Chapter 3: Matrices

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 $a_{13}, a_{21}, a_{33}, a_{24}, a_{23}$

Solution:
(i) Take a quick look at the matrix — you can count 3 rows running horizontally and 4 columns running vertically.
Therefore, 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 in it.
(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 order it can have? What, if it has 13 elements?

Solution:
Recall that a matrix of order $m \times n$ contains exactly $m n$ elements. So, to figure out every possible order for a matrix with 24 elements, we need to list all ordered pairs of natural numbers that multiply together to give 24.
The ordered pairs are: $(1,24),(24,1),(2,12),(12,2),(3,8),(8,3),(4,6)$, and $(6,4)$
Hence, the possible orders of a matrix having 24 elements are:
$1 \times 24,24 \times 1,2 \times 12,12 \times 2,3 \times 8,8 \times 3,4 \times 6$, and $6 \times 4$
Since 13 is a prime number, the only ordered pairs of natural numbers whose product is 13 are $(1,13)$ and $(13,1)$.
Hence, the possible orders of a matrix having 13 elements are $1 \times 13$ and $13 \times 1$.

3. If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

Solution:
Recall that a matrix of order $m \times n$ contains exactly $m n$ elements. So, to figure out every possible order for a matrix with 18 elements, we need to list all ordered pairs of natural numbers that multiply together to give 18.
The ordered pairs are: $(1,18),(18,1),(2,9),(9,2),(3,6$,$)$, and $(6,3)$
Hence, the possible orders of a matrix having 18 elements are:
$1 \times 18,18 \times 1,2 \times 9,9 \times 2,3 \times 6$, and $6 \times 3$
Since 5 is a prime number, the only ordered pairs of natural numbers whose product is 5 are $(1,5)$ and $(5,1)$.
Hence, the possible orders of a matrix having 5 elements are $1 \times 5$ and $5 \times 1$.

4. Construct a $2 \times 2$ matrix, $A=[a_ij]$, whose elements are given by
(i) $a_{i j}=\frac{(i+j)^2}{2}$
(ii) $a_{i j}=\frac{i}{j}$
(iii) $a_{i j}=\frac{(i+2 j)^2}{2}$

Solution:
In general, a $2 \times 2$ matrix is given by $A=\begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix}$
(i) $a_{i j}=\frac{(i+j)^2}{2}$, $i=1,2$ and $j=1,2$
therefore $a_{11}=\frac{(1+1)^2}{2}=\frac{(2)^2}{2}=\frac{4}{2}=2$
$a_{12}=\frac{(1+2)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2}$
$a_{21}=\frac{(2+1)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2}$
$a_{22}=\frac{(2+2)^2}{2}=\frac{(4)^2}{2}=\frac{16}{2}=8$
Therefore, the required matrix is $A=\begin{bmatrix}2 & \frac{9}{2} \\ \frac{9}{2} & 8\end{bmatrix}$
(ii) $a_{i j}=\frac{i}{j}$, $i=1,2$ and $j=1,2$
therefore $a_{11}=\frac{1}{1}=1$
$a_{12}=\frac{1}{2}$
$a_{21}=\frac{2}{1}=2$
$a_{22}=\frac{2}{2}=1$
Therefore, the required matrix is $A=\begin{bmatrix}1 & \frac{1}{2} \\ 2 & 1\end{bmatrix}$
{ (iii) } $a_{i j}=\frac{(i+2 j)^2}{2}$, $i=1,2$ and $j=1,2$
therefore $a_{11}=\frac{(1+2(1))^2}{2}=\frac{(3)^2}{2}=\frac{9}{2}$
$a_{12}=\frac{(1+2(2))^2}{2}=\frac{(5)^2}{2}=\frac{25}{2}$
$a_{21}=\frac{(2+2(1))^2}{2}=\frac{(4)^2}{2}=\frac{16}{2}=8$
$a_{22}=\frac{(2+2(2))^2}{2}=\frac{(6)^2}{2}=\frac{36}{2}=18$
Therefore, the required matrix is $A=\begin{bmatrix}
\frac{9}{2} & \frac{25}{2} \\ 8 & 18 \end{bmatrix}$

5. Construct a $3 \times 4$ matrix, whose elements are given by
(i) $a_{i j}=\frac{1}{2}|-3 i+j|$
(ii) $a_{i j}=2 i-j$

Solution:
To begin, let’s write out the general structure of a $3 \times 4$ matrix: $A=\left[\begin{array}{llll}a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34}\end{array}\right]$
(i) $\quad a_{i j}=\frac{1}{2}|-3 i+j|, i=1,2,3$ and $j=1,2,3,4$
Now we substitute each valid combination of $i$ and $j$ into the formula to find every element one by one:
$a_{11}=\frac{1}{2}|-3 \times 1+1|=\frac{1}{2}|-3+1|=\frac{1}{2}|-2|=\frac{2}{2}=1$
$a_{21}=\frac{1}{2}|-3 \times 2+1|=\frac{1}{2}|-6+1|=\frac{1}{2}|-5|=\frac{5}{2}$
$a_{31}=\frac{1}{2}|-3 \times 3+1|=\frac{1}{2}|-9+1|=\frac{1}{2}|-8|=\frac{8}{2}=4$
$a_{12}=\frac{1}{2}|-3 \times 1+2|=\frac{1}{2}|-3+2|=\frac{1}{2}|-1|=\frac{1}{2}$
$a_{22}=\frac{1}{2}|-3 \times 2+2|=\frac{1}{2}|-6+2|=\frac{1}{2}|-4|=\frac{4}{2}=2$
$a_{32}=\frac{1}{2}|-3 \times 3+2|=\frac{1}{2}|-9+2|=\frac{1}{2}|-7|=\frac{7}{2}$
$a_{13}=\frac{1}{2}|-3 \times 1+3|=\frac{1}{2}|-3+3|=0$
$a_{23}=\frac{1}{2}|-3 \times 2+3|=\frac{1}{2}|-6+3|=\frac{1}{2}|-3|=\frac{3}{2}$
$a_{33}=\frac{1}{2}|-3 \times 3+3|=\frac{1}{2}|-9+3|=\frac{1}{2}|-6|=\frac{6}{2}=3$
$a_{14}=\frac{1}{2}|-3 \times 1+4|=\frac{1}{2}|-3+4|=\frac{1}{2}|1|=\frac{1}{2}$
$a_{24}=\frac{1}{2}|-3 \times 2+4|=\frac{1}{2}|-6+4|=\frac{1}{2}|-2|=\frac{2}{2}=1$
$a_{34}=\frac{1}{2}|-3 \times 3+4|=\frac{1}{2}|-9+4|=\frac{1}{2}|-5|=\frac{5}{2}$
Therefore, the required matrix is $A=\left[\begin{array}{llll}1 & \frac{1}{2} & 0 & \frac{1}{2} \\
\frac{5}{2} & \frac{1}{2} & 0 & \frac{1}{2} \\
4 & \frac{7}{2} & 3 & \frac{5}{2}
\end{array}\right]$
(ii) $a_{i j}=2 i-j, i=1,2,3$ and $j=1,2,3,4$
Again, we plug in each combination of $i$ and $j$ into the formula step by step:
therefore
$a_{11}=2 \times 1-1=2-1=1$
$a_{21}=2 \times 2-1=4-1=3$
$a_{31}=2 \times 3-1=6-1=5$
$a_{12}=2 \times 1-2=2-2=0$
$a_{22}=2 \times 2-2=4-2=2$
$a_{32}=2 \times 3-2=6-2=4$
$a_{13}=2 \times 1-3=2-3=-1$
$a_{23}=2 \times 2-3=4-3=1$
$a_{33}=2 \times 3-3=6-3=3$
$a_{14}=2 \times 1-4=2-4=-2$
$a_{24}=2 \times 2-4=4-4=0$
$a_{34}=2 \times 3-4=6-4=2$
Therefore, the required matrix is $A=\begin{bmatrix}1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2\end{bmatrix}$

6. Find the value of $x, y$, and $z$ from the following equation:
(i) $\left[\begin{array}{ll}4 & 3 \\ x & 5\end{array}\right]=\left[\begin{array}{ll}y & z \\ 1 & 5\end{array}\right]$ (ii) $\left[\begin{array}{ll}x+y & 2 \\ 5+z & x y\end{array}\right]=\left[\begin{array}{ll}6 & 2 \\ 5 & 8\end{array}\right]$
(iii) $\left[\begin{array}{c}x+y+z \\ x+z \\ y+z\end{array}\right]=\left[\begin{array}{l}9 \\ 5 \\ 7\end{array}\right]$

Solution:
(i) $\left[\begin{array}{ll}4 & 3 \\ x & 5\end{array}\right]=\left[\begin{array}{ll}y & z \\ 1 & 5\end{array}\right]$
Two matrices are equal only when every element in the same position matches across both matrices. Matching up the corresponding positions gives us:
$x=1, y=4$, and $z=3$
(ii) $\left[\begin{array}{ll}x+y & 2 \\ 5+z & x y\end{array}\right]=\left[\begin{array}{ll}6 & 2 \\ 5 & 8\end{array}\right]$
Two matrices are equal only when every element in the same position matches across both matrices. Matching up the corresponding positions gives us:
$x+y=6, x y=8,5+z=5$
Now, $5+z=5 \Rightarrow z=0$
To find x and y, we use the useful algebraic identity:
$(x-y)^{2}=(x+y)^{2}-4 x y$
$\Rightarrow(x-y)^{2}=36-32=4$
$\Rightarrow x-y= \pm 2$
Now, when $x-y=2$ and $x+y=6$, we get $x=4$ and $y=2$
When $x-y=-2$ and $x+y=6$, we get $x=2$ and $y=4$
$\therefore x=4, y=2$, and $z=0$ or $x=2, y=4$, and $z=0$
(iii) $\left[\begin{array}{c}x+y+z \\ x+z \\ y+z\end{array}\right]=\left[\begin{array}{l}9 \\ 5 \\ 7\end{array}\right]$
Since the two matrices are equal, their elements at each position must be identical. Setting the corresponding elements equal to each other gives us the following system of equations:
$x+y+z=9 … (1)$,
$x+z=5 … (2)$,
$y+z=7 … (3)$
From (1) and (2), we substitute equation (2) directly into equation (1):
$y+5=9$
$\Rightarrow y=4$
Then, from (3), we substitute the value of y:
$4+z=7$
$\Rightarrow z=3$
$\therefore x+z=5$
$\Rightarrow x=2$
$\therefore x=2, y=4$, and $z=3$

7. Find the value of $a, b, c$, and $d$ from the equation:
$\left[\begin{array}{ll}a-b & 2 a+c \\ 2 a-b & 3 c+d\end{array}\right]=\left[\begin{array}{ll}-1 & 5 \\ 0 & 13\end{array}\right]$

Solution:
$\left[\begin{array}{ll}a-b & 2 a+c \\ 2 a-b & 3 c+d\end{array}\right]=\left[\begin{array}{ll}-1 & 5 \\ 0 & 13\end{array}\right]$
Since the two matrices are equal, their elements at each position must be identical. Setting the corresponding elements equal to each other gives us the following system of equations:
$a-b=-1 … (1)$,
$2 a-b=0 … (2)$,
$2a+c=5 … (3)$,
$3c+d=13 … (4)$
From (2), we can directly express b in terms of a:
$b=2 a$
Substituting this into (1), we get:
$a-2 a=-1$
$\Rightarrow a=1$
$\Rightarrow b=2$
Now, substituting the value of a into (3):
$2 \times 1+c=5$
$\Rightarrow c=3$
Finally, substituting the value of c into (4):
$3 \times 3+d=13$
$\Rightarrow 9+d=13 \Rightarrow d=4$
$\therefore a=1, b=2, c=3$, and $d=4$

8. $A=\left[a_{i j}\right]_{m \times n}$ is a square matrix, if
(A) $mn$
(C) $m=n$
(D) None of these

Solution:
The correct answer is C.
A matrix earns the title “square matrix” only when its number of rows and number of columns are exactly the same.
Therefore, $A=\left[a_{i j}\right]_{m \times n}$ is a square matrix, if $m=n$.

9. Which of the given values of $x$ and $y$ make the following pair of matrices equal
$\left[\begin{array}{ll}3 x+7 & 5 \\ y+1 & 2-3 x\end{array}\right]=\left[\begin{array}{ll}0 & y-2 \\ 8 & 4\end{array}\right]$
(A) $x=\frac{-1}{3}, y=7$
(B) Not possible to find
(C) $y=7, x=\frac{-2}{3}$
(D) $x=\frac{-1}{3}, y=\frac{-2}{3}$

Solution:
The correct answer is B.
Starting with the given equation $\left[\begin{array}{ll}3 x+7 & 5 \\ y+1 & 2-3 x\end{array}\right]=\left[\begin{array}{ll}0 & y-2 \\ 8 & 4\end{array}\right]$, we match up the element in each position and solve:
$3 x+7=0 \Rightarrow x=-\frac{7}{3}$
$5=y-2 \Rightarrow y=7$
$y+1=8 \Rightarrow y=7$
$2-3 x=4 \Rightarrow x=-\frac{2}{3}$
Notice the contradiction: the first equation gives $x=-\frac{7}{3}$, while the fourth equation gives $x=-\frac{2}{3}$. A variable cannot hold two different values at the same time, so no consistent solution exists.
Hence, it is not possible to find the values of $x$ and $y$ for which the given matrices are equal.

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 correct answer is D.
A $3 \times 3$ matrix has a total of 9 positions, and each position can independently hold one of just two values: 0 or 1.
Since every one of the 9 positions has 2 independent choices, we apply the multiplication principle across all positions.
Therefore, by the multiplication principle, the required number of possible matrices is $2^{9} =512$