\over2}=\frac{4}{2}=2\)
\((a_{12})={(1+2)^2\over2}=\frac{9}{2}\)
\((a_{21})={(2+1)^2\over2}=\frac{9}{2}\)
\((a_{22})={(2+2)^2\over2}=\frac{16}{2}=8\)
Hence required matrix is
\[A= \begin{bmatrix}2&\frac{9}{2}\\ \frac{9}{2}&8\\\end{bmatrix}\]
(ii)\((a_{ij})=\frac{i}{j}\)
Therefore, \((a_{11})={1\over1}=1\)
\((a_{12})={1\over2}=\frac{1}{2}\)
\((a_{21})={2\over1}=2\)
\((a_{22})={2\over2}=1\)
Hence required matrix is
\[A= \begin{bmatrix}1&\frac{1}{2}\\ 2&1\\\end{bmatrix}\]
(iii)\((a_{ij})={(i+2j)^2\over2}\)
Therefore, \((a_{11})={(1+2\times1)^2\over2}=\frac{3^2}{2}=\frac{9}{2}\)
\((a_{12})={(1+2\times2)^2\over2}=\frac{5^2}{2}=\frac{25}{2}\)
\((a_{21})={(2+2\times1)^2\over1}=\frac{4^2}{1}=8\)
\((a_{22})={(2+2\times2)^2\over2}=\frac{6^2}{2}=\frac{36}{2}=18\)
Hence required matrix is
\[A= \begin{bmatrix}\frac{9}{2}&\frac{25}{2}\\ 8&18\\\end{bmatrix}\]
\[A= \begin{bmatrix}\frac{9}{2}&\frac{25}{2}\\ 8&18\\\end{bmatrix}\]
5. Construct a \(3\times4\) matrix, A=[\(a_{ij}\)], whose elemnts are given by:
(i) \((a_{ij})={1\over2}\lvert-3i+j\rvert \) (ii) \((a_{ij})=2i-j\)
To construct a \(3\times4\) matrix we need the elements \(a_{11},a_{12},a_{13},a_{14},a_{21},a_{22},a_{23},a_{24},a_{31},a_{32},a_{33},a_{34},\).
Since a \(3\times4\) matrix A is given as
\[A= \begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\a_{31}&a_{32}&a_{33}&a_{34}\\\end{bmatrix}\]
(i) \((a_{ij})={1\over2}\lvert-3i+j\rvert\)
Therefore,\((a_{11})={1\over2}\lvert-3\times1+1\rvert\) = \({1\over2}\lvert-2\rvert\)=\(\frac{1}{2}\times2=1\)
\((a_{12})={1\over2}\lvert-3\times1+2\rvert\) =\({1\over2}\lvert-1\rvert\)=\(\frac{1}{2}\times1=\frac{1}{2}\)
\((a_{13})={1\over2}\lvert-3\times1+3\rvert\) =\({1\over2}\lvert0\rvert\)=\(\frac{1}{2}\times0=0\)
\((a_{14})={1\over2}\lvert-3\times1+4\rvert\) =\({1\over2}\lvert1\rvert\)=\(\frac{1}{2}\times1=\frac{1}{2}\)
\((a_{21})={1\over2}\lvert-3\times2+1\rvert\) =\({1\over2}\lvert-5\rvert\)=\(\frac{1}{2}\times5=\frac{5}{2}\)
\((a_{22})={1\over2}\lvert-3\times2+2\rvert\) =\({1\over2}\lvert-4\rvert\)=\(\frac{1}{2}\times4=2\)
\((a_{23})={1\over2}\lvert-3\times2+3\rvert\) =\({1\over2}\lvert-3\rvert\)=\(\frac{1}{2}\times3=\frac{3}{2}\)
\((a_{24})={1\over2}\lvert-3\times2+4\rvert\) =\({1\over2}\lvert-2\rvert\)=\(\frac{1}{2}\times2=\frac{2}{2}=1\)
\((a_{31})={1\over2}\lvert-3\times3+1\rvert\) =\({1\over2}\lvert-8\rvert\)=\(\frac{1}{2}\times8=\frac{8}{2}=4\)
\((a_{32})={1\over2}\lvert-3\times3+2\rvert\) =\({1\over2}\lvert-7\rvert\)=\(\frac{1}{2}\times7=\frac{7}{2}\)
\((a_{33})={1\over2}\lvert-3\times3+3\rvert\) =\({1\over2}\lvert-6\rvert\)=\(\frac{1}{2}\times6=\frac{6}{2}=3\)
\((a_{34})={1\over2}\lvert-3\times3+4\rvert\) =\({1\over2}\lvert-5\rvert\)=\(\frac{1}{2}\times5=\frac{5}{2}\)
Hence the required matrix is
\[A= \begin{bmatrix}1&\frac{1}{2}&0&\frac{1}{2}\\ \frac{5}{2}&2&\frac{3}{2}&1\\4&\frac{7}{2}&3&\frac{5}{2}\\\end{bmatrix}\]
(ii) \((a_{ij})=2i-j\)
Therefore, \((a_{11})=2\times1-2=1\)
\((a_{12})=2\times1-2=0\)
(i) \((a_{ij})={1\over2}\lvert-3i+j\rvert \) (ii) \((a_{ij})=2i-j\)
To construct a \(3\times4\) matrix we need the elements \(a_{11},a_{12},a_{13},a_{14},a_{21},a_{22},a_{23},a_{24},a_{31},a_{32},a_{33},a_{34},\).
Since a \(3\times4\) matrix A is given as
\[A= \begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\a_{31}&a_{32}&a_{33}&a_{34}\\\end{bmatrix}\]
(i) \((a_{ij})={1\over2}\lvert-3i+j\rvert\)
Therefore,\((a_{11})={1\over2}\lvert-3\times1+1\rvert\) = \({1\over2}\lvert-2\rvert\)=\(\frac{1}{2}\times2=1\)
\((a_{12})={1\over2}\lvert-3\times1+2\rvert\) =\({1\over2}\lvert-1\rvert\)=\(\frac{1}{2}\times1=\frac{1}{2}\)
\((a_{13})={1\over2}\lvert-3\times1+3\rvert\) =\({1\over2}\lvert0\rvert\)=\(\frac{1}{2}\times0=0\)
\((a_{14})={1\over2}\lvert-3\times1+4\rvert\) =\({1\over2}\lvert1\rvert\)=\(\frac{1}{2}\times1=\frac{1}{2}\)
\((a_{21})={1\over2}\lvert-3\times2+1\rvert\) =\({1\over2}\lvert-5\rvert\)=\(\frac{1}{2}\times5=\frac{5}{2}\)
\((a_{22})={1\over2}\lvert-3\times2+2\rvert\) =\({1\over2}\lvert-4\rvert\)=\(\frac{1}{2}\times4=2\)
\((a_{23})={1\over2}\lvert-3\times2+3\rvert\) =\({1\over2}\lvert-3\rvert\)=\(\frac{1}{2}\times3=\frac{3}{2}\)
\((a_{24})={1\over2}\lvert-3\times2+4\rvert\) =\({1\over2}\lvert-2\rvert\)=\(\frac{1}{2}\times2=\frac{2}{2}=1\)
\((a_{31})={1\over2}\lvert-3\times3+1\rvert\) =\({1\over2}\lvert-8\rvert\)=\(\frac{1}{2}\times8=\frac{8}{2}=4\)
\((a_{32})={1\over2}\lvert-3\times3+2\rvert\) =\({1\over2}\lvert-7\rvert\)=\(\frac{1}{2}\times7=\frac{7}{2}\)
\((a_{33})={1\over2}\lvert-3\times3+3\rvert\) =\({1\over2}\lvert-6\rvert\)=\(\frac{1}{2}\times6=\frac{6}{2}=3\)
\((a_{34})={1\over2}\lvert-3\times3+4\rvert\) =\({1\over2}\lvert-5\rvert\)=\(\frac{1}{2}\times5=\frac{5}{2}\)
Hence the required matrix is
\[A= \begin{bmatrix}1&\frac{1}{2}&0&\frac{1}{2}\\ \frac{5}{2}&2&\frac{3}{2}&1\\4&\frac{7}{2}&3&\frac{5}{2}\\\end{bmatrix}\]
(ii) \((a_{ij})=2i-j\)
Therefore, \((a_{11})=2\times1-2=1\)
\((a_{12})=2\times1-2=0\)
\((a_{13})=2\times1-3=-1\)
\((a_{14})=2\times1-4=-2\)
\((a_{21})=2\times2-1=3\)
\((a_{14})=2\times1-4=-2\)
\((a_{21})=2\times2-1=3\)
\((a_{22})=2\times2-2=2\)
\((a_{23})=2\times2-3=1\)
\((a_{24})=2\times2-4=0\)
\((a_{31})=2\times3-1=5\)
\((a_{32})=2\times3-2=4\)
\((a_{33})=2\times3-3=3\)
\((a_{34})=2\times3-4=2\)
Hence the required matrix is
\[A= \begin{bmatrix}1&0&-1&-2\\ 3&2&1&0\\5&4&3&2\\ \end{bmatrix}\]
6. Find the values of x,y and z from the following equations:
(i)\(\begin{bmatrix}4&3\\x&5\\ \end{bmatrix}=\begin{bmatrix} y&z\\1&5\\ \end{bmatrix}\) (ii) \(\begin{bmatrix}x+y&2\\5+z&xy\\ \end{bmatrix}==\begin{bmatrix}6&2\\5&8\\ \end{bmatrix}\) (iii) \(\begin{bmatrix}x+y+z\\x+z\\y+z\\ \end{bmatrix}=\begin{bmatrix}9\\5\\7\\ \end{bmatrix}\)
Sol. (i) By equality of two matrices, equating the corresponding element we get
4=y 3=z and x=1
(ii) Equating the corresponding element by the equality of matrices we have
\begin{align}
x+y =6 &&\text{--------------}\tag1\\
5+z=5 &&\text{--------------}\tag2\\
xy=8 &&\text{--------------}\tag3\\
\end{align}
From (2) we get z=0
From (1) we get y=6-x
putting the value of y in (3) we get
\begin{align}
x(6-x)=8\\
6x-x^2=8\\
x^2-6x+8=0\\
x^2 -4x-2x+8=0\\
x(x-4)-2(x-4)=0\\
(x-2)(x-4)=0\\
\implies x=4\\or, x=2\\
\end{align}
If x=4 , y=2 and if x=2 y =4 .
So we have two possible solutions
\[A= \begin{bmatrix}1&0&-1&-2\\ 3&2&1&0\\5&4&3&2\\ \end{bmatrix}\]
6. Find the values of x,y and z from the following equations:
(i)\(\begin{bmatrix}4&3\\x&5\\ \end{bmatrix}=\begin{bmatrix} y&z\\1&5\\ \end{bmatrix}\) (ii) \(\begin{bmatrix}x+y&2\\5+z&xy\\ \end{bmatrix}==\begin{bmatrix}6&2\\5&8\\ \end{bmatrix}\) (iii) \(\begin{bmatrix}x+y+z\\x+z\\y+z\\ \end{bmatrix}=\begin{bmatrix}9\\5\\7\\ \end{bmatrix}\)
Sol. (i) By equality of two matrices, equating the corresponding element we get
4=y 3=z and x=1
(ii) Equating the corresponding element by the equality of matrices we have
\begin{align}
x+y =6 &&\text{--------------}\tag1\\
5+z=5 &&\text{--------------}\tag2\\
xy=8 &&\text{--------------}\tag3\\
\end{align}
From (2) we get z=0
From (1) we get y=6-x
putting the value of y in (3) we get
\begin{align}
x(6-x)=8\\
6x-x^2=8\\
x^2-6x+8=0\\
x^2 -4x-2x+8=0\\
x(x-4)-2(x-4)=0\\
(x-2)(x-4)=0\\
\implies x=4\\or, x=2\\
\end{align}
If x=4 , y=2 and if x=2 y =4 .
So we have two possible solutions
x=4, y=2, z=0 or x=2, y=4, z=0
(iii) Equating the corresponding elements by the equality of matrices we have
\begin{align}
x+y+z=9&&\tag1\\
x+z=5&&\tag2\\
y+z=7&&\tag3\\
\end{align}
Subtracting (2) from (1) we get y= 4
Subtracting (3) from (1) we get x=2
Putting the value of y in (3) we get z=3.
7. Find the value of a,b,c and d from the equation:
\[\begin{bmatrix}a-b&2a+c\\2a-b&3c+d\\ \end{bmatrix}=\begin{bmatrix} -1&5\\0&13\\ \end{bmatrix}\]
Sol. Equating the corresponding elements by the equality of matrices we get
\begin{align}
a-b=-1&&\tag1\\
2a+c=5&&\tag2\\
2a-b=0&&\tag3\\
3c+d=13&&\tag4\\
\end{align}
From (3) we get 2a=b. Putting this value of b in (1) we get a-2a =-1 \(\implies\)-a = -1\(\implies\) a = 1. So b=2a=2\(\times1\)=2.
Putting a = 1 in (2) we get 2+c = 5\(\implies\) c = 5-2\(\implies\) c=3.
Putting c =3 in (4) we get 9 + d = 13\(\implies\) d=4
Hence a = 1, b = 2, c = 3, d = 4.
8. A=\([a_{ij}]_{m\timesn}\) is a square matrix if
(A) m\(\lt\)n (B) m\(\gt\)n (C) m = n (D) None of these
Sol. A aquare matrix has equal no. of rows and columns . m is the no of rows and n is the no. of columns .So, m = n . The correct anser is option (C)
9. Which of the given values of x and y make the following pair of matrices equal
\(\begin{bmatrix}3x + 7&5\\y+1&2-3x\\ \end{bmatrix}, \begin{bmatrix} 0&y-2\\8&4\\ \end{bmatrix}\)
(A) x = \(frac{-1}{3}\), y = 7 (B) Not possible to find (C) y = 7, x = \(\frac{-2}{3}\) (D) \(x=\frac{-1}{3}, y = \frac{-2}{3}\)
Sol. Equating corresponding elements by the equality of matrices we get y=7 but for x we get two different values at the same time x=\(\frac{-7}{3}\) ans x= \(\frac{2}{3}\) which is absurd . So the correct option is (B)
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
Sol. For a 3\(\times\)3 we have a total of 9 elements . Each element can have two entries either 0 or 1. So there is two ways of filling one element. Hence, total no. of ways of filling 9 elements is \(2^9=512 ways\). So, 512 different matrices are possible. The correct answer is option (D).
x+y+z=9&&\tag1\\
x+z=5&&\tag2\\
y+z=7&&\tag3\\
\end{align}
Subtracting (2) from (1) we get y= 4
Subtracting (3) from (1) we get x=2
Putting the value of y in (3) we get z=3.
7. Find the value of a,b,c and d from the equation:
\[\begin{bmatrix}a-b&2a+c\\2a-b&3c+d\\ \end{bmatrix}=\begin{bmatrix} -1&5\\0&13\\ \end{bmatrix}\]
Sol. Equating the corresponding elements by the equality of matrices we get
\begin{align}
a-b=-1&&\tag1\\
2a+c=5&&\tag2\\
2a-b=0&&\tag3\\
3c+d=13&&\tag4\\
\end{align}
From (3) we get 2a=b. Putting this value of b in (1) we get a-2a =-1 \(\implies\)-a = -1\(\implies\) a = 1. So b=2a=2\(\times1\)=2.
Putting a = 1 in (2) we get 2+c = 5\(\implies\) c = 5-2\(\implies\) c=3.
Putting c =3 in (4) we get 9 + d = 13\(\implies\) d=4
Hence a = 1, b = 2, c = 3, d = 4.
8. A=\([a_{ij}]_{m\timesn}\) is a square matrix if
(A) m\(\lt\)n (B) m\(\gt\)n (C) m = n (D) None of these
Sol. A aquare matrix has equal no. of rows and columns . m is the no of rows and n is the no. of columns .So, m = n . The correct anser is option (C)
9. Which of the given values of x and y make the following pair of matrices equal
\(\begin{bmatrix}3x + 7&5\\y+1&2-3x\\ \end{bmatrix}, \begin{bmatrix} 0&y-2\\8&4\\ \end{bmatrix}\)
(A) x = \(frac{-1}{3}\), y = 7 (B) Not possible to find (C) y = 7, x = \(\frac{-2}{3}\) (D) \(x=\frac{-1}{3}, y = \frac{-2}{3}\)
Sol. Equating corresponding elements by the equality of matrices we get y=7 but for x we get two different values at the same time x=\(\frac{-7}{3}\) ans x= \(\frac{2}{3}\) which is absurd . So the correct option is (B)
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
Sol. For a 3\(\times\)3 we have a total of 9 elements . Each element can have two entries either 0 or 1. So there is two ways of filling one element. Hence, total no. of ways of filling 9 elements is \(2^9=512 ways\). So, 512 different matrices are possible. The correct answer is option (D).
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