Vardaan Learning Institute

Construct a $3 \times 4$ matrix whose elements are given by $a_{ij} = \frac{1}{2}|-3i + j|$.
Answer:
For $i=1$: $a_{11} = 1, a_{12} = \frac{1}{2}, a_{13} = 0, a_{14} = \frac{1}{2}$
For $i=2$: $a_{21} = \frac{5}{2}, a_{22} = 2, a_{23} = \frac{3}{2}, a_{24} = 1$
For $i=3$: $a_{31} = 4, a_{32} = \frac{7}{2}, a_{33} = 3, a_{34} = \frac{5}{2}$
Matrix $A = \begin{bmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2} \end{bmatrix}$.

Find the total number of possible matrices of order $3 \times 3$ with each entry being $0$ or $1$.
Answer: A $3 \times 3$ matrix has 9 elements. Each element can be either $0$ or $1$ (2 choices). Total possible matrices = $2^9 = 512$.

If a matrix has 24 elements, list all the possible orders it can have. What if it has 13 elements?
Answer: For 24 elements: $1 \times 24$, $24 \times 1$, $2 \times 12$, $12 \times 2$, $3 \times 8$, $8 \times 3$, $4 \times 6$, $6 \times 4$. For 13 elements (since 13 is prime): $1 \times 13$ and $13 \times 1$.

Construct a $2 \times 2$ matrix $A = [a_{ij}]$ whose elements are defined by $a_{ij} = e^{2ix} \sin(jx)$.
Answer: $A = \begin{bmatrix} e^{2(1)x}\sin(1x) & e^{2(1)x}\sin(2x) \\ e^{2(2)x}\sin(1x) & e^{2(2)x}\sin(2x) \end{bmatrix} = \begin{bmatrix} e^{2x}\sin x & e^{2x}\sin 2x \\ e^{4x}\sin x & e^{4x}\sin 2x \end{bmatrix}$.

Prove that the matrix $A = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}$ is an idempotent matrix.
Answer: We need to show $A^2 = A$.
$A^2 = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}$
$a_{11} = 4+2-4 = 2$; $a_{12} = -4-6+8 = -2$; $a_{13} = -8-8+12 = -4$
Continuing multiplication, we get exactly matrix $A$. Hence, $A^2 = A$.

Show that $B = \begin{bmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{bmatrix}$ is nilpotent of index 2.
Answer: We must show $B^2 = O$.
$B^2 = \begin{bmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{bmatrix} \begin{bmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{bmatrix}$.
$R_1 \cdot C_1 = (1)(1) + (-3)(-1) + (-4)(1) = 1 + 3 - 4 = 0$.
Computing all terms similarly gives $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$. Since $B^2 = O$ and $B \neq O$, it is nilpotent of index 2.

Find $x, y, z$ if the matrix $A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$ is an orthogonal matrix (i.e., $AA' = I$).
Answer: $AA' = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Row 1 $\times$ Col 1: $4y^2 + z^2 = 1$
Row 2 $\times$ Col 2: $x^2 + y^2 + z^2 = 1$
Row 3 $\times$ Col 3: $x^2 + y^2 + z^2 = 1$
Row 1 $\times$ Col 2: $2y^2 - z^2 = 0 \Rightarrow z^2 = 2y^2$.
Substitute in first eq: $4y^2 + 2y^2 = 1 \Rightarrow 6y^2 = 1 \Rightarrow y = \pm \frac{1}{\sqrt{6}}$.
Then $z^2 = 2(\frac{1}{6}) = \frac{1}{3} \Rightarrow z = \pm \frac{1}{\sqrt{3}}$.
Using $x^2 + y^2 + z^2 = 1 \Rightarrow x^2 + \frac{1}{6} + \frac{1}{3} = 1 \Rightarrow x^2 + \frac{1}{2} = 1 \Rightarrow x = \pm \frac{1}{\sqrt{2}}$.

If $A$ is a square matrix and $k$ is a scalar, express the trace of $kA$ in terms of the trace of $A$.
Answer: The diagonal elements of $kA$ are $ka_{11}, ka_{22}, \dots, ka_{nn}$.
$\text{Trace}(kA) = ka_{11} + ka_{22} + \dots + ka_{nn} = k(a_{11} + a_{22} + \dots + a_{nn}) = k \cdot \text{Trace}(A)$.

Find $x, y, a, b$ if $\begin{bmatrix} 2x-3y & a-b \\ 2a+b & x+4y \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 5 & 6 \end{bmatrix}$.
Answer: Equate corresponding elements:
(1) $2x - 3y = 1$ and (2) $x + 4y = 6$. Multiply (2) by 2: $2x + 8y = 12$. Subtract (1): $11y = 11 \Rightarrow y = 1$. Substitute into (2): $x + 4(1) = 6 \Rightarrow x = 2$.
(3) $a - b = -2$ and (4) $2a + b = 5$. Add them: $3a = 3 \Rightarrow a = 1$. Substitute into (3): $1 - b = -2 \Rightarrow b = 3$.
Final values: $x=2, y=1, a=1, b=3$.

Find $x$ and $y$ if $2 \begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$.
Answer: $\begin{bmatrix} 2+y & 6 \\ 1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$.
Equating: $2+y = 5 \Rightarrow y = 3$.
$2x+2 = 8 \Rightarrow 2x = 6 \Rightarrow x = 3$.

Solve for $x, y, z,$ and $t$: $2\begin{bmatrix} x & z \\ y & t \end{bmatrix} + 3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = 3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}$.
Answer: $\begin{bmatrix} 2x+3 & 2z-3 \\ 2y & 2t+6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}$.
$2x+3=9 \Rightarrow 2x=6 \Rightarrow x=3$.
$2z-3=15 \Rightarrow 2z=18 \Rightarrow z=9$.
$2y=12 \Rightarrow y=6$.
$2t+6=18 \Rightarrow 2t=12 \Rightarrow t=6$.

If $A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$, $B = \begin{bmatrix} 2 & 0 \\ 1 & 3 \end{bmatrix}$, and $C = \begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix}$, compute $(AB)C$ and verify that it equals $A(BC)$.
Answer: $AB = \begin{bmatrix} 1(2)+2(1) & 1(0)+2(3) \\ 2(2)+1(1) & 2(0)+1(3) \end{bmatrix} = \begin{bmatrix} 4 & 6 \\ 5 & 3 \end{bmatrix}$.
$(AB)C = \begin{bmatrix} 4 & 6 \\ 5 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 16 & 22 \\ 11 & 14 \end{bmatrix}$.
$BC = \begin{bmatrix} 2 & 0 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 7 & 10 \end{bmatrix}$.
$A(BC) = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 2 & 2 \\ 7 & 10 \end{bmatrix} = \begin{bmatrix} 16 & 22 \\ 11 & 14 \end{bmatrix}$.
Hence, $(AB)C = A(BC)$. Verified.

Let $f(x) = x^2 - 5x + 6$. Find $f(A)$ if $A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}$.
Answer: $f(A) = A^2 - 5A + 6I$.
$A^2 = A \cdot A = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix}$.
$5A = \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix}$, $6I = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}$.
$A^2 - 5A + 6I = \begin{bmatrix} 5-10+6 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+6 & 5-15+0 \\ 0-5+0 & -1-(-5)+0 & -2-0+6 \end{bmatrix} = \begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}$.

Find a matrix $X$ such that $X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$.
Answer: Order of $X$ must be $2 \times 2$. Let $X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} a+4b & 2a+5b & 3a+6b \\ c+4d & 2c+5d & 3c+6d \end{bmatrix}$.
Equating: $a+4b = -7$ and $2a+5b = -8$. Solving these yields $b = 2$, $a = -15$ Wait, recalculate: Multiply $a+4b=-7$ by 2 $\Rightarrow 2a+8b=-14$. Subtract $2a+5b=-8$ $\Rightarrow 3b = -6 \Rightarrow b = -2$. Then $a - 8 = -7 \Rightarrow a = 1$.
Similarly, $c+4d = 2$ and $2c+5d = 4$. Solving gives $d = 0$, $c = 2$.
$X = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}$.

If $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$, show that $A^2 = \begin{bmatrix} \cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{bmatrix}$.
Answer: $A^2 = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$.
$a_{11} = \cos^2 \alpha - \sin^2 \alpha = \cos 2\alpha$.
$a_{12} = -\cos\alpha\sin\alpha - \sin\alpha\cos\alpha = -2\sin\alpha\cos\alpha = -\sin 2\alpha$.
$a_{21} = \sin\alpha\cos\alpha + \cos\alpha\sin\alpha = 2\sin\alpha\cos\alpha = \sin 2\alpha$.
$a_{22} = -\sin^2 \alpha + \cos^2 \alpha = \cos 2\alpha$. Proved.

Find $x$ if $\begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} = O$.
Answer: $\begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix} = \begin{bmatrix} 1+2x+15 & 3+5x+3 & 2+x+2 \end{bmatrix} = \begin{bmatrix} 2x+16 & 5x+6 & x+4 \end{bmatrix}$.
Now, $\begin{bmatrix} 2x+16 & 5x+6 & x+4 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} = 0$.
$(2x+16)(1) + (5x+6)(2) + (x+4)(x) = 0 \Rightarrow 2x+16 + 10x+12 + x^2+4x = 0$.
$x^2 + 16x + 28 = 0 \Rightarrow (x+14)(x+2) = 0 \Rightarrow x = -2$ or $x = -14$.

Verify that $(AB)' = B'A'$ for $A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$.
Answer: $AB = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix} \Rightarrow (AB)' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$.
$B'A' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$. Verified.

If $A = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$, verify that $A'A = I$.
Answer: $A' = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$.
$A'A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix} \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix} = \begin{bmatrix} \sin^2\alpha + \cos^2\alpha & \sin\alpha\cos\alpha - \cos\alpha\sin\alpha \\ \cos\alpha\sin\alpha - \sin\alpha\cos\alpha & \cos^2\alpha + \sin^2\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.

For any square matrix $A$ with real numbers, prove that $(A+A')$ is a symmetric matrix.
Answer: Let $P = A + A'$. We must prove $P' = P$.
$P' = (A + A')' = A' + (A')'$ (Using $(X+Y)' = X' + Y'$).
Since $(A')' = A$, we have $P' = A' + A = A + A' = P$. Thus, $P$ is symmetric.

Using the properties of transpose, prove that $(A^n)' = (A')^n$ for any positive integer $n$.
Answer: We know the reversal law: $(AB)' = B'A'$.
For $n=2$, $(A^2)' = (AA)' = A'A' = (A')^2$.
Assume it is true for $n=k$, so $(A^k)' = (A')^k$.
For $n=k+1$, $(A^{k+1})' = (A^k \cdot A)' = A' \cdot (A^k)' = A' \cdot (A')^k = (A')^{k+1}$.
By Principle of Mathematical Induction, it holds true for all $n \in \mathbb{N}$.

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

Show that all positive odd integral powers of a skew-symmetric matrix are skew-symmetric.
Answer: Let $A$ be a skew-symmetric matrix, so $A' = -A$.
Let $n$ be a positive odd integer. Consider $(A^n)'$.
$(A^n)' = (A')^n = (-A)^n$. Since $n$ is odd, $(-A)^n = -A^n$.
Because $(A^n)' = -(A^n)$, the matrix $A^n$ is skew-symmetric.

Let $A$ and $B$ be symmetric matrices. Prove that $AB$ is symmetric if and only if $A$ and $B$ commute (i.e., $AB = BA$).
Answer: Given $A'=A$ and $B'=B$.
Assume $AB$ is symmetric, then $(AB)' = AB$. But by reversal law $(AB)' = B'A' = BA$. So $AB = BA$ (they commute).
Conversely, assume $AB=BA$. We need to show $AB$ is symmetric. $(AB)' = B'A' = BA = AB$. Thus, it is symmetric.

If $A$ and $B$ are symmetric matrices of the same order, prove that $AB + BA$ is a symmetric matrix.
Answer: Let $P = AB + BA$. We must prove $P' = P$.
$P' = (AB + BA)' = (AB)' + (BA)' = B'A' + A'B'$.
Since $A, B$ are symmetric, $A'=A$ and $B'=B$.
Thus, $P' = BA + AB = AB + BA = P$. Proved.

Using elementary transformations, find the inverse of the matrix $A = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}$.
Answer: Write $A = IA \Rightarrow \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}A$.
$R_2 \rightarrow R_2 - 2R_1 \Rightarrow \begin{bmatrix} 1 & 2 \\ 0 & -5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}A$.
$R_2 \rightarrow \frac{-1}{5}R_2 \Rightarrow \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2/5 & -1/5 \end{bmatrix}A$.
$R_1 \rightarrow R_1 - 2R_2 \Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 - 4/5 & 0 - (-2/5) \\ 2/5 & -1/5 \end{bmatrix}A = \begin{bmatrix} 1/5 & 2/5 \\ 2/5 & -1/5 \end{bmatrix}A$.
Inverse is $\frac{1}{5}\begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}$.

Obtain the inverse of the matrix $A = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$ using elementary row operations.
Answer: $A = IA \Rightarrow \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}A$.
$R_1 \leftrightarrow R_2 \Rightarrow \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}A$.
$R_2 \rightarrow R_2 - 2R_1 \Rightarrow \begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & -2 \end{bmatrix}A$.
$R_2 \rightarrow -R_2 \Rightarrow \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 2 \end{bmatrix}A$.
$R_1 \rightarrow R_1 - R_2 \Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}A$.
Inverse is $\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}$.

By using elementary column operations, find the inverse of $A = \begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix}$.
Answer: For column ops, $A = AI$. $\begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix} = A\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
$C_1 \rightarrow C_1 + 2C_2 \Rightarrow \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = A\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$.
$C_2 \rightarrow C_2 + C_1 \Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} = A\begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix}$.
$C_2 \rightarrow \frac{1}{2}C_2 \Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = A\begin{bmatrix} 1 & 1/2 \\ 2 & 3/2 \end{bmatrix}$.
Inverse is $\begin{bmatrix} 1 & 1/2 \\ 2 & 3/2 \end{bmatrix}$.

Prove that the inverse of an invertible symmetric matrix is a symmetric matrix.
Answer: Let $A$ be a symmetric invertible matrix, so $A' = A$. We need to prove $(A^{-1})' = A^{-1}$.
By definition, $A A^{-1} = I$. Taking transpose on both sides: $(A A^{-1})' = I' = I$.
By reversal law, $(A^{-1})' A' = I$.
Since $A' = A$, we have $(A^{-1})' A = I$.
Multiplying both sides by $A^{-1}$ on the right: $(A^{-1})' A A^{-1} = I A^{-1} \Rightarrow (A^{-1})' I = A^{-1} \Rightarrow (A^{-1})' = A^{-1}$. Proved.