Class 12 NCERT Solutions
Chapter 3: Matrices
Master matrix operations, symmetric properties, and elementary transformations for solving linear systems with our step-by-step logic.
Exercise 3.1
1. In the matrix \(A=\left[\begin{array}{cccc}2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17\end{array}\right]\), write:
(i) The order of the matrix
(ii) The number of elements
(iii) Write the elements \(a_{13}, a_{21}, a_{33}, a_{24}, a_{23}\)
(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) The matrix has 3 rows and 4 columns, so the order is \(3 \times 4\).
(ii) Number of elements \(= 3 \times 4 = 12\).
(iii) \(a_{13}=19,\ a_{21}=35,\ a_{33}=-5,\ a_{24}=12,\ a_{23}=\dfrac{5}{2}\)
(i) The matrix has 3 rows and 4 columns, so the order is \(3 \times 4\).
(ii) Number of elements \(= 3 \times 4 = 12\).
(iii) \(a_{13}=19,\ a_{21}=35,\ a_{33}=-5,\ a_{24}=12,\ a_{23}=\dfrac{5}{2}\)
2. If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
Solution:
A matrix of order \(m \times n\) has \(mn\) elements. For 24 elements, all ordered pairs \((m,n)\) with \(mn=24\) are:
\((1,24),(24,1),(2,12),(12,2),(3,8),(8,3),(4,6),(6,4)\)
Hence possible orders: \(1\times24,\ 24\times1,\ 2\times12,\ 12\times2,\ 3\times8,\ 8\times3,\ 4\times6,\ 6\times4\).
Since 13 is prime, the only pairs are \((1,13)\) and \((13,1)\).
Hence possible orders: \(1\times13\) and \(13\times1\).
A matrix of order \(m \times n\) has \(mn\) elements. For 24 elements, all ordered pairs \((m,n)\) with \(mn=24\) are:
\((1,24),(24,1),(2,12),(12,2),(3,8),(8,3),(4,6),(6,4)\)
Hence possible orders: \(1\times24,\ 24\times1,\ 2\times12,\ 12\times2,\ 3\times8,\ 8\times3,\ 4\times6,\ 6\times4\).
Since 13 is prime, the only pairs are \((1,13)\) and \((13,1)\).
Hence possible orders: \(1\times13\) and \(13\times1\).
3. If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
Solution:
For 18 elements, ordered pairs \((m,n)\) with \(mn=18\):
\((1,18),(18,1),(2,9),(9,2),(3,6),(6,3)\)
Hence possible orders: \(1\times18,\ 18\times1,\ 2\times9,\ 9\times2,\ 3\times6,\ 6\times3\).
Since 5 is prime, the only pairs are \((1,5)\) and \((5,1)\).
Hence possible orders: \(1\times5\) and \(5\times1\).
For 18 elements, ordered pairs \((m,n)\) with \(mn=18\):
\((1,18),(18,1),(2,9),(9,2),(3,6),(6,3)\)
Hence possible orders: \(1\times18,\ 18\times1,\ 2\times9,\ 9\times2,\ 3\times6,\ 6\times3\).
Since 5 is prime, the only pairs are \((1,5)\) and \((5,1)\).
Hence possible orders: \(1\times5\) and \(5\times1\).
4. Construct a \(2 \times 2\) matrix \(A=[a_{ij}]\), whose elements are given by
(i) \(a_{ij}=\dfrac{(i+j)^2}{2}\)
(ii) \(a_{ij}=\dfrac{i}{j}\)
(iii) \(a_{ij}=\dfrac{(i+2j)^2}{2}\)
(i) \(a_{ij}=\dfrac{(i+j)^2}{2}\) (ii) \(a_{ij}=\dfrac{i}{j}\) (iii) \(a_{ij}=\dfrac{(i+2j)^2}{2}\)
Solution:
In general, \(A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\).
(i) \(a_{ij}=\dfrac{(i+j)^2}{2}\):
\(a_{11}=\dfrac{4}{2}=2,\quad a_{12}=\dfrac{9}{2},\quad a_{21}=\dfrac{9}{2},\quad a_{22}=\dfrac{16}{2}=8\)
\(\therefore A=\begin{bmatrix}2&\frac{9}{2}\\\frac{9}{2}&8\end{bmatrix}\)
(ii) \(a_{ij}=\dfrac{i}{j}\):
\(a_{11}=1,\quad a_{12}=\dfrac{1}{2},\quad a_{21}=2,\quad a_{22}=1\)
\(\therefore A=\begin{bmatrix}1&\frac{1}{2}\\2&1\end{bmatrix}\)
(iii) \(a_{ij}=\dfrac{(i+2j)^2}{2}\):
\(a_{11}=\dfrac{9}{2},\quad a_{12}=\dfrac{25}{2},\quad a_{21}=8,\quad a_{22}=18\)
\(\therefore A=\begin{bmatrix}\frac{9}{2}&\frac{25}{2}\\8&18\end{bmatrix}\)
In general, \(A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\).
(i) \(a_{ij}=\dfrac{(i+j)^2}{2}\):
\(a_{11}=\dfrac{4}{2}=2,\quad a_{12}=\dfrac{9}{2},\quad a_{21}=\dfrac{9}{2},\quad a_{22}=\dfrac{16}{2}=8\)
\(\therefore A=\begin{bmatrix}2&\frac{9}{2}\\\frac{9}{2}&8\end{bmatrix}\)
(ii) \(a_{ij}=\dfrac{i}{j}\):
\(a_{11}=1,\quad a_{12}=\dfrac{1}{2},\quad a_{21}=2,\quad a_{22}=1\)
\(\therefore A=\begin{bmatrix}1&\frac{1}{2}\\2&1\end{bmatrix}\)
(iii) \(a_{ij}=\dfrac{(i+2j)^2}{2}\):
\(a_{11}=\dfrac{9}{2},\quad a_{12}=\dfrac{25}{2},\quad a_{21}=8,\quad a_{22}=18\)
\(\therefore A=\begin{bmatrix}\frac{9}{2}&\frac{25}{2}\\8&18\end{bmatrix}\)
5. Construct a \(3\times4\) matrix, whose elements are given by
(i) \(a_{ij}=\dfrac{1}{2}|-3i+j|\)
(ii) \(a_{ij}=2i-j\)
(i) \(a_{ij}=\dfrac{1}{2}|-3i+j|\) (ii) \(a_{ij}=2i-j\)
Solution:
General \(3\times4\) structure: \(A=\left[\begin{array}{llll}a_{11}&a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\a_{31}&a_{32}&a_{33}&a_{34}\end{array}\right]\)
(i) \(a_{ij}=\dfrac{1}{2}|-3i+j|\):
\(a_{11}=1,\ a_{12}=\dfrac{1}{2},\ a_{13}=0,\ a_{14}=\dfrac{1}{2}\)
\(a_{21}=\dfrac{5}{2},\ a_{22}=2,\ a_{23}=\dfrac{3}{2},\ a_{24}=1\)
\(a_{31}=4,\ a_{32}=\dfrac{7}{2},\ a_{33}=3,\ a_{34}=\dfrac{5}{2}\)
\(\therefore A=\left[\begin{array}{cccc}1&\frac{1}{2}&0&\frac{1}{2}\\\frac{5}{2}&2&\frac{3}{2}&1\\4&\frac{7}{2}&3&\frac{5}{2}\end{array}\right]\)
(ii) \(a_{ij}=2i-j\):
\(a_{11}=1,\ a_{12}=0,\ a_{13}=-1,\ a_{14}=-2\)
\(a_{21}=3,\ a_{22}=2,\ a_{23}=1,\ a_{24}=0\)
\(a_{31}=5,\ a_{32}=4,\ a_{33}=3,\ a_{34}=2\)
\(\therefore A=\begin{bmatrix}1&0&-1&-2\\3&2&1&0\\5&4&3&2\end{bmatrix}\)
General \(3\times4\) structure: \(A=\left[\begin{array}{llll}a_{11}&a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\a_{31}&a_{32}&a_{33}&a_{34}\end{array}\right]\)
(i) \(a_{ij}=\dfrac{1}{2}|-3i+j|\):
\(a_{11}=1,\ a_{12}=\dfrac{1}{2},\ a_{13}=0,\ a_{14}=\dfrac{1}{2}\)
\(a_{21}=\dfrac{5}{2},\ a_{22}=2,\ a_{23}=\dfrac{3}{2},\ a_{24}=1\)
\(a_{31}=4,\ a_{32}=\dfrac{7}{2},\ a_{33}=3,\ a_{34}=\dfrac{5}{2}\)
\(\therefore A=\left[\begin{array}{cccc}1&\frac{1}{2}&0&\frac{1}{2}\\\frac{5}{2}&2&\frac{3}{2}&1\\4&\frac{7}{2}&3&\frac{5}{2}\end{array}\right]\)
(ii) \(a_{ij}=2i-j\):
\(a_{11}=1,\ a_{12}=0,\ a_{13}=-1,\ a_{14}=-2\)
\(a_{21}=3,\ a_{22}=2,\ a_{23}=1,\ a_{24}=0\)
\(a_{31}=5,\ a_{32}=4,\ a_{33}=3,\ a_{34}=2\)
\(\therefore A=\begin{bmatrix}1&0&-1&-2\\3&2&1&0\\5&4&3&2\end{bmatrix}\)
6. Find the value of \(x, y\), and \(z\) from the following equations:
(i) \(\left[\begin{array}{ll}4&3\\x&5\end{array}\right]=\left[\begin{array}{ll}y&z\\1&5\end{array}\right]\)
(ii) \(\left[\begin{array}{ll}x+y&2\\5+z&xy\end{array}\right]=\left[\begin{array}{ll}6&2\\5&8\end{array}\right]\)
(iii) \(\left[\begin{array}{c}x+y+z\\x+z\\y+z\end{array}\right]=\left[\begin{array}{c}9\\5\\7\end{array}\right]\)
(i) \(\left[\begin{array}{ll}4&3\\x&5\end{array}\right]=\left[\begin{array}{ll}y&z\\1&5\end{array}\right]\)
(ii) \(\left[\begin{array}{ll}x+y&2\\5+z&xy\end{array}\right]=\left[\begin{array}{ll}6&2\\5&8\end{array}\right]\)
(iii) \(\left[\begin{array}{c}x+y+z\\x+z\\y+z\end{array}\right]=\left[\begin{array}{c}9\\5\\7\end{array}\right]\)
Solution:
(i) Matching corresponding elements: \(x=1,\ y=4,\ z=3\).
(ii) Matching elements: \(x+y=6,\ xy=8,\ 5+z=5\Rightarrow z=0\).
Using \((x-y)^2=(x+y)^2-4xy=36-32=4\Rightarrow x-y=\pm2\).
When \(x-y=2\): \(x=4,y=2\). When \(x-y=-2\): \(x=2,y=4\).
\(\therefore x=4,y=2,z=0\) or \(x=2,y=4,z=0\).
(iii) System: \(x+y+z=9\) …(1), \(x+z=5\) …(2), \(y+z=7\) …(3).
From (1) and (2): \(y+5=9\Rightarrow y=4\).
From (3): \(4+z=7\Rightarrow z=3\).
From (2): \(x=5-3=2\).
\(\therefore x=2,\ y=4,\ z=3\).
(i) Matching corresponding elements: \(x=1,\ y=4,\ z=3\).
(ii) Matching elements: \(x+y=6,\ xy=8,\ 5+z=5\Rightarrow z=0\).
Using \((x-y)^2=(x+y)^2-4xy=36-32=4\Rightarrow x-y=\pm2\).
When \(x-y=2\): \(x=4,y=2\). When \(x-y=-2\): \(x=2,y=4\).
\(\therefore x=4,y=2,z=0\) or \(x=2,y=4,z=0\).
(iii) System: \(x+y+z=9\) …(1), \(x+z=5\) …(2), \(y+z=7\) …(3).
From (1) and (2): \(y+5=9\Rightarrow y=4\).
From (3): \(4+z=7\Rightarrow z=3\).
From (2): \(x=5-3=2\).
\(\therefore x=2,\ y=4,\ z=3\).
7. Find the value of \(a,b,c\) and \(d\) from the equation:
\(\left[\begin{array}{ll}a-b&2a+c\\2a-b&3c+d\end{array}\right]=\left[\begin{array}{ll}-1&5\\0&13\end{array}\right]\)
Solution:
Matching elements: \(a-b=-1\) …(1), \(2a-b=0\) …(2), \(2a+c=5\) …(3), \(3c+d=13\) …(4).
From (2): \(b=2a\). Substituting in (1): \(a-2a=-1\Rightarrow a=1\Rightarrow b=2\).
From (3): \(2+c=5\Rightarrow c=3\).
From (4): \(9+d=13\Rightarrow d=4\).
\(\therefore a=1,\ b=2,\ c=3,\ d=4\).
Matching elements: \(a-b=-1\) …(1), \(2a-b=0\) …(2), \(2a+c=5\) …(3), \(3c+d=13\) …(4).
From (2): \(b=2a\). Substituting in (1): \(a-2a=-1\Rightarrow a=1\Rightarrow b=2\).
From (3): \(2+c=5\Rightarrow c=3\).
From (4): \(9+d=13\Rightarrow d=4\).
\(\therefore a=1,\ b=2,\ c=3,\ d=4\).
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
(A) \(m < n\) (B) \(m > n\) (C) \(m=n\) (D) None of these
Solution: (C)
A matrix is called a square matrix when its number of rows equals its number of columns, i.e., \(m=n\).
A matrix is called a square matrix when its number of rows equals its number of columns, i.e., \(m=n\).
9. Which of the given values of \(x\) and \(y\) make the following pair of matrices equal?
\(\left[\begin{array}{ll}3x+7&5\\y+1&2-3x\end{array}\right]=\left[\begin{array}{ll}0&y-2\\8&4\end{array}\right]\)
(A) \(x=\frac{-1}{3},y=7\) (B) Not possible to find (C) \(y=7,x=\frac{-2}{3}\) (D) \(x=\frac{-1}{3},y=\frac{-2}{3}\)
(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: (B)
Matching elements:
\(3x+7=0\Rightarrow x=-\dfrac{7}{3}\)
\(2-3x=4\Rightarrow x=-\dfrac{2}{3}\)
These give contradictory values of \(x\), so it is not possible to find consistent values.
Matching elements:
\(3x+7=0\Rightarrow x=-\dfrac{7}{3}\)
\(2-3x=4\Rightarrow x=-\dfrac{2}{3}\)
These give contradictory values of \(x\), so it is not possible to find consistent values.
10. The number of all possible matrices of order \(3\times3\) with each entry 0 or 1 is:
(A) 27 (B) 18 (C) 81 (D) 512
(A) 27 (B) 18 (C) 81 (D) 512
Solution: (D)
A \(3\times3\) matrix has 9 entries; each can independently be 0 or 1 (2 choices).
Total number of matrices \(= 2^9 = 512\).
A \(3\times3\) matrix has 9 entries; each can independently be 0 or 1 (2 choices).
Total number of matrices \(= 2^9 = 512\).
Exercise 3.2
1. Let \(A=\begin{bmatrix}2&4\\3&2\end{bmatrix},\ B=\begin{bmatrix}1&3\\-2&5\end{bmatrix},\ C=\begin{bmatrix}-2&5\\3&4\end{bmatrix}\). Find:
(i) \(A+B\) (ii) \(A-B\) (iii) \(3A-C\) (iv) \(AB\) (v) \(BA\)
(i) \(A+B\) (ii) \(A-B\) (iii) \(3A-C\) (iv) \(AB\) (v) \(BA\)
Solution:
(i) \(A+B=\begin{bmatrix}2+1&4+3\\3-2&2+5\end{bmatrix}=\begin{bmatrix}3&7\\1&7\end{bmatrix}\)
(ii) \(A-B=\begin{bmatrix}2-1&4-3\\3-(-2)&2-5\end{bmatrix}=\begin{bmatrix}1&1\\5&-3\end{bmatrix}\)
(iii) \(3A-C=\begin{bmatrix}6&12\\9&6\end{bmatrix}-\begin{bmatrix}-2&5\\3&4\end{bmatrix}=\begin{bmatrix}8&7\\6&2\end{bmatrix}\)
(iv) \(AB=\begin{bmatrix}2&4\\3&2\end{bmatrix}\begin{bmatrix}1&3\\-2&5\end{bmatrix}=\begin{bmatrix}2-8&6+20\\3-4&9+10\end{bmatrix}=\begin{bmatrix}-6&26\\-1&19\end{bmatrix}\)
(v) \(BA=\begin{bmatrix}1&3\\-2&5\end{bmatrix}\begin{bmatrix}2&4\\3&2\end{bmatrix}=\begin{bmatrix}2+9&4+6\\-4+15&-8+10\end{bmatrix}=\begin{bmatrix}11&10\\11&2\end{bmatrix}\)
(i) \(A+B=\begin{bmatrix}2+1&4+3\\3-2&2+5\end{bmatrix}=\begin{bmatrix}3&7\\1&7\end{bmatrix}\)
(ii) \(A-B=\begin{bmatrix}2-1&4-3\\3-(-2)&2-5\end{bmatrix}=\begin{bmatrix}1&1\\5&-3\end{bmatrix}\)
(iii) \(3A-C=\begin{bmatrix}6&12\\9&6\end{bmatrix}-\begin{bmatrix}-2&5\\3&4\end{bmatrix}=\begin{bmatrix}8&7\\6&2\end{bmatrix}\)
(iv) \(AB=\begin{bmatrix}2&4\\3&2\end{bmatrix}\begin{bmatrix}1&3\\-2&5\end{bmatrix}=\begin{bmatrix}2-8&6+20\\3-4&9+10\end{bmatrix}=\begin{bmatrix}-6&26\\-1&19\end{bmatrix}\)
(v) \(BA=\begin{bmatrix}1&3\\-2&5\end{bmatrix}\begin{bmatrix}2&4\\3&2\end{bmatrix}=\begin{bmatrix}2+9&4+6\\-4+15&-8+10\end{bmatrix}=\begin{bmatrix}11&10\\11&2\end{bmatrix}\)
2. Compute the following:
(i) \(\begin{bmatrix}a&b\\-b&a\end{bmatrix}+\begin{bmatrix}a&b\\b&a\end{bmatrix}\)
(ii) \(\begin{bmatrix}a^2+b^2&b^2+c^2\\a^2+c^2&a^2+b^2\end{bmatrix}+\begin{bmatrix}2ab&2bc\\-2ac&-2ab\end{bmatrix}\)
(iii) \(\begin{bmatrix}-1&4&-6\\8&5&16\\2&8&5\end{bmatrix}+\begin{bmatrix}12&7&6\\8&0&5\\3&2&4\end{bmatrix}\)
(iv) \(\begin{bmatrix}\cos^2 x&\sin^2 x\\\sin^2 x&\cos^2 x\end{bmatrix}+\begin{bmatrix}\sin^2 x&\cos^2 x\\\cos^2 x&\sin^2 x\end{bmatrix}\)
(i) \(\begin{bmatrix}a&b\\-b&a\end{bmatrix}+\begin{bmatrix}a&b\\b&a\end{bmatrix}\)
(ii) \(\begin{bmatrix}a^2+b^2&b^2+c^2\\a^2+c^2&a^2+b^2\end{bmatrix}+\begin{bmatrix}2ab&2bc\\-2ac&-2ab\end{bmatrix}\)
(iii) \(\begin{bmatrix}-1&4&-6\\8&5&16\\2&8&5\end{bmatrix}+\begin{bmatrix}12&7&6\\8&0&5\\3&2&4\end{bmatrix}\)
(iv) \(\begin{bmatrix}\cos^2 x&\sin^2 x\\\sin^2 x&\cos^2 x\end{bmatrix}+\begin{bmatrix}\sin^2 x&\cos^2 x\\\cos^2 x&\sin^2 x\end{bmatrix}\)
Solution:
(i) \(\begin{bmatrix}2a&2b\\0&2a\end{bmatrix}\)
(ii) \(\begin{bmatrix}a^2+b^2+2ab&b^2+c^2+2bc\\a^2+c^2-2ac&a^2+b^2-2ab\end{bmatrix}=\begin{bmatrix}(a+b)^2&(b+c)^2\\(a-c)^2&(a-b)^2\end{bmatrix}\)
(iii) \(\begin{bmatrix}11&11&0\\16&5&21\\5&10&9\end{bmatrix}\)
(iv) \(\begin{bmatrix}\cos^2x+\sin^2x&\sin^2x+\cos^2x\\\sin^2x+\cos^2x&\cos^2x+\sin^2x\end{bmatrix}=\begin{bmatrix}1&1\\1&1\end{bmatrix}\quad(\because\sin^2x+\cos^2x=1)\)
(i) \(\begin{bmatrix}2a&2b\\0&2a\end{bmatrix}\)
(ii) \(\begin{bmatrix}a^2+b^2+2ab&b^2+c^2+2bc\\a^2+c^2-2ac&a^2+b^2-2ab\end{bmatrix}=\begin{bmatrix}(a+b)^2&(b+c)^2\\(a-c)^2&(a-b)^2\end{bmatrix}\)
(iii) \(\begin{bmatrix}11&11&0\\16&5&21\\5&10&9\end{bmatrix}\)
(iv) \(\begin{bmatrix}\cos^2x+\sin^2x&\sin^2x+\cos^2x\\\sin^2x+\cos^2x&\cos^2x+\sin^2x\end{bmatrix}=\begin{bmatrix}1&1\\1&1\end{bmatrix}\quad(\because\sin^2x+\cos^2x=1)\)
3. Compute the indicated products:
(i) \(\begin{bmatrix}a&b\\-b&a\end{bmatrix}\begin{bmatrix}a&-b\\b&a\end{bmatrix}\)
(ii) \(\begin{bmatrix}1\\2\\3\end{bmatrix}\begin{bmatrix}2&3&4\end{bmatrix}\)
(iii) \(\begin{bmatrix}1&-2\\2&3\end{bmatrix}\begin{bmatrix}1&2&3\\2&3&1\end{bmatrix}\)
(iv) \(\begin{bmatrix}2&3&4\\3&4&5\\4&5&6\end{bmatrix}\begin{bmatrix}1&-3&5\\0&2&4\\3&0&5\end{bmatrix}\)
(v) \(\begin{bmatrix}2&1\\3&2\\-1&1\end{bmatrix}\begin{bmatrix}1&0&1\\-1&2&1\end{bmatrix}\)
(vi) \(\begin{bmatrix}3&-1&3\\-1&0&2\end{bmatrix}\begin{bmatrix}2&-3\\1&0\\3&1\end{bmatrix}\)
(i) \(\begin{bmatrix}a&b\\-b&a\end{bmatrix}\begin{bmatrix}a&-b\\b&a\end{bmatrix}\)
(ii) \(\begin{bmatrix}1\\2\\3\end{bmatrix}\begin{bmatrix}2&3&4\end{bmatrix}\)
(iii) \(\begin{bmatrix}1&-2\\2&3\end{bmatrix}\begin{bmatrix}1&2&3\\2&3&1\end{bmatrix}\)
(iv) \(\begin{bmatrix}2&3&4\\3&4&5\\4&5&6\end{bmatrix}\begin{bmatrix}1&-3&5\\0&2&4\\3&0&5\end{bmatrix}\)
(v) \(\begin{bmatrix}2&1\\3&2\\-1&1\end{bmatrix}\begin{bmatrix}1&0&1\\-1&2&1\end{bmatrix}\)
(vi) \(\begin{bmatrix}3&-1&3\\-1&0&2\end{bmatrix}\begin{bmatrix}2&-3\\1&0\\3&1\end{bmatrix}\)
Solution:
(i) \(\begin{bmatrix}a^2+b^2&-ab+ab\\-ab+ab&b^2+a^2\end{bmatrix}=\begin{bmatrix}a^2+b^2&0\\0&a^2+b^2\end{bmatrix}\)
(ii) \(\begin{bmatrix}2&3&4\\4&6&8\\6&9&12\end{bmatrix}\)
(iii) \(\begin{bmatrix}1-4&2-6&3-2\\2+6&4+9&6+3\end{bmatrix}=\begin{bmatrix}-3&-4&1\\8&13&9\end{bmatrix}\)
(iv) \(\begin{bmatrix}2+0+12&-6+6+0&10+12+20\\3+0+15&-9+8+0&15+16+25\\4+0+18&-12+10+0&20+20+30\end{bmatrix}=\begin{bmatrix}14&0&42\\18&-1&56\\22&-2&70\end{bmatrix}\)
(v) \(\begin{bmatrix}2-1&0+2&2+1\\3-2&0+4&3+2\\-1-1&0+2&-1+1\end{bmatrix}=\begin{bmatrix}1&2&3\\1&4&5\\-2&2&0\end{bmatrix}\)
(vi) \(\begin{bmatrix}6-1+9&-9-0+3\\-2+0+6&3+0+2\end{bmatrix}=\begin{bmatrix}14&-6\\4&5\end{bmatrix}\)
(i) \(\begin{bmatrix}a^2+b^2&-ab+ab\\-ab+ab&b^2+a^2\end{bmatrix}=\begin{bmatrix}a^2+b^2&0\\0&a^2+b^2\end{bmatrix}\)
(ii) \(\begin{bmatrix}2&3&4\\4&6&8\\6&9&12\end{bmatrix}\)
(iii) \(\begin{bmatrix}1-4&2-6&3-2\\2+6&4+9&6+3\end{bmatrix}=\begin{bmatrix}-3&-4&1\\8&13&9\end{bmatrix}\)
(iv) \(\begin{bmatrix}2+0+12&-6+6+0&10+12+20\\3+0+15&-9+8+0&15+16+25\\4+0+18&-12+10+0&20+20+30\end{bmatrix}=\begin{bmatrix}14&0&42\\18&-1&56\\22&-2&70\end{bmatrix}\)
(v) \(\begin{bmatrix}2-1&0+2&2+1\\3-2&0+4&3+2\\-1-1&0+2&-1+1\end{bmatrix}=\begin{bmatrix}1&2&3\\1&4&5\\-2&2&0\end{bmatrix}\)
(vi) \(\begin{bmatrix}6-1+9&-9-0+3\\-2+0+6&3+0+2\end{bmatrix}=\begin{bmatrix}14&-6\\4&5\end{bmatrix}\)
4. If \(A=\begin{bmatrix}1&2&-3\\5&0&2\\1&-1&1\end{bmatrix},\ B=\begin{bmatrix}3&-1&2\\4&2&5\\2&0&3\end{bmatrix},\ C=\begin{bmatrix}4&1&2\\0&3&2\\1&-2&3\end{bmatrix}\), compute \((A+B)\) and \((B-C)\). Verify that \(A+(B-C)=(A+B)-C\).
Solution:
\(A+B=\begin{bmatrix}4&1&-1\\9&2&7\\3&-1&4\end{bmatrix}\)
\(B-C=\begin{bmatrix}-1&-2&0\\4&-1&3\\1&2&0\end{bmatrix}\)
\(A+(B-C)=\begin{bmatrix}1&2&-3\\5&0&2\\1&-1&1\end{bmatrix}+\begin{bmatrix}-1&-2&0\\4&-1&3\\1&2&0\end{bmatrix}=\begin{bmatrix}0&0&-3\\9&-1&5\\2&1&1\end{bmatrix}\)
\((A+B)-C=\begin{bmatrix}4&1&-1\\9&2&7\\3&-1&4\end{bmatrix}-\begin{bmatrix}4&1&2\\0&3&2\\1&-2&3\end{bmatrix}=\begin{bmatrix}0&0&-3\\9&-1&5\\2&1&1\end{bmatrix}\)
Both sides are equal, so \(A+(B-C)=(A+B)-C\) is verified.
\(A+B=\begin{bmatrix}4&1&-1\\9&2&7\\3&-1&4\end{bmatrix}\)
\(B-C=\begin{bmatrix}-1&-2&0\\4&-1&3\\1&2&0\end{bmatrix}\)
\(A+(B-C)=\begin{bmatrix}1&2&-3\\5&0&2\\1&-1&1\end{bmatrix}+\begin{bmatrix}-1&-2&0\\4&-1&3\\1&2&0\end{bmatrix}=\begin{bmatrix}0&0&-3\\9&-1&5\\2&1&1\end{bmatrix}\)
\((A+B)-C=\begin{bmatrix}4&1&-1\\9&2&7\\3&-1&4\end{bmatrix}-\begin{bmatrix}4&1&2\\0&3&2\\1&-2&3\end{bmatrix}=\begin{bmatrix}0&0&-3\\9&-1&5\\2&1&1\end{bmatrix}\)
Both sides are equal, so \(A+(B-C)=(A+B)-C\) is verified.
5. If \(A=\begin{bmatrix}\frac{2}{3}&1&\frac{5}{3}\\\frac{1}{3}&\frac{2}{3}&\frac{4}{3}\\\frac{7}{3}&2&\frac{2}{3}\end{bmatrix}\) and \(B=\begin{bmatrix}\frac{2}{5}&\frac{3}{5}&1\\\frac{1}{5}&\frac{2}{5}&\frac{4}{5}\\\frac{7}{5}&\frac{6}{5}&\frac{2}{5}\end{bmatrix}\), compute \(3A-5B\).
Solution:
\(3A=\begin{bmatrix}2&3&5\\1&2&4\\7&6&2\end{bmatrix},\quad 5B=\begin{bmatrix}2&3&5\\1&2&4\\7&6&2\end{bmatrix}\)
\(\therefore 3A-5B=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\)
\(3A=\begin{bmatrix}2&3&5\\1&2&4\\7&6&2\end{bmatrix},\quad 5B=\begin{bmatrix}2&3&5\\1&2&4\\7&6&2\end{bmatrix}\)
\(\therefore 3A-5B=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\)
6. Simplify \(\cos\theta\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}+\sin\theta\begin{bmatrix}\sin\theta&-\cos\theta\\\cos\theta&\sin\theta\end{bmatrix}\)
Solution:
\(=\begin{bmatrix}\cos^2\theta&\cos\theta\sin\theta\\-\sin\theta\cos\theta&\cos^2\theta\end{bmatrix}+\begin{bmatrix}\sin^2\theta&-\sin\theta\cos\theta\\\sin\theta\cos\theta&\sin^2\theta\end{bmatrix}\)
\(=\begin{bmatrix}\cos^2\theta+\sin^2\theta&0\\0&\cos^2\theta+\sin^2\theta\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\quad(\because\cos^2\theta+\sin^2\theta=1)\)
\(=\begin{bmatrix}\cos^2\theta&\cos\theta\sin\theta\\-\sin\theta\cos\theta&\cos^2\theta\end{bmatrix}+\begin{bmatrix}\sin^2\theta&-\sin\theta\cos\theta\\\sin\theta\cos\theta&\sin^2\theta\end{bmatrix}\)
\(=\begin{bmatrix}\cos^2\theta+\sin^2\theta&0\\0&\cos^2\theta+\sin^2\theta\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\quad(\because\cos^2\theta+\sin^2\theta=1)\)
7. Find \(X\) and \(Y\), if
(i) \(X+Y=\begin{bmatrix}7&0\\2&5\end{bmatrix}\) and \(X-Y=\begin{bmatrix}3&0\\0&3\end{bmatrix}\)
(ii) \(2X+3Y=\begin{bmatrix}2&3\\4&0\end{bmatrix}\) and \(3X+2Y=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}\)
(i) \(X+Y=\begin{bmatrix}7&0\\2&5\end{bmatrix}\) and \(X-Y=\begin{bmatrix}3&0\\0&3\end{bmatrix}\)
(ii) \(2X+3Y=\begin{bmatrix}2&3\\4&0\end{bmatrix}\) and \(3X+2Y=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}\)
Solution:
(i) Adding the two equations: \(2X=\begin{bmatrix}10&0\\2&8\end{bmatrix}\Rightarrow X=\begin{bmatrix}5&0\\1&4\end{bmatrix}\)
Substituting back: \(Y=\begin{bmatrix}7&0\\2&5\end{bmatrix}-\begin{bmatrix}5&0\\1&4\end{bmatrix}=\begin{bmatrix}2&0\\1&1\end{bmatrix}\)
(ii) Multiply first equation by 2: \(4X+6Y=\begin{bmatrix}4&6\\8&0\end{bmatrix}\) …(5)
Multiply second equation by 3: \(9X+6Y=\begin{bmatrix}6&-6\\-3&15\end{bmatrix}\) …(6)
Subtracting (5) from (6): \(-5X=\begin{bmatrix}-2&12\\11&-15\end{bmatrix}\Rightarrow X=\begin{bmatrix}\frac{2}{5}&-\frac{12}{5}\\-\frac{11}{5}&3\end{bmatrix}\)
Then: \(Y=\begin{bmatrix}\frac{2}{5}&\frac{13}{5}\\\frac{14}{5}&-2\end{bmatrix}\)
(i) Adding the two equations: \(2X=\begin{bmatrix}10&0\\2&8\end{bmatrix}\Rightarrow X=\begin{bmatrix}5&0\\1&4\end{bmatrix}\)
Substituting back: \(Y=\begin{bmatrix}7&0\\2&5\end{bmatrix}-\begin{bmatrix}5&0\\1&4\end{bmatrix}=\begin{bmatrix}2&0\\1&1\end{bmatrix}\)
(ii) Multiply first equation by 2: \(4X+6Y=\begin{bmatrix}4&6\\8&0\end{bmatrix}\) …(5)
Multiply second equation by 3: \(9X+6Y=\begin{bmatrix}6&-6\\-3&15\end{bmatrix}\) …(6)
Subtracting (5) from (6): \(-5X=\begin{bmatrix}-2&12\\11&-15\end{bmatrix}\Rightarrow X=\begin{bmatrix}\frac{2}{5}&-\frac{12}{5}\\-\frac{11}{5}&3\end{bmatrix}\)
Then: \(Y=\begin{bmatrix}\frac{2}{5}&\frac{13}{5}\\\frac{14}{5}&-2\end{bmatrix}\)
8. Find \(X\), if \(Y=\begin{bmatrix}3&2\\1&4\end{bmatrix}\) and \(2X+Y=\begin{bmatrix}1&0\\-3&2\end{bmatrix}\).
Solution:
\(2X=\begin{bmatrix}1&0\\-3&2\end{bmatrix}-\begin{bmatrix}3&2\\1&4\end{bmatrix}=\begin{bmatrix}-2&-2\\-4&-2\end{bmatrix}\)
\(\therefore X=\begin{bmatrix}-1&-1\\-2&-1\end{bmatrix}\)
\(2X=\begin{bmatrix}1&0\\-3&2\end{bmatrix}-\begin{bmatrix}3&2\\1&4\end{bmatrix}=\begin{bmatrix}-2&-2\\-4&-2\end{bmatrix}\)
\(\therefore X=\begin{bmatrix}-1&-1\\-2&-1\end{bmatrix}\)
9. Find \(x\) and \(y\), if \(2\begin{bmatrix}1&3\\0&x\end{bmatrix}+\begin{bmatrix}y&0\\1&2\end{bmatrix}=\begin{bmatrix}5&6\\1&8\end{bmatrix}\).
Solution:
\(\begin{bmatrix}2+y&6\\1&2x+2\end{bmatrix}=\begin{bmatrix}5&6\\1&8\end{bmatrix}\)
\(2+y=5\Rightarrow y=3\)
\(2x+2=8\Rightarrow x=3\)
\(\therefore x=3,\ y=3\)
\(\begin{bmatrix}2+y&6\\1&2x+2\end{bmatrix}=\begin{bmatrix}5&6\\1&8\end{bmatrix}\)
\(2+y=5\Rightarrow y=3\)
\(2x+2=8\Rightarrow x=3\)
\(\therefore x=3,\ y=3\)
10. Solve for \(x,y,z\) and \(t\) if \(2\begin{bmatrix}x&z\\y&t\end{bmatrix}+3\begin{bmatrix}1&-1\\0&2\end{bmatrix}=3\begin{bmatrix}3&5\\4&6\end{bmatrix}\).
Solution:
\(\begin{bmatrix}2x+3&2z-3\\2y&2t+6\end{bmatrix}=\begin{bmatrix}9&15\\12&18\end{bmatrix}\)
\(2x+3=9\Rightarrow x=3\)
\(2y=12\Rightarrow y=6\)
\(2z-3=15\Rightarrow z=9\)
\(2t+6=18\Rightarrow t=6\)
\(\therefore x=3,\ y=6,\ z=9,\ t=6\)
\(\begin{bmatrix}2x+3&2z-3\\2y&2t+6\end{bmatrix}=\begin{bmatrix}9&15\\12&18\end{bmatrix}\)
\(2x+3=9\Rightarrow x=3\)
\(2y=12\Rightarrow y=6\)
\(2z-3=15\Rightarrow z=9\)
\(2t+6=18\Rightarrow t=6\)
\(\therefore x=3,\ y=6,\ z=9,\ t=6\)
11. If \(x\begin{bmatrix}2\\3\end{bmatrix}+y\begin{bmatrix}-1\\1\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}\), find values of \(x\) and \(y\).
Solution:
\(\begin{bmatrix}2x-y\\3x+y\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}\Rightarrow 2x-y=10\) and \(3x+y=5\).
Adding: \(5x=15\Rightarrow x=3\).
Then \(y=5-9=-4\).
\(\therefore x=3,\ y=-4\)
\(\begin{bmatrix}2x-y\\3x+y\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}\Rightarrow 2x-y=10\) and \(3x+y=5\).
Adding: \(5x=15\Rightarrow x=3\).
Then \(y=5-9=-4\).
\(\therefore x=3,\ y=-4\)
12. Given \(3\begin{bmatrix}x&y\\z&w\end{bmatrix}=\begin{bmatrix}x&6\\-1&2w\end{bmatrix}+\begin{bmatrix}4&x+y\\z+w&3\end{bmatrix}\), find \(x,y,z,w\).
Solution:
\(\begin{bmatrix}3x&3y\\3z&3w\end{bmatrix}=\begin{bmatrix}x+4&6+x+y\\-1+z+w&2w+3\end{bmatrix}\)
\(3x=x+4\Rightarrow x=2\)
\(3y=6+x+y=8\Rightarrow y=4\)
\(3w=2w+3\Rightarrow w=3\)
\(3z=-1+z+w=2\Rightarrow z=1\)
\(\therefore x=2,\ y=4,\ z=1,\ w=3\)
\(\begin{bmatrix}3x&3y\\3z&3w\end{bmatrix}=\begin{bmatrix}x+4&6+x+y\\-1+z+w&2w+3\end{bmatrix}\)
\(3x=x+4\Rightarrow x=2\)
\(3y=6+x+y=8\Rightarrow y=4\)
\(3w=2w+3\Rightarrow w=3\)
\(3z=-1+z+w=2\Rightarrow z=1\)
\(\therefore x=2,\ y=4,\ z=1,\ w=3\)
13. If \(F(x)=\begin{bmatrix}\cos x&-\sin x&0\\\sin x&\cos x&0\\0&0&1\end{bmatrix}\), show that \(F(x)F(y)=F(x+y)\).
Solution:
\(F(x)F(y)=\begin{bmatrix}\cos x&-\sin x&0\\\sin x&\cos x&0\\0&0&1\end{bmatrix}\begin{bmatrix}\cos y&-\sin y&0\\\sin y&\cos y&0\\0&0&1\end{bmatrix}\)
\(=\begin{bmatrix}\cos x\cos y-\sin x\sin y&-\cos x\sin y-\sin x\cos y&0\\\sin x\cos y+\cos x\sin y&-\sin x\sin y+\cos x\cos y&0\\0&0&1\end{bmatrix}\)
Applying angle addition identities:
\(=\begin{bmatrix}\cos(x+y)&-\sin(x+y)&0\\\sin(x+y)&\cos(x+y)&0\\0&0&1\end{bmatrix}=F(x+y)\)
\(\therefore F(x)F(y)=F(x+y)\)
\(F(x)F(y)=\begin{bmatrix}\cos x&-\sin x&0\\\sin x&\cos x&0\\0&0&1\end{bmatrix}\begin{bmatrix}\cos y&-\sin y&0\\\sin y&\cos y&0\\0&0&1\end{bmatrix}\)
\(=\begin{bmatrix}\cos x\cos y-\sin x\sin y&-\cos x\sin y-\sin x\cos y&0\\\sin x\cos y+\cos x\sin y&-\sin x\sin y+\cos x\cos y&0\\0&0&1\end{bmatrix}\)
Applying angle addition identities:
\(=\begin{bmatrix}\cos(x+y)&-\sin(x+y)&0\\\sin(x+y)&\cos(x+y)&0\\0&0&1\end{bmatrix}=F(x+y)\)
\(\therefore F(x)F(y)=F(x+y)\)
14. Show that
(i) \(\begin{bmatrix}5&-1\\6&7\end{bmatrix}\begin{bmatrix}2&1\\3&4\end{bmatrix}\neq\begin{bmatrix}2&1\\3&4\end{bmatrix}\begin{bmatrix}5&-1\\6&7\end{bmatrix}\)
(ii) \(\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}\neq\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}\)
(i) \(\begin{bmatrix}5&-1\\6&7\end{bmatrix}\begin{bmatrix}2&1\\3&4\end{bmatrix}\neq\begin{bmatrix}2&1\\3&4\end{bmatrix}\begin{bmatrix}5&-1\\6&7\end{bmatrix}\)
(ii) \(\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}\neq\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}\)
Solution:
(i) \(\begin{bmatrix}5&-1\\6&7\end{bmatrix}\begin{bmatrix}2&1\\3&4\end{bmatrix}=\begin{bmatrix}7&1\\33&34\end{bmatrix}\)
\(\begin{bmatrix}2&1\\3&4\end{bmatrix}\begin{bmatrix}5&-1\\6&7\end{bmatrix}=\begin{bmatrix}16&5\\39&25\end{bmatrix}\)
Since the results differ, the inequality is shown. ✓
(ii) \(\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}=\begin{bmatrix}5&8&14\\0&-1&1\\-1&0&1\end{bmatrix}\)
\(\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}=\begin{bmatrix}-1&-1&-3\\1&0&0\\6&11&6\end{bmatrix}\)
Since the results differ, the inequality is shown. ✓
(i) \(\begin{bmatrix}5&-1\\6&7\end{bmatrix}\begin{bmatrix}2&1\\3&4\end{bmatrix}=\begin{bmatrix}7&1\\33&34\end{bmatrix}\)
\(\begin{bmatrix}2&1\\3&4\end{bmatrix}\begin{bmatrix}5&-1\\6&7\end{bmatrix}=\begin{bmatrix}16&5\\39&25\end{bmatrix}\)
Since the results differ, the inequality is shown. ✓
(ii) \(\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}=\begin{bmatrix}5&8&14\\0&-1&1\\-1&0&1\end{bmatrix}\)
\(\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}=\begin{bmatrix}-1&-1&-3\\1&0&0\\6&11&6\end{bmatrix}\)
Since the results differ, the inequality is shown. ✓
15. Find \(A^2-5A+6I\) if \(A=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}\).
Solution:
\(A^2=\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}\)
\(A^2-5A+6I=\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}-\begin{bmatrix}10&0&5\\10&5&15\\5&-5&0\end{bmatrix}+\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}=\begin{bmatrix}1&-1&-3\\-1&-1&-10\\-5&4&4\end{bmatrix}\)
\(A^2=\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}\)
\(A^2-5A+6I=\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}-\begin{bmatrix}10&0&5\\10&5&15\\5&-5&0\end{bmatrix}+\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}=\begin{bmatrix}1&-1&-3\\-1&-1&-10\\-5&4&4\end{bmatrix}\)
16. If \(A=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\), prove that \(A^3-6A^2+7A+2I=O\).
Solution:
\(A^2=\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}\)
\(A^3=A^2\cdot A=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}\)
\(A^3-6A^2+7A+2I\)
\(=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}-\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix}+\begin{bmatrix}7&0&14\\0&14&7\\14&0&21\end{bmatrix}+\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}\)
\(=\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix}-\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix}=O\) ✓
\(A^2=\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}\)
\(A^3=A^2\cdot A=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}\)
\(A^3-6A^2+7A+2I\)
\(=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}-\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix}+\begin{bmatrix}7&0&14\\0&14&7\\14&0&21\end{bmatrix}+\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}\)
\(=\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix}-\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix}=O\) ✓
17. If \(A=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\) and \(I=\begin{bmatrix}1&0\\0&1\end{bmatrix}\), find \(k\) so that \(A^2=kA-2I\).
Solution:
\(A^2=\begin{bmatrix}9-8&-6+4\\12-8&-8+4\end{bmatrix}=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}\)
Setting \(A^2=kA-2I\):
\(\begin{bmatrix}1&-2\\4&-4\end{bmatrix}=\begin{bmatrix}3k-2&-2k\\4k&-2k-2\end{bmatrix}\)
From \(3k-2=1\Rightarrow k=1\).
\(A^2=\begin{bmatrix}9-8&-6+4\\12-8&-8+4\end{bmatrix}=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}\)
Setting \(A^2=kA-2I\):
\(\begin{bmatrix}1&-2\\4&-4\end{bmatrix}=\begin{bmatrix}3k-2&-2k\\4k&-2k-2\end{bmatrix}\)
From \(3k-2=1\Rightarrow k=1\).
18. If \(A=\begin{bmatrix}0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}&0\end{bmatrix}\) and \(I\) is the identity matrix of order 2, show that \(I+A=(I-A)\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}\).
Solution:
L.H.S.: \(I+A=\begin{bmatrix}1&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}&1\end{bmatrix}\) …(1)
R.H.S.: \((I-A)\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}=\begin{bmatrix}1&\tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2}&1\end{bmatrix}\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}\)
After expanding using \(\cos\alpha=1-2\sin^2\frac{\alpha}{2}\), \(\sin\alpha=2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}\) and simplifying:
\(=\begin{bmatrix}1&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}&1\end{bmatrix}\) …(2)
From (1) and (2), L.H.S. = R.H.S. ✓
L.H.S.: \(I+A=\begin{bmatrix}1&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}&1\end{bmatrix}\) …(1)
R.H.S.: \((I-A)\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}=\begin{bmatrix}1&\tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2}&1\end{bmatrix}\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}\)
After expanding using \(\cos\alpha=1-2\sin^2\frac{\alpha}{2}\), \(\sin\alpha=2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}\) and simplifying:
\(=\begin{bmatrix}1&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}&1\end{bmatrix}\) …(2)
From (1) and (2), L.H.S. = R.H.S. ✓
19. A trust fund has Rs 30,000 to invest in two bonds paying 5% and 7% per year. Using matrix multiplication, find how to divide Rs 30,000 to obtain an annual interest of (a) Rs 1,800 (b) Rs 2,000.
Solution:
Let Rs \(x\) be invested in the first bond; Rs \((30000-x)\) in the second.
(a) \(\begin{bmatrix}x&(30000-x)\end{bmatrix}\begin{bmatrix}\frac{5}{100}\\\frac{7}{100}\end{bmatrix}=1800\)
\(\Rightarrow 5x+210000-7x=180000\Rightarrow x=15000\)
Invest Rs 15,000 in each bond.
(b) \(\begin{bmatrix}x&(30000-x)\end{bmatrix}\begin{bmatrix}\frac{5}{100}\\\frac{7}{100}\end{bmatrix}=2000\)
\(\Rightarrow 5x+210000-7x=200000\Rightarrow x=5000\)
Invest Rs 5,000 in the first bond and Rs 25,000 in the second.
Let Rs \(x\) be invested in the first bond; Rs \((30000-x)\) in the second.
(a) \(\begin{bmatrix}x&(30000-x)\end{bmatrix}\begin{bmatrix}\frac{5}{100}\\\frac{7}{100}\end{bmatrix}=1800\)
\(\Rightarrow 5x+210000-7x=180000\Rightarrow x=15000\)
Invest Rs 15,000 in each bond.
(b) \(\begin{bmatrix}x&(30000-x)\end{bmatrix}\begin{bmatrix}\frac{5}{100}\\\frac{7}{100}\end{bmatrix}=2000\)
\(\Rightarrow 5x+210000-7x=200000\Rightarrow x=5000\)
Invest Rs 5,000 in the first bond and Rs 25,000 in the second.
20. A bookshop has 10 dozen chemistry, 8 dozen physics, and 10 dozen economics books with selling prices Rs 80, Rs 60, and Rs 40 each. Find the total amount the bookshop will receive using matrix algebra.
Solution:
\(12\begin{bmatrix}10&8&10\end{bmatrix}\begin{bmatrix}80\\60\\40\end{bmatrix}=12[800+480+400]=12\times1680=\text{Rs }20160\)
\(12\begin{bmatrix}10&8&10\end{bmatrix}\begin{bmatrix}80\\60\\40\end{bmatrix}=12[800+480+400]=12\times1680=\text{Rs }20160\)
21. \(X,Y,Z,W,P\) are matrices of order \(2\times n,3\times k,2\times p,n\times3,p\times k\). The restriction on \(n,k,p\) so that \(PY+WY\) is defined:
A. \(k=3,p=n\) B. \(k\) arbitrary, \(p=2\) C. \(p\) arbitrary, \(k=3\) D. \(k=2,p=3\)
A. \(k=3,p=n\) B. \(k\) arbitrary, \(p=2\) C. \(p\) arbitrary, \(k=3\) D. \(k=2,p=3\)
Solution: (A)
\(PY\) (\(p\times k\) times \(3\times k\)) is defined if \(k=3\); result is \(p\times k\).
\(WY\) (\(n\times3\) times \(3\times k\)) is always defined; result is \(n\times k\).
For addition, orders must match: \(p=n\).
Hence \(k=3\) and \(p=n\).
\(PY\) (\(p\times k\) times \(3\times k\)) is defined if \(k=3\); result is \(p\times k\).
\(WY\) (\(n\times3\) times \(3\times k\)) is always defined; result is \(n\times k\).
For addition, orders must match: \(p=n\).
Hence \(k=3\) and \(p=n\).
22. If \(n=p\), then the order of matrix \(7X-5Z\) is:
A. \(p\times2\) B. \(2\times n\) C. \(n\times3\) D. \(p\times n\)
A. \(p\times2\) B. \(2\times n\) C. \(n\times3\) D. \(p\times n\)
Solution: (B)
\(X\) is \(2\times n\), so \(7X\) is \(2\times n\). \(Z\) is \(2\times p=2\times n\) (since \(n=p\)), so \(5Z\) is \(2\times n\).
Therefore \(7X-5Z\) is of order \(2\times n\).
\(X\) is \(2\times n\), so \(7X\) is \(2\times n\). \(Z\) is \(2\times p=2\times n\) (since \(n=p\)), so \(5Z\) is \(2\times n\).
Therefore \(7X-5Z\) is of order \(2\times n\).
Exercise 3.3
1. Find the transpose of each of the following matrices:
(i) \(\begin{bmatrix}5\\\frac{1}{2}\\-1\end{bmatrix}\)
(ii) \(\begin{bmatrix}1&-1\\2&3\end{bmatrix}\)
(iii) \(\begin{bmatrix}-1&5&6\\\sqrt{3}&5&6\\2&3&-1\end{bmatrix}\)
(i) \(\begin{bmatrix}5\\\frac{1}{2}\\-1\end{bmatrix}\) (ii) \(\begin{bmatrix}1&-1\\2&3\end{bmatrix}\) (iii) \(\begin{bmatrix}-1&5&6\\\sqrt{3}&5&6\\2&3&-1\end{bmatrix}\)
Solution:
(i) \(A^{\mathrm{T}}=\begin{bmatrix}5&\frac{1}{2}&-1\end{bmatrix}\)
(ii) \(A^{\mathrm{T}}=\begin{bmatrix}1&2\\-1&3\end{bmatrix}\)
(iii) \(A^{\mathrm{T}}=\begin{bmatrix}-1&\sqrt{3}&2\\5&5&3\\6&6&-1\end{bmatrix}\)
(i) \(A^{\mathrm{T}}=\begin{bmatrix}5&\frac{1}{2}&-1\end{bmatrix}\)
(ii) \(A^{\mathrm{T}}=\begin{bmatrix}1&2\\-1&3\end{bmatrix}\)
(iii) \(A^{\mathrm{T}}=\begin{bmatrix}-1&\sqrt{3}&2\\5&5&3\\6&6&-1\end{bmatrix}\)
2. If \(A=\begin{bmatrix}-1&2&3\\5&7&9\\-2&1&1\end{bmatrix}\) and \(B=\begin{bmatrix}-4&1&-5\\1&2&0\\1&3&1\end{bmatrix}\), verify that
(i) \((A+B)^{\prime}=A^{\prime}+B^{\prime}\) (ii) \((A-B)^{\prime}=A^{\prime}-B^{\prime}\)
(i) \((A+B)^{\prime}=A^{\prime}+B^{\prime}\) (ii) \((A-B)^{\prime}=A^{\prime}-B^{\prime}\)
Solution:
\(A^{\prime}=\begin{bmatrix}-1&5&-2\\2&7&1\\3&9&1\end{bmatrix},\quad B^{\prime}=\begin{bmatrix}-4&1&1\\1&2&3\\-5&0&1\end{bmatrix}\)
(i) \(A+B=\begin{bmatrix}-5&3&-2\\6&9&9\\-1&4&2\end{bmatrix}\Rightarrow(A+B)^{\prime}=\begin{bmatrix}-5&6&-1\\3&9&4\\-2&9&2\end{bmatrix}\)
\(A^{\prime}+B^{\prime}=\begin{bmatrix}-5&6&-1\\3&9&4\\-2&9&2\end{bmatrix}\) ✓
(ii) \(A-B=\begin{bmatrix}3&1&8\\4&5&9\\-3&-2&0\end{bmatrix}\Rightarrow(A-B)^{\prime}=\begin{bmatrix}3&4&-3\\1&5&-2\\8&9&0\end{bmatrix}\)
\(A^{\prime}-B^{\prime}=\begin{bmatrix}3&4&-3\\1&5&-2\\8&9&0\end{bmatrix}\) ✓
\(A^{\prime}=\begin{bmatrix}-1&5&-2\\2&7&1\\3&9&1\end{bmatrix},\quad B^{\prime}=\begin{bmatrix}-4&1&1\\1&2&3\\-5&0&1\end{bmatrix}\)
(i) \(A+B=\begin{bmatrix}-5&3&-2\\6&9&9\\-1&4&2\end{bmatrix}\Rightarrow(A+B)^{\prime}=\begin{bmatrix}-5&6&-1\\3&9&4\\-2&9&2\end{bmatrix}\)
\(A^{\prime}+B^{\prime}=\begin{bmatrix}-5&6&-1\\3&9&4\\-2&9&2\end{bmatrix}\) ✓
(ii) \(A-B=\begin{bmatrix}3&1&8\\4&5&9\\-3&-2&0\end{bmatrix}\Rightarrow(A-B)^{\prime}=\begin{bmatrix}3&4&-3\\1&5&-2\\8&9&0\end{bmatrix}\)
\(A^{\prime}-B^{\prime}=\begin{bmatrix}3&4&-3\\1&5&-2\\8&9&0\end{bmatrix}\) ✓
3. If \(A^{\prime}=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix}\) and \(B=\begin{bmatrix}-1&2&1\\1&2&3\end{bmatrix}\), verify that
(i) \((A+B)^{\prime}=A^{\prime}+B^{\prime}\) (ii) \((A-B)^{\prime}=A^{\prime}-B^{\prime}\)
(i) \((A+B)^{\prime}=A^{\prime}+B^{\prime}\) (ii) \((A-B)^{\prime}=A^{\prime}-B^{\prime}\)
Solution:
Since \(A^{\prime}\) is given, \(A=(A^{\prime})^{\prime}=\begin{bmatrix}3&-1&0\\4&2&1\end{bmatrix}\)
\(B^{\prime}=\begin{bmatrix}-1&1\\2&2\\1&3\end{bmatrix}\)
(i) \(A+B=\begin{bmatrix}2&1&1\\5&4&4\end{bmatrix}\Rightarrow(A+B)^{\prime}=\begin{bmatrix}2&5\\1&4\\1&4\end{bmatrix}\)
\(A^{\prime}+B^{\prime}=\begin{bmatrix}2&5\\1&4\\1&4\end{bmatrix}\) ✓
(ii) \(A-B=\begin{bmatrix}4&-3&-1\\3&0&-2\end{bmatrix}\Rightarrow(A-B)^{\prime}=\begin{bmatrix}4&3\\-3&0\\-1&-2\end{bmatrix}\)
\(A^{\prime}-B^{\prime}=\begin{bmatrix}4&3\\-3&0\\-1&-2\end{bmatrix}\) ✓
Since \(A^{\prime}\) is given, \(A=(A^{\prime})^{\prime}=\begin{bmatrix}3&-1&0\\4&2&1\end{bmatrix}\)
\(B^{\prime}=\begin{bmatrix}-1&1\\2&2\\1&3\end{bmatrix}\)
(i) \(A+B=\begin{bmatrix}2&1&1\\5&4&4\end{bmatrix}\Rightarrow(A+B)^{\prime}=\begin{bmatrix}2&5\\1&4\\1&4\end{bmatrix}\)
\(A^{\prime}+B^{\prime}=\begin{bmatrix}2&5\\1&4\\1&4\end{bmatrix}\) ✓
(ii) \(A-B=\begin{bmatrix}4&-3&-1\\3&0&-2\end{bmatrix}\Rightarrow(A-B)^{\prime}=\begin{bmatrix}4&3\\-3&0\\-1&-2\end{bmatrix}\)
\(A^{\prime}-B^{\prime}=\begin{bmatrix}4&3\\-3&0\\-1&-2\end{bmatrix}\) ✓
4. If \(A^{\prime}=\begin{bmatrix}-2&3\\1&2\end{bmatrix}\) and \(B=\begin{bmatrix}-1&0\\1&2\end{bmatrix}\), find \((A+2B)^{\prime}\).
Solution:
\(A=(A^{\prime})^{\prime}=\begin{bmatrix}-2&1\\3&2\end{bmatrix}\)
\(A+2B=\begin{bmatrix}-2&1\\3&2\end{bmatrix}+\begin{bmatrix}-2&0\\2&4\end{bmatrix}=\begin{bmatrix}-4&1\\5&6\end{bmatrix}\)
\(\therefore(A+2B)^{\prime}=\begin{bmatrix}-4&5\\1&6\end{bmatrix}\)
\(A=(A^{\prime})^{\prime}=\begin{bmatrix}-2&1\\3&2\end{bmatrix}\)
\(A+2B=\begin{bmatrix}-2&1\\3&2\end{bmatrix}+\begin{bmatrix}-2&0\\2&4\end{bmatrix}=\begin{bmatrix}-4&1\\5&6\end{bmatrix}\)
\(\therefore(A+2B)^{\prime}=\begin{bmatrix}-4&5\\1&6\end{bmatrix}\)
5. For matrices \(A\) and \(B\), verify that \((AB)^{\prime}=B^{\prime}A^{\prime}\) where
(i) \(A=\begin{bmatrix}1\\-4\\3\end{bmatrix},\ B=\begin{bmatrix}-1&2&1\end{bmatrix}\)
(ii) \(A=\begin{bmatrix}0\\1\\2\end{bmatrix},\ B=\begin{bmatrix}1&5&7\end{bmatrix}\)
(i) \(A=\begin{bmatrix}1\\-4\\3\end{bmatrix},\ B=\begin{bmatrix}-1&2&1\end{bmatrix}\) (ii) \(A=\begin{bmatrix}0\\1\\2\end{bmatrix},\ B=\begin{bmatrix}1&5&7\end{bmatrix}\)
Solution:
(i) \(AB=\begin{bmatrix}-1&2&1\\4&-8&-4\\-3&6&3\end{bmatrix}\Rightarrow(AB)^{\prime}=\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix}\)
\(B^{\prime}A^{\prime}=\begin{bmatrix}-1\\2\\1\end{bmatrix}\begin{bmatrix}1&-4&3\end{bmatrix}=\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix}\) ✓
(ii) \(AB=\begin{bmatrix}0&0&0\\1&5&7\\2&10&14\end{bmatrix}\Rightarrow(AB)^{\prime}=\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}\)
\(B^{\prime}A^{\prime}=\begin{bmatrix}1\\5\\7\end{bmatrix}\begin{bmatrix}0&1&2\end{bmatrix}=\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}\) ✓
(i) \(AB=\begin{bmatrix}-1&2&1\\4&-8&-4\\-3&6&3\end{bmatrix}\Rightarrow(AB)^{\prime}=\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix}\)
\(B^{\prime}A^{\prime}=\begin{bmatrix}-1\\2\\1\end{bmatrix}\begin{bmatrix}1&-4&3\end{bmatrix}=\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix}\) ✓
(ii) \(AB=\begin{bmatrix}0&0&0\\1&5&7\\2&10&14\end{bmatrix}\Rightarrow(AB)^{\prime}=\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}\)
\(B^{\prime}A^{\prime}=\begin{bmatrix}1\\5\\7\end{bmatrix}\begin{bmatrix}0&1&2\end{bmatrix}=\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}\) ✓
6. If (i) \(A=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}\), verify that \(A^{\prime}A=I\)
(ii) \(A=\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}\), verify that \(A^{\prime}A=I\)
(ii) \(A=\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}\), verify that \(A^{\prime}A=I\)
Solution:
(i) \(A^{\prime}=\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}\)
\(A^{\prime}A=\begin{bmatrix}\cos^2\alpha+\sin^2\alpha&0\\0&\sin^2\alpha+\cos^2\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I\) ✓
(ii) \(A^{\prime}=\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}\)
\(A^{\prime}A=\begin{bmatrix}\sin^2\alpha+\cos^2\alpha&0\\0&\cos^2\alpha+\sin^2\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I\) ✓
(i) \(A^{\prime}=\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}\)
\(A^{\prime}A=\begin{bmatrix}\cos^2\alpha+\sin^2\alpha&0\\0&\sin^2\alpha+\cos^2\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I\) ✓
(ii) \(A^{\prime}=\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}\)
\(A^{\prime}A=\begin{bmatrix}\sin^2\alpha+\cos^2\alpha&0\\0&\cos^2\alpha+\sin^2\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I\) ✓
7. (i) Show that \(A=\begin{bmatrix}1&-1&5\\-1&2&1\\5&1&3\end{bmatrix}\) is symmetric.
(ii) Show that \(A=\begin{bmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{bmatrix}\) is skew-symmetric.
(ii) Show that \(A=\begin{bmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{bmatrix}\) is skew-symmetric.
Solution:
(i) \(A^{\prime}=\begin{bmatrix}1&-1&5\\-1&2&1\\5&1&3\end{bmatrix}=A\Rightarrow A^{\prime}=A\). Hence symmetric. ✓
(ii) \(A^{\prime}=\begin{bmatrix}0&-1&1\\1&0&-1\\-1&1&0\end{bmatrix}=-\begin{bmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{bmatrix}=-A\Rightarrow A^{\prime}=-A\). Hence skew-symmetric. ✓
(i) \(A^{\prime}=\begin{bmatrix}1&-1&5\\-1&2&1\\5&1&3\end{bmatrix}=A\Rightarrow A^{\prime}=A\). Hence symmetric. ✓
(ii) \(A^{\prime}=\begin{bmatrix}0&-1&1\\1&0&-1\\-1&1&0\end{bmatrix}=-\begin{bmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{bmatrix}=-A\Rightarrow A^{\prime}=-A\). Hence skew-symmetric. ✓
8. For the matrix \(A=\begin{bmatrix}1&5\\6&7\end{bmatrix}\), verify that (i) \((A+A^{\prime})\) is symmetric (ii) \((A-A^{\prime})\) is skew-symmetric.
Solution:
\(A^{\prime}=\begin{bmatrix}1&6\\5&7\end{bmatrix}\)
(i) \(A+A^{\prime}=\begin{bmatrix}2&11\\11&14\end{bmatrix}\Rightarrow(A+A^{\prime})^{\prime}=\begin{bmatrix}2&11\\11&14\end{bmatrix}=A+A^{\prime}\). Symmetric ✓
(ii) \(A-A^{\prime}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\Rightarrow(A-A^{\prime})^{\prime}=\begin{bmatrix}0&1\\-1&0\end{bmatrix}=-(A-A^{\prime})\). Skew-symmetric ✓
\(A^{\prime}=\begin{bmatrix}1&6\\5&7\end{bmatrix}\)
(i) \(A+A^{\prime}=\begin{bmatrix}2&11\\11&14\end{bmatrix}\Rightarrow(A+A^{\prime})^{\prime}=\begin{bmatrix}2&11\\11&14\end{bmatrix}=A+A^{\prime}\). Symmetric ✓
(ii) \(A-A^{\prime}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\Rightarrow(A-A^{\prime})^{\prime}=\begin{bmatrix}0&1\\-1&0\end{bmatrix}=-(A-A^{\prime})\). Skew-symmetric ✓
9. Find \(\frac{1}{2}(A+A^{\prime})\) and \(\frac{1}{2}(A-A^{\prime})\), when \(A=\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}\).
Solution:
\(A^{\prime}=\begin{bmatrix}0&-a&-b\\a&0&-c\\b&c&0\end{bmatrix}\)
\(A+A^{\prime}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\Rightarrow\frac{1}{2}(A+A^{\prime})=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\)
\(A-A^{\prime}=\begin{bmatrix}0&2a&2b\\-2a&0&2c\\-2b&-2c&0\end{bmatrix}\Rightarrow\frac{1}{2}(A-A^{\prime})=\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}\)
\(A^{\prime}=\begin{bmatrix}0&-a&-b\\a&0&-c\\b&c&0\end{bmatrix}\)
\(A+A^{\prime}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\Rightarrow\frac{1}{2}(A+A^{\prime})=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\)
\(A-A^{\prime}=\begin{bmatrix}0&2a&2b\\-2a&0&2c\\-2b&-2c&0\end{bmatrix}\Rightarrow\frac{1}{2}(A-A^{\prime})=\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}\)
10. Express as the sum of a symmetric and a skew-symmetric matrix:
(i) \(\begin{bmatrix}3&5\\1&-1\end{bmatrix}\)
(ii) \(\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}\)
(iii) \(\begin{bmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{bmatrix}\)
(iv) \(\begin{bmatrix}1&5\\-1&2\end{bmatrix}\)
(i) \(\begin{bmatrix}3&5\\1&-1\end{bmatrix}\) (ii) \(\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}\) (iii) \(\begin{bmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{bmatrix}\) (iv) \(\begin{bmatrix}1&5\\-1&2\end{bmatrix}\)
Solution:
For any square matrix \(A\): \(A=P+Q\) where \(P=\frac{1}{2}(A+A^{\prime})\) (symmetric) and \(Q=\frac{1}{2}(A-A^{\prime})\) (skew-symmetric).
(i) \(A=\begin{bmatrix}3&5\\1&-1\end{bmatrix},\ A^{\prime}=\begin{bmatrix}3&1\\5&-1\end{bmatrix}\)
\(P=\frac{1}{2}\begin{bmatrix}6&6\\6&-2\end{bmatrix}=\begin{bmatrix}3&3\\3&-1\end{bmatrix},\quad Q=\frac{1}{2}\begin{bmatrix}0&4\\-4&0\end{bmatrix}=\begin{bmatrix}0&2\\-2&0\end{bmatrix}\)
Verification: \(P+Q=\begin{bmatrix}3&5\\1&-1\end{bmatrix}=A\) ✓
(ii) Since \(A=A^{\prime}\) (already symmetric): \(P=A=\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix},\quad Q=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\)
(iii) \(A=\begin{bmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{bmatrix},\ A^{\prime}=\begin{bmatrix}3&-2&-4\\3&-2&-5\\-1&1&2\end{bmatrix}\)
\(P=\begin{bmatrix}3&\frac{1}{2}&-\frac{5}{2}\\\frac{1}{2}&-2&-2\\-\frac{5}{2}&-2&2\end{bmatrix},\quad Q=\begin{bmatrix}0&\frac{5}{2}&\frac{3}{2}\\-\frac{5}{2}&0&3\\-\frac{3}{2}&-3&0\end{bmatrix}\)
(iv) \(A=\begin{bmatrix}1&5\\-1&2\end{bmatrix},\ A^{\prime}=\begin{bmatrix}1&-1\\5&2\end{bmatrix}\)
\(P=\begin{bmatrix}1&2\\2&2\end{bmatrix},\quad Q=\begin{bmatrix}0&3\\-3&0\end{bmatrix}\)
Verification: \(P+Q=\begin{bmatrix}1&5\\-1&2\end{bmatrix}=A\) ✓
For any square matrix \(A\): \(A=P+Q\) where \(P=\frac{1}{2}(A+A^{\prime})\) (symmetric) and \(Q=\frac{1}{2}(A-A^{\prime})\) (skew-symmetric).
(i) \(A=\begin{bmatrix}3&5\\1&-1\end{bmatrix},\ A^{\prime}=\begin{bmatrix}3&1\\5&-1\end{bmatrix}\)
\(P=\frac{1}{2}\begin{bmatrix}6&6\\6&-2\end{bmatrix}=\begin{bmatrix}3&3\\3&-1\end{bmatrix},\quad Q=\frac{1}{2}\begin{bmatrix}0&4\\-4&0\end{bmatrix}=\begin{bmatrix}0&2\\-2&0\end{bmatrix}\)
Verification: \(P+Q=\begin{bmatrix}3&5\\1&-1\end{bmatrix}=A\) ✓
(ii) Since \(A=A^{\prime}\) (already symmetric): \(P=A=\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix},\quad Q=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\)
(iii) \(A=\begin{bmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{bmatrix},\ A^{\prime}=\begin{bmatrix}3&-2&-4\\3&-2&-5\\-1&1&2\end{bmatrix}\)
\(P=\begin{bmatrix}3&\frac{1}{2}&-\frac{5}{2}\\\frac{1}{2}&-2&-2\\-\frac{5}{2}&-2&2\end{bmatrix},\quad Q=\begin{bmatrix}0&\frac{5}{2}&\frac{3}{2}\\-\frac{5}{2}&0&3\\-\frac{3}{2}&-3&0\end{bmatrix}\)
(iv) \(A=\begin{bmatrix}1&5\\-1&2\end{bmatrix},\ A^{\prime}=\begin{bmatrix}1&-1\\5&2\end{bmatrix}\)
\(P=\begin{bmatrix}1&2\\2&2\end{bmatrix},\quad Q=\begin{bmatrix}0&3\\-3&0\end{bmatrix}\)
Verification: \(P+Q=\begin{bmatrix}1&5\\-1&2\end{bmatrix}=A\) ✓
11. If \(A,B\) are symmetric matrices of same order, then \(AB-BA\) is a
A. Skew-symmetric matrix B. Symmetric matrix C. Zero matrix D. Identity matrix
A. Skew-symmetric matrix B. Symmetric matrix C. Zero matrix D. Identity matrix
Solution: (A)
Since \(A^{\prime}=A\) and \(B^{\prime}=B\):
\((AB-BA)^{\prime}=(AB)^{\prime}-(BA)^{\prime}=B^{\prime}A^{\prime}-A^{\prime}B^{\prime}=BA-AB=-(AB-BA)\)
Hence \(AB-BA\) is skew-symmetric.
Since \(A^{\prime}=A\) and \(B^{\prime}=B\):
\((AB-BA)^{\prime}=(AB)^{\prime}-(BA)^{\prime}=B^{\prime}A^{\prime}-A^{\prime}B^{\prime}=BA-AB=-(AB-BA)\)
Hence \(AB-BA\) is skew-symmetric.
12. If \(A=\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}\), then \(A+A^{\prime}=I\) if the value of \(\alpha\) is
A. \(\frac{\pi}{6}\) B. \(\frac{\pi}{3}\) C. \(\pi\) D. \(\frac{3\pi}{2}\)
A. \(\frac{\pi}{6}\) B. \(\frac{\pi}{3}\) C. \(\pi\) D. \(\frac{3\pi}{2}\)
Solution: (B)
\(A+A^{\prime}=\begin{bmatrix}2\cos\alpha&0\\0&2\cos\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
\(\Rightarrow 2\cos\alpha=1\Rightarrow\cos\alpha=\dfrac{1}{2}=\cos\dfrac{\pi}{3}\Rightarrow\alpha=\dfrac{\pi}{3}\)
\(A+A^{\prime}=\begin{bmatrix}2\cos\alpha&0\\0&2\cos\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
\(\Rightarrow 2\cos\alpha=1\Rightarrow\cos\alpha=\dfrac{1}{2}=\cos\dfrac{\pi}{3}\Rightarrow\alpha=\dfrac{\pi}{3}\)
Exercise 3.4
1. Matrices \(A\) and \(B\) will be inverse of each other only if
A. \(AB=BA\) B. \(AB=BA=O\) C. \(AB=O,BA=I\) D. \(AB=BA=I\)
A. \(AB=BA\) B. \(AB=BA=O\) C. \(AB=O,BA=I\) D. \(AB=BA=I\)
Solution: (D)
By definition, if \(B\) is the inverse of \(A\), then \(AB=BA=I\). Thus matrices \(A\) and \(B\) are inverses of each other only if \(AB=BA=I\).
By definition, if \(B\) is the inverse of \(A\), then \(AB=BA=I\). Thus matrices \(A\) and \(B\) are inverses of each other only if \(AB=BA=I\).
Miscellaneous Exercise
1. If \(A\) and \(B\) are symmetric matrices, prove that \(AB-BA\) is a skew-symmetric matrix.
Solution:
Since \(A^{\prime}=A\) and \(B^{\prime}=B\):
\((AB-BA)^{\prime}=(AB)^{\prime}-(BA)^{\prime}=B^{\prime}A^{\prime}-A^{\prime}B^{\prime}=BA-AB=-(AB-BA)\)
\(\therefore(AB-BA)^{\prime}=-(AB-BA)\)
Hence \(AB-BA\) is a skew-symmetric matrix. ✓
Since \(A^{\prime}=A\) and \(B^{\prime}=B\):
\((AB-BA)^{\prime}=(AB)^{\prime}-(BA)^{\prime}=B^{\prime}A^{\prime}-A^{\prime}B^{\prime}=BA-AB=-(AB-BA)\)
\(\therefore(AB-BA)^{\prime}=-(AB-BA)\)
Hence \(AB-BA\) is a skew-symmetric matrix. ✓
2. Show that \(B^{\prime}AB\) is symmetric or skew-symmetric according as \(A\) is symmetric or skew-symmetric.
Solution:
Case 1: \(A\) is symmetric, so \(A^{\prime}=A\).
\((B^{\prime}AB)^{\prime}=B^{\prime}A^{\prime}(B^{\prime})^{\prime}=B^{\prime}AB\)
\(\therefore B^{\prime}AB\) is symmetric.
Case 2: \(A\) is skew-symmetric, so \(A^{\prime}=-A\).
\((B^{\prime}AB)^{\prime}=B^{\prime}A^{\prime}B=B^{\prime}(-A)B=-B^{\prime}AB\)
\(\therefore B^{\prime}AB\) is skew-symmetric.
Hence \(B^{\prime}AB\) inherits the symmetric or skew-symmetric nature of \(A\). ✓
Case 1: \(A\) is symmetric, so \(A^{\prime}=A\).
\((B^{\prime}AB)^{\prime}=B^{\prime}A^{\prime}(B^{\prime})^{\prime}=B^{\prime}AB\)
\(\therefore B^{\prime}AB\) is symmetric.
Case 2: \(A\) is skew-symmetric, so \(A^{\prime}=-A\).
\((B^{\prime}AB)^{\prime}=B^{\prime}A^{\prime}B=B^{\prime}(-A)B=-B^{\prime}AB\)
\(\therefore B^{\prime}AB\) is skew-symmetric.
Hence \(B^{\prime}AB\) inherits the symmetric or skew-symmetric nature of \(A\). ✓
4. For what values of \(x\),
\(\begin{bmatrix}1&2&1\end{bmatrix}\begin{bmatrix}1&2&0\\2&0&1\\1&0&2\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix}=O\)?
Solution:
\(\begin{bmatrix}1&2&1\end{bmatrix}\begin{bmatrix}1&2&0\\2&0&1\\1&0&2\end{bmatrix}=\begin{bmatrix}6&2&4\end{bmatrix}\)
\(\begin{bmatrix}6&2&4\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix}=[0+4+4x]=[4+4x]=O\)
\(\Rightarrow 4+4x=0\Rightarrow x=-1\)
\(\begin{bmatrix}1&2&1\end{bmatrix}\begin{bmatrix}1&2&0\\2&0&1\\1&0&2\end{bmatrix}=\begin{bmatrix}6&2&4\end{bmatrix}\)
\(\begin{bmatrix}6&2&4\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix}=[0+4+4x]=[4+4x]=O\)
\(\Rightarrow 4+4x=0\Rightarrow x=-1\)
5. If \(A=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\), show that \(A^2-5A+7I=O\).
Solution:
\(A^2=\begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}\)
\(A^2-5A+7I=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O\) ✓
\(A^2=\begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}\)
\(A^2-5A+7I=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O\) ✓
6. Find \(x\), if \(\begin{bmatrix}x&-5&-1\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix}=O\).
Solution:
\(\begin{bmatrix}x-2&-10&2x-8\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix}=O\)
\(\Rightarrow x(x-2)-40+2x-8=0\)
\(\Rightarrow x^2-48=0\Rightarrow x=\pm4\sqrt{3}\)
\(\begin{bmatrix}x-2&-10&2x-8\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix}=O\)
\(\Rightarrow x(x-2)-40+2x-8=0\)
\(\Rightarrow x^2-48=0\Rightarrow x=\pm4\sqrt{3}\)
7. A manufacturer sells products \(x,y,z\) in two markets with annual sales:
Market I: 10000, 2000, 18000 Market II: 6000, 20000, 8000.
(a) If unit prices are Rs 2.50, Rs 1.50, Rs 1.00, find total revenue in each market.
(b) If unit costs are Rs 2.00, Rs 1.00, Rs 0.50, find gross profit.
Market I: 10000, 2000, 18000 Market II: 6000, 20000, 8000.
(a) If unit prices are Rs 2.50, Rs 1.50, Rs 1.00, find total revenue in each market.
(b) If unit costs are Rs 2.00, Rs 1.00, Rs 0.50, find gross profit.
Solution:
(a) Revenue Market I: \(\begin{bmatrix}10000&2000&18000\end{bmatrix}\begin{bmatrix}2.50\\1.50\\1.00\end{bmatrix}=25000+3000+18000=\text{Rs }46000\)
Revenue Market II: \(\begin{bmatrix}6000&20000&8000\end{bmatrix}\begin{bmatrix}2.50\\1.50\\1.00\end{bmatrix}=15000+30000+8000=\text{Rs }53000\)
(b) Cost Market I: \(\begin{bmatrix}10000&2000&18000\end{bmatrix}\begin{bmatrix}2.00\\1.00\\0.50\end{bmatrix}=20000+2000+9000=\text{Rs }31000\)
Gross profit Market I: Rs \(46000-31000=\text{Rs }15000\)
Cost Market II: \(\begin{bmatrix}6000&20000&8000\end{bmatrix}\begin{bmatrix}2.00\\1.00\\0.50\end{bmatrix}=12000+20000+4000=\text{Rs }36000\)
Gross profit Market II: Rs \(53000-36000=\text{Rs }17000\)
(a) Revenue Market I: \(\begin{bmatrix}10000&2000&18000\end{bmatrix}\begin{bmatrix}2.50\\1.50\\1.00\end{bmatrix}=25000+3000+18000=\text{Rs }46000\)
Revenue Market II: \(\begin{bmatrix}6000&20000&8000\end{bmatrix}\begin{bmatrix}2.50\\1.50\\1.00\end{bmatrix}=15000+30000+8000=\text{Rs }53000\)
(b) Cost Market I: \(\begin{bmatrix}10000&2000&18000\end{bmatrix}\begin{bmatrix}2.00\\1.00\\0.50\end{bmatrix}=20000+2000+9000=\text{Rs }31000\)
Gross profit Market I: Rs \(46000-31000=\text{Rs }15000\)
Cost Market II: \(\begin{bmatrix}6000&20000&8000\end{bmatrix}\begin{bmatrix}2.00\\1.00\\0.50\end{bmatrix}=12000+20000+4000=\text{Rs }36000\)
Gross profit Market II: Rs \(53000-36000=\text{Rs }17000\)
8. Find the matrix \(X\) so that \(X\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}\).
Solution:
\(X\) must be \(2\times2\). Let \(X=\begin{bmatrix}a&c\\b&d\end{bmatrix}\).
\(\begin{bmatrix}a+4c&2a+5c&3a+6c\\b+4d&2b+5d&3b+6d\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}\)
From \(a+4c=-7\) and \(2a+5c=-8\): solving gives \(c=-2,\ a=1\).
From \(b+4d=2\) and \(2b+5d=4\): solving gives \(d=0,\ b=2\).
\(\therefore X=\begin{bmatrix}1&-2\\2&0\end{bmatrix}\)
\(X\) must be \(2\times2\). Let \(X=\begin{bmatrix}a&c\\b&d\end{bmatrix}\).
\(\begin{bmatrix}a+4c&2a+5c&3a+6c\\b+4d&2b+5d&3b+6d\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}\)
From \(a+4c=-7\) and \(2a+5c=-8\): solving gives \(c=-2,\ a=1\).
From \(b+4d=2\) and \(2b+5d=4\): solving gives \(d=0,\ b=2\).
\(\therefore X=\begin{bmatrix}1&-2\\2&0\end{bmatrix}\)
9. If \(A=\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}\) is such that \(A^2=I\), then:
A. \(1+\alpha^2+\beta\gamma=0\) B. \(1-\alpha^2+\beta\gamma=0\) C. \(1-\alpha^2-\beta\gamma=0\) D. \(1+\alpha^2-\beta\gamma=0\)
A. \(1+\alpha^2+\beta\gamma=0\) B. \(1-\alpha^2+\beta\gamma=0\) C. \(1-\alpha^2-\beta\gamma=0\) D. \(1+\alpha^2-\beta\gamma=0\)
Solution: (C)
\(A^2=\begin{bmatrix}\alpha^2+\beta\gamma&0\\0&\beta\gamma+\alpha^2\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
\(\Rightarrow\alpha^2+\beta\gamma=1\Rightarrow1-\alpha^2-\beta\gamma=0\)
\(A^2=\begin{bmatrix}\alpha^2+\beta\gamma&0\\0&\beta\gamma+\alpha^2\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
\(\Rightarrow\alpha^2+\beta\gamma=1\Rightarrow1-\alpha^2-\beta\gamma=0\)
10. If the matrix \(A\) is both symmetric and skew-symmetric, then:
A. \(A\) is a diagonal matrix B. \(A\) is a zero matrix C. \(A\) is a square matrix D. None of these
A. \(A\) is a diagonal matrix B. \(A\) is a zero matrix C. \(A\) is a square matrix D. None of these
Solution: (B)
If \(A^{\prime}=A\) and \(A^{\prime}=-A\), then \(A=-A\Rightarrow2A=O\Rightarrow A=O\).
Therefore \(A\) must be the zero matrix.
If \(A^{\prime}=A\) and \(A^{\prime}=-A\), then \(A=-A\Rightarrow2A=O\Rightarrow A=O\).
Therefore \(A\) must be the zero matrix.
11. If \(A\) is a square matrix such that \(A^2=A\), then \((I+A)^3-7A\) is equal to:
A. \(A\) B. \(I-A\) C. \(I\) D. \(3A\)
A. \(A\) B. \(I-A\) C. \(I\) D. \(3A\)
Solution: (C)
\((I+A)^3-7A=I+A^3+3A+3A^2-7A\)
\(=I+A^2\cdot A+3A+3A^2-7A\)
\(=I+A\cdot A+3A+3A-7A\quad(\because A^2=A)\)
\(=I+A^2-A=I+A-A=I\)
\((I+A)^3-7A=I+A^3+3A+3A^2-7A\)
\(=I+A^2\cdot A+3A+3A^2-7A\)
\(=I+A\cdot A+3A+3A-7A\quad(\because A^2=A)\)
\(=I+A^2-A=I+A-A=I\)
