Vardaan Learning Institute

Write the order of the matrix $A = \begin{bmatrix} \sqrt{2} & 5 & -1 \\ 3 & 0 & 4 \\ 1 & 8 & \frac{1}{2} \end{bmatrix}$.
Answer: $3 \times 3$ (3 rows and 3 columns)

If a matrix has $12$ elements, list all possible orders it can have.
Answer: $1 \times 12, 12 \times 1, 2 \times 6, 6 \times 2, 3 \times 4, 4 \times 3$

If a matrix has $5$ elements, what are the possible orders it can have?
Answer: $1 \times 5$ and $5 \times 1$

In the matrix $A = \begin{bmatrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{bmatrix}$, find the elements $a_{13}, a_{21}, a_{33}, a_{24}$.
Answer: $a_{13} = 19$, $a_{21} = 35$, $a_{33} = -5$, $a_{24} = 12$

Construct a $2 \times 2$ matrix $A = [a_{ij}]$ whose elements are given by $a_{ij} = \frac{(i+j)^2}{2}$.
Answer: $A = \begin{bmatrix} \frac{(1+1)^2}{2} & \frac{(1+2)^2}{2} \\ \frac{(2+1)^2}{2} & \frac{(2+2)^2}{2} \end{bmatrix} = \begin{bmatrix} 2 & \frac{9}{2} \\ \frac{9}{2} & 8 \end{bmatrix}$

Construct a $2 \times 3$ matrix whose elements are given by $a_{ij} = i - 2j$.
Answer: $A = \begin{bmatrix} (1-2) & (1-4) & (1-6) \\ (2-2) & (2-4) & (2-6) \end{bmatrix} = \begin{bmatrix} -1 & -3 & -5 \\ 0 & -2 & -4 \end{bmatrix}$

Construct a $3 \times 2$ matrix whose elements are given by $a_{ij} = \frac{1}{2} |i - 3j|$.
Answer: $A = \begin{bmatrix} \frac{1}{2}|1-3| & \frac{1}{2}|1-6| \\ \frac{1}{2}|2-3| & \frac{1}{2}|2-6| \\ \frac{1}{2}|3-3| & \frac{1}{2}|3-6| \end{bmatrix} = \begin{bmatrix} 1 & \frac{5}{2} \\ \frac{1}{2} & 2 \\ 0 & \frac{3}{2} \end{bmatrix}$

For a square matrix of order $n$, what is the relationship between the indices $i$ and $j$ for the elements lying on its principal diagonal?
Answer: $i = j$

Classify the matrix $A = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}$ as diagonal, scalar, or identity matrix.
Answer: Scalar Matrix (It is also diagonal, but "scalar" is the most specific and accurate classification here).

Give an example of a row matrix which is also a column matrix.
Answer: A matrix of order $1 \times 1$. For example, $\begin{bmatrix} 5 \end{bmatrix}$.

Find the trace of the matrix $B = \begin{bmatrix} -1 & 5 & 6 \\ 2 & 3 & 1 \\ 4 & -2 & 8 \end{bmatrix}$.
Answer: Trace = sum of diagonal elements = $-1 + 3 + 8 = 10$.

If $A$ is a scalar matrix of order 3 and Trace$(A) = 15$, find the matrix $A$.
Answer: Let diagonal elements be $k, k, k$. Trace = $3k = 15 \Rightarrow k = 5$. Thus, $A = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$.

Show that the matrix $A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is an idempotent matrix.
Answer: For idempotent, $A^2 = A$. Here, $A^2 = I \times I = I = A$. Hence proved.

Verify if the matrix $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ is nilpotent. If yes, what is its index?
Answer: $A^2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$. Yes, it is nilpotent. The index is $2$.

A matrix $A$ is called involutory if $A^2 = I$. Verify if $A = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$ is involutory.
Answer: $A^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$. Yes, it is involutory.

What is the condition for a square matrix $A$ to be orthogonal?
Answer: $AA' = A'A = I$

Find the values of $x, y$ and $z$ if $\begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix} = \begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}$.
Answer: By comparing corresponding elements: $y=4, z=3, x=1$.

Find $x, y, z$ from the equation: $\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}$.
Answer: $5+z = 5 \Rightarrow z = 0$. $x+y = 6$ and $xy = 8$. Solving these gives $x=2, y=4$ or $x=4, y=2$.

Find the values of $a, b, c$ and $d$ from the equation: $\begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}$.
Answer: $a-b = -1$, $2a-b = 0 \Rightarrow a = 1, b = 2$. $2a+c = 5 \Rightarrow 2(1)+c=5 \Rightarrow c=3$. $3c+d = 13 \Rightarrow 9+d=13 \Rightarrow d=4$.

If $\begin{bmatrix} x+y+z \\ x+z \\ y+z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}$, find the values of $x, y,$ and $z$.
Answer: Substitute $y+z=7$ into first eq: $x+7=9 \Rightarrow x=2$. Then $x+z=5 \Rightarrow 2+z=5 \Rightarrow z=3$. Finally $y+3=7 \Rightarrow y=4$.

Find the value of $x$ and $y$ if $2 \begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$.
Answer: $2x+3 = 7 \Rightarrow 2x=4 \Rightarrow x=2$. $2(y-3)+2 = 14 \Rightarrow 2y-6+2 = 14 \Rightarrow 2y=18 \Rightarrow y=9$.

Determine $x$ and $y$ such that $\begin{bmatrix} 2x-y \\ x+2y \end{bmatrix} = \begin{bmatrix} 5 \\ 5 \end{bmatrix}$.
Answer: Eq 1: $2x-y = 5$, Eq 2: $x+2y = 5$. Multiply Eq 1 by 2: $4x-2y=10$. Add to Eq 2: $5x=15 \Rightarrow x=3$. Substitute $x=3$: $3+2y=5 \Rightarrow y=1$.

Are the matrices $A = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ equal? Give a reason.
Answer: No. Their orders are different. Matrix $A$ is $1 \times 3$ while matrix $B$ is $3 \times 1$.

Given $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$, find $A+B$.
Answer: $\begin{bmatrix} 2+1 & 4+3 \\ 3-2 & 2+5 \end{bmatrix} = \begin{bmatrix} 3 & 7 \\ 1 & 7 \end{bmatrix}$

Given the same matrices $A$ and $B$ from Q24, find $A-B$.
Answer: $\begin{bmatrix} 2-1 & 4-3 \\ 3-(-2) & 2-5 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix}$

If $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$ and $C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$, find $3A - C$.
Answer: $\begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix} - \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix}$

Compute the product: $\begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$.
Answer: $\begin{bmatrix} a(a)+b(b) & a(-b)+b(a) \\ -b(a)+a(b) & -b(-b)+a(a) \end{bmatrix} = \begin{bmatrix} a^2+b^2 & 0 \\ 0 & a^2+b^2 \end{bmatrix}$

Compute the product: $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \end{bmatrix}$.
Answer: $\begin{bmatrix} 1(2) & 1(3) & 1(4) \\ 2(2) & 2(3) & 2(4) \\ 3(2) & 3(3) & 3(4) \end{bmatrix} = \begin{bmatrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{bmatrix}$

Compute the product: $\begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}$.
Answer: $\begin{bmatrix} 1-4 & 2-6 & 3-2 \\ 2+6 & 4+9 & 6+3 \end{bmatrix} = \begin{bmatrix} -3 & -4 & 1 \\ 8 & 13 & 9 \end{bmatrix}$

If $A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 2 \\ 4 \end{bmatrix}$, evaluate $AB$. Can you evaluate $BA$?
Answer: $AB = \begin{bmatrix} 1(2) - 1(4) \\ 2(2) + 3(4) \end{bmatrix} = \begin{bmatrix} -2 \\ 16 \end{bmatrix}$. No, $BA$ is not defined (orders are $2 \times 1$ and $2 \times 2$, inner dimensions don't match).

Find matrix $X$ if $Y = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$ and $2X + Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$.
Answer: $2X = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix}$. Thus $X = \begin{bmatrix} -1 & -1 \\ -2 & -1 \end{bmatrix}$.

If $X + Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}$ and $X - Y = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$, find matrices $X$ and $Y$.
Answer: Adding the two eq: $2X = \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix} \Rightarrow X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix}$. Subtracting: $2Y = \begin{bmatrix} 4 & 0 \\ 2 & 2 \end{bmatrix} \Rightarrow Y = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$.

Find the transpose of the matrix $A = \begin{bmatrix} 5 \\ \frac{1}{2} \\ -1 \end{bmatrix}$.
Answer: $A' = \begin{bmatrix} 5 & \frac{1}{2} & -1 \end{bmatrix}$

Find the transpose of the matrix $B = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$.
Answer: $B' = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}$

If $A = \begin{bmatrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}$, verify that $(A+B)' = A' + B'$.
Answer: $A+B = \begin{bmatrix} -5 & 6 & 1 \\ \sqrt{3}+1 & 7 & 6 \\ 3 & 6 & 0 \end{bmatrix} \Rightarrow (A+B)' = \begin{bmatrix} -5 & \sqrt{3}+1 & 3 \\ 6 & 7 & 6 \\ 1 & 6 & 0 \end{bmatrix}$. Computing $A' + B'$ yields the same result.

If $A' = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix}$, find $(A-B)'$.
Answer: $(A-B)' = A' - B'$. Here $B' = \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix}$. Thus, $A' - B' = \begin{bmatrix} 3-(-1) & 4-1 \\ -1-2 & 2-2 \\ 0-1 & 1-3 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}$.

If $A = \begin{bmatrix} -2 \\ 4 \\ 5 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 3 & -6 \end{bmatrix}$, verify the reversal law: $(AB)' = B'A'$.
Answer: $AB = \begin{bmatrix} -2 & -6 & 12 \\ 4 & 12 & -24 \\ 5 & 15 & -30 \end{bmatrix}$. $(AB)' = \begin{bmatrix} -2 & 4 & 5 \\ -6 & 12 & 15 \\ 12 & -24 & -30 \end{bmatrix}$. $B'A' = \begin{bmatrix} 1 \\ 3 \\ -6 \end{bmatrix} \begin{bmatrix} -2 & 4 & 5 \end{bmatrix}$ which yields the same matrix. Verified.

If $A = \begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix}$, show that $(3A)' = 3A'$.
Answer: $3A = \begin{bmatrix} 6 & -3 \\ 12 & 9 \end{bmatrix} \Rightarrow (3A)' = \begin{bmatrix} 6 & 12 \\ -3 & 9 \end{bmatrix}$. Also, $A' = \begin{bmatrix} 2 & 4 \\ -1 & 3 \end{bmatrix} \Rightarrow 3A' = \begin{bmatrix} 6 & 12 \\ -3 & 9 \end{bmatrix}$. Verified.

If $A$ is a matrix of order $m \times n$, what is the order of $A'$?
Answer: $n \times m$

Show that the matrix $A = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix}$ is a symmetric matrix.
Answer: $A' = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix}$. Since $A' = A$, the matrix is symmetric.

Show that the matrix $A = \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix}$ is a skew-symmetric matrix.
Answer: $A' = \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix} = - \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix} = -A$. Hence, it is skew-symmetric.

For the matrix $A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$, verify that $(A + A')$ is a symmetric matrix.
Answer: $A' = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}$. Let $P = A + A' = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}$. $P' = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix} = P$. Verified.

For the same matrix $A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$, verify that $(A - A')$ is a skew-symmetric matrix.
Answer: Let $Q = A - A' = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. $Q' = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = -Q$. Verified.

Find the values of $a$ and $b$ if the matrix $\begin{bmatrix} 0 & a \\ -3 & 0 \end{bmatrix}$ is a skew-symmetric matrix.
Answer: For skew-symmetric, $A' = -A$. $\begin{bmatrix} 0 & -3 \\ a & 0 \end{bmatrix} = \begin{bmatrix} 0 & -a \\ 3 & 0 \end{bmatrix}$. Comparing gives $a = 3$.

Find the value of $x$ for which the matrix $A = \begin{bmatrix} 2 & x \\ 4 & 5 \end{bmatrix}$ is symmetric.
Answer: For symmetric, $A = A'$. $\begin{bmatrix} 2 & x \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ x & 5 \end{bmatrix}$. By equality, $x = 4$.

Express the matrix $A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.
Answer: $P = \frac{1}{2}(A+A') = \frac{1}{2} \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}$. $Q = \frac{1}{2}(A-A') = \frac{1}{2} \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$. Thus, $A = P + Q = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$.

If $A$ and $B$ are symmetric matrices of the same order, what can be said about $AB - BA$?
Answer: $(AB-BA)' = (AB)' - (BA)' = B'A' - A'B' = BA - AB = -(AB - BA)$. It is a skew-symmetric matrix.

Apply the elementary row operation $R_1 \leftrightarrow R_2$ on the matrix $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$.
Answer: Interchange Row 1 and Row 2. The result is $\begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix}$.

Apply the operation $C_2 \rightarrow C_2 - 2C_1$ on the matrix $B = \begin{bmatrix} 1 & 2 \\ 2 & 5 \end{bmatrix}$.
Answer: Column 1 stays same. Column 2 changes: $C_{12} = 2 - 2(1) = 0$, $C_{22} = 5 - 2(2) = 1$. Result is $\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$.

Apply the operation $R_2 \rightarrow R_2 - 3R_1$ to the matrix $\begin{bmatrix} 1 & 2 \\ 3 & 7 \end{bmatrix}$.
Answer: Result is $\begin{bmatrix} 1 & 2 \\ 3-3(1) & 7-3(2) \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$.

If $A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$, show that $AB = BA = I$.
Answer: $AB = \begin{bmatrix} 2(2)+3(-1) & 2(-3)+3(2) \\ 1(2)+2(-1) & 1(-3)+2(2) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$. $BA$ also equals $I$. Verified.

From Q51, what can we conclude about the relationship between matrix $A$ and matrix $B$?
Answer: Matrix $B$ is the inverse of matrix $A$ (and vice versa). i.e., $B = A^{-1}$.

Write the initial matrix equation required to find the inverse of $A = \begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix}$ using elementary row operations.
Answer: We write $A = IA \Rightarrow \begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} A$.

Does the inverse of every square matrix exist? (Yes/No)
Answer: No. The inverse only exists if the matrix is non-singular.

If $A$ and $B$ are invertible matrices of the same order, simplify $(AB)^{-1}$.
Answer: $B^{-1}A^{-1}$