Vardaan Learning Institute

Let $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and let $P(x) = x^2 - 5x - 2$. Show that $P(A) = O$. Hence, use this result to compute $A^3$. Solution: First, calculate $A^2 = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 1+6 & 2+8 \\ 3+12 & 6+16 \end{bmatrix} = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix}$.
$P(A) = A^2 - 5A - 2I = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix} - \begin{bmatrix} 5 & 10 \\ 15 & 20 \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.
Since $A^2 - 5A - 2I = O$, we multiply by $A$ to get $A^3 - 5A^2 - 2A = O \Rightarrow A^3 = 5A^2 + 2A$.
$A^3 = 5\begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix} + 2\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 35 & 50 \\ 75 & 110 \end{bmatrix} + \begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix} = \begin{bmatrix} 37 & 54 \\ 81 & 118 \end{bmatrix}$.

If $A$ and $B$ are skew-symmetric matrices of the same order, prove that the matrix $AB$ is symmetric if and only if $A$ and $B$ commute (i.e., $AB = BA$). Solution: Given $A' = -A$ and $B' = -B$.
Let's take the transpose of $AB$: $(AB)' = B'A' = (-B)(-A) = BA$.
Forward Proof: Assume $AB$ is symmetric. Then $(AB)' = AB$. From above, $(AB)' = BA$. Therefore, $BA = AB$, meaning they commute.
Reverse Proof: Assume $A$ and $B$ commute, so $AB = BA$. We need to show $(AB)' = AB$. We know $(AB)' = BA$. Since $BA = AB$, we get $(AB)' = AB$. Thus $AB$ is symmetric.

A square matrix $A$ is said to be orthogonal if $AA' = A'A = I$. Prove that if $A$ is an orthogonal matrix, then $A^{-1} = A'$. Solution: By the definition of an orthogonal matrix, we have $AA' = I$.
We know that for any invertible matrix $A$, its inverse $A^{-1}$ satisfies the property $A A^{-1} = I$.
Comparing the two equations: $A A' = A A^{-1}$.
Multiplying both sides by $A^{-1}$ from the left: $A^{-1}(AA') = A^{-1}(A A^{-1}) \Rightarrow (A^{-1}A)A' = (A^{-1}A)A^{-1} \Rightarrow I A' = I A^{-1} \Rightarrow A' = A^{-1}$. Proved.

Let $A = \begin{bmatrix} \cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{bmatrix}$. Prove by the Principle of Mathematical Induction that $A^n = \begin{bmatrix} \cos n\theta & i\sin n\theta \\ i\sin n\theta & \cos n\theta \end{bmatrix}$ for all $n \in \mathbb{N}$, where $i^2 = -1$. Solution: Step 1 (Base Case): For $n=1$, $A^1 = \begin{bmatrix} \cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{bmatrix}$, which is true.
Step 2 (Assumption): Let it be true for $n=k$, so $A^k = \begin{bmatrix} \cos k\theta & i\sin k\theta \\ i\sin k\theta & \cos k\theta \end{bmatrix}$.
Step 3 (Induction): For $n=k+1$, $A^{k+1} = A^k \cdot A = \begin{bmatrix} \cos k\theta & i\sin k\theta \\ i\sin k\theta & \cos k\theta \end{bmatrix} \begin{bmatrix} \cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{bmatrix}$.
First element $= \cos k\theta \cos \theta + i^2 \sin k\theta \sin \theta$. Since $i^2 = -1$, this is $\cos k\theta \cos \theta - \sin k\theta \sin \theta = \cos(k+1)\theta$.
Second element $= i\cos k\theta \sin \theta + i\sin k\theta \cos \theta = i\sin(k+1)\theta$.
Continuing this, we get $\begin{bmatrix} \cos(k+1)\theta & i\sin(k+1)\theta \\ i\sin(k+1)\theta & \cos(k+1)\theta \end{bmatrix}$. Hence, proved by PMI.

Let $A$ and $B$ be symmetric matrices of order $n$. Prove that the expression $A(BA) + B(AB)$ yields a symmetric matrix, while $A(BA) - B(AB)$ yields a skew-symmetric matrix. Solution: Given $A' = A$ and $B' = B$. Let $P = A(BA) + B(AB) = ABA + BAB$.
$P' = (ABA + BAB)' = (ABA)' + (BAB)' = A'B'A' + B'A'B'$.
Since matrices are symmetric, $P' = ABA + BAB = P$. Hence, $P$ is symmetric.
Let $Q = ABA - BAB$.
$Q' = (ABA - BAB)' = (ABA)' - (BAB)' = A'B'A' - B'A'B' = ABA - BAB = -(BAB - ABA) = -Q$.
Since $Q' = -Q$, the expression yields a skew-symmetric matrix.

Let $A$ be a nilpotent matrix of index 2 (meaning $A^2 = O$). Prove that the inverse of the matrix $(I - A)$ is given by $(I + A)$. Solution: To prove that $(I+A)$ is the inverse of $(I-A)$, we must show their product is the identity matrix $I$.
Multiply them: $(I - A)(I + A) = I \cdot I + I \cdot A - A \cdot I - A \cdot A$.
$= I^2 + A - A - A^2$.
We know $I^2 = I$ and given $A^2 = O$.
$= I + O - O = I$.
Since $(I - A)(I + A) = I$, by definition, $(I - A)^{-1} = I + A$.

If $A$ is an idempotent matrix (i.e., $A^2 = A$), compute the simplified value of $(I+A)^3 - 7A$. Solution: Expand $(I+A)^3$ using the binomial formula (since $I$ and $A$ commute):
$(I+A)^3 = I^3 + 3I^2A + 3IA^2 + A^3$.
We know $I^3 = I$, $I^2A = A$, $IA^2 = A^2$.
Given $A^2 = A$. This also means $A^3 = A^2 \cdot A = A \cdot A = A^2 = A$.
Substituting these values back: $(I+A)^3 = I + 3A + 3A + A = I + 7A$.
Therefore, the expression $(I+A)^3 - 7A = (I + 7A) - 7A = I$.

Find the matrix $A$ if it satisfies the equation: $\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$. Solution: Let $X = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$ and $Y = \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix}$. The equation is $X A Y = I$.
To isolate $A$, multiply by $X^{-1}$ on the left and $Y^{-1}$ on the right: $A = X^{-1} I Y^{-1} = X^{-1} Y^{-1}$.
We find inverses: $X^{-1} = \frac{1}{(4-3)}\begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}$.
$Y^{-1} = \frac{1}{(9-10)}\begin{bmatrix} -3 & -2 \\ -5 & -3 \end{bmatrix} = -1\begin{bmatrix} -3 & -2 \\ -5 & -3 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix}$.
$A = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix} = \begin{bmatrix} (6-5) & (4-3) \\ (-9+10) & (-6+6) \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$.

If $A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$, find all possible real values of $\alpha$ and $\beta$ such that $(\alpha I + \beta A)^2 = A$. Solution: First, compute $A^2 = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I$.
Expand LHS: $(\alpha I + \beta A)^2 = \alpha^2 I^2 + \beta^2 A^2 + 2\alpha\beta A = \alpha^2 I + \beta^2 (-I) + 2\alpha\beta A = (\alpha^2 - \beta^2)I + 2\alpha\beta A$.
Equate LHS to RHS: $(\alpha^2 - \beta^2)I + 2\alpha\beta A = A$.
Comparing coefficients of $I$ and $A$:
1) $\alpha^2 - \beta^2 = 0 \Rightarrow \alpha^2 = \beta^2 \Rightarrow \alpha = \beta$ or $\alpha = -\beta$.
2) $2\alpha\beta = 1$.
If $\alpha = \beta$, then $2\alpha^2 = 1 \Rightarrow \alpha^2 = 1/2 \Rightarrow \alpha = \pm \frac{1}{\sqrt{2}}$. Hence $\beta = \pm \frac{1}{\sqrt{2}}$ (same signs).
If $\alpha = -\beta$, then $-2\alpha^2 = 1 \Rightarrow \alpha^2 = -1/2$, which has no real solutions.
Thus, the real values are $(\alpha, \beta) = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ or $\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$.

Prove that if $A$ and $B$ are square matrices such that $AB = BA$, then for any positive integer $n \ge 1$, $AB^n = B^nA$. Solution: We will use Mathematical Induction.
Base Step ($n=1$): $AB^1 = B^1A \Rightarrow AB = BA$. This is given true.
Inductive Step: Assume it is true for $n = k$, i.e., $AB^k = B^kA$.
We must prove it for $n = k+1$. Consider LHS $= AB^{k+1}$.
$AB^{k+1} = A(B^k B) = (AB^k)B$.
Using the assumption, substitute $AB^k = B^kA$: $\Rightarrow (B^kA)B = B^k(AB)$.
Using the given $AB = BA$: $\Rightarrow B^k(BA) = (B^kB)A = B^{k+1}A = \text{RHS}$.
By PMI, $AB^n = B^nA$ holds for all $n \ge 1$.

Prove that for any two matrices $A$ and $B$ of order $n \times n$, $\text{Trace}(AB) = \text{Trace}(BA)$. Use this property to show that there exist no matrices $A$ and $B$ such that $AB - BA = I$. Solution: Let $A = [a_{ij}]$ and $B = [b_{ij}]$.
$\text{Trace}(AB) = \sum_{i=1}^n (AB)_{ii} = \sum_{i=1}^n \left( \sum_{j=1}^n a_{ij}b_{ji} \right)$.
Since scalars commute, $a_{ij}b_{ji} = b_{ji}a_{ij}$. Rearranging the sums: $\sum_{j=1}^n \left( \sum_{i=1}^n b_{ji}a_{ij} \right) = \sum_{j=1}^n (BA)_{jj} = \text{Trace}(BA)$.
Now, assume $AB - BA = I$.
Taking the trace on both sides: $\text{Trace}(AB - BA) = \text{Trace}(I)$.
By trace properties, $\text{Trace}(AB) - \text{Trace}(BA) = n$ (since trace of $I_{n}$ is $n$).
$0 = n$. Since $n \ge 1$ for matrices to exist, $0 = n$ is a contradiction. Hence, no such matrices exist.

Let $A = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}$. Find a general formula for $A^n$, and then evaluate the matrix limit $\lim_{n \to \infty} \frac{1}{n} A^n$. Solution: Calculate early powers to find a pattern:
$A^2 = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2a \\ 0 & 1 \end{bmatrix}$.
$A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 2a \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3a \\ 0 & 1 \end{bmatrix}$.
By observation, $A^n = \begin{bmatrix} 1 & na \\ 0 & 1 \end{bmatrix}$.
Now evaluate the limit: $\lim_{n \to \infty} \frac{1}{n} A^n = \lim_{n \to \infty} \begin{bmatrix} 1/n & (na)/n \\ 0 & 1/n \end{bmatrix} = \lim_{n \to \infty} \begin{bmatrix} 1/n & a \\ 0 & 1/n \end{bmatrix}$.
As $n \to \infty$, $1/n \to 0$. Thus, the limit is $\begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix}$.

If $f(x) = x^2 - 4x + 7$, and $A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix}$, show that $f(A) = O$. Using this polynomial equation, compute the exact matrix for $A^5$ without repeatedly multiplying matrices. Solution: $A^2 = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix}$.
$f(A) = A^2 - 4A + 7I = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} - \begin{bmatrix} 8 & 12 \\ -4 & 8 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.
From this, $A^2 = 4A - 7I$.
To find $A^5$, we reduce the power iteratively:
$A^4 = (A^2)^2 = (4A - 7I)^2 = 16A^2 - 56A + 49I = 16(4A - 7I) - 56A + 49I = 64A - 112I - 56A + 49I = 8A - 63I$.
$A^5 = A \cdot A^4 = A(8A - 63I) = 8A^2 - 63A = 8(4A - 7I) - 63A = 32A - 56I - 63A = -31A - 56I$.
$A^5 = -31\begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} - \begin{bmatrix} 56 & 0 \\ 0 & 56 \end{bmatrix} = \begin{bmatrix} -62 & -93 \\ 31 & -62 \end{bmatrix} - \begin{bmatrix} 56 & 0 \\ 0 & 56 \end{bmatrix} = \begin{bmatrix} -118 & -93 \\ 31 & -118 \end{bmatrix}$.

A matrix $X$ has $(a+b)$ rows and $(a+2)$ columns, while the matrix $Y$ has $(b+1)$ rows and $(a+3)$ columns. If both the products $XY$ and $YX$ are well-defined, determine the values of $a$ and $b$. Solution: Order of $X = (a+b) \times (a+2)$. Order of $Y = (b+1) \times (a+3)$.
For product $XY$ to be defined, the number of columns in $X$ must equal the number of rows in $Y$:
$\Rightarrow a+2 = b+1 \Rightarrow a - b = -1$ --- (Eq 1)
For product $YX$ to be defined, the number of columns in $Y$ must equal the number of rows in $X$:
$\Rightarrow a+3 = a+b \Rightarrow b = 3$.
Substitute $b = 3$ into Eq 1: $a - 3 = -1 \Rightarrow a = 2$.
Thus, $a = 2, b = 3$.

Let $A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$. Find the inverse of matrix $A$ strictly by using elementary row transformations. Solution: Write $A = IA$: $\begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A$.
$R_1 \leftrightarrow R_2$: $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} A$.
$R_3 \rightarrow R_3 - 3R_1$: $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -5 & -8 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1 \end{bmatrix} A$.
$R_3 \rightarrow R_3 + 5R_2$: $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 5 & -3 & 1 \end{bmatrix} A$.
$R_3 \rightarrow \frac{1}{2}R_3$: $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 5/2 & -3/2 & 1/2 \end{bmatrix} A$.
$R_2 \rightarrow R_2 - 2R_3$: $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{bmatrix} A$.
$R_1 \rightarrow R_1 - 3R_3$: $\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -15/2 & 11/2 & -3/2 \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{bmatrix} A$.
$R_1 \rightarrow R_1 - 2R_2$: $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1/2 & -1/2 & 1/2 \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{bmatrix} A$.
The inverse is $A^{-1} = \frac{1}{2} \begin{bmatrix} 1 & -1 & 1 \\ -8 & 6 & -2 \\ 5 & -3 & 1 \end{bmatrix}$.