Q.Simplify $\cos\theta \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta \begin{bmatrix} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{bmatrix}$.

Concept. If $\{v_1,v_2,v_3\}$ is a basis of $\mathbb{R}^3$, then $\operatorname{rank}(M) = \dim\operatorname{span}\{Mv_1, Mv_2, Mv_3\}$ for any $3\times3$ matrix $M$ — we never need $A$ itself.

Step 1 — the given vectors form a basis. $v_1=(0,1,1)^T$, $v_2=(1,1,1)^T$, $v_3=(1,1,0)^T$:

Step 2 — images under $A-I$. Using the given actions of $A$:

• $(A-I)v_1 = (2,0,6)^T-(0,1,1)^T = (2,-1,5)^T$
• $(A-I)v_2 = (0,0,0)^T-(1,1,1)^T = (-1,-1,-1)^T$
• $(A-I)v_3 = (1,0,0)^T-(1,1,0)^T = (0,-1,0)^T$

Step 3 — test independence of the images.

$\det\begin{pmatrix}2&-1&0\\-1&-1&-1\\5&-1&0\end{pmatrix}$ …

Expand along the first row

Evaluate each bracket:

• $2\big[(k^2-1)-1\big] = 2(k^2-2) = 2k^2 - 4$
• $-2k\big[k-1\big] = -2k^2 + 2k$
• $1\big[1 - 2k + 2\big] = 3 - 2k$

Add them …

The Cayley–Hamilton theorem says every square matrix satisfies its own characteristic equation. That means if we find the characteristic polynomial of $A$, we can write $A^3$ in terms of lower powers, and then manipulate to get $A^{-1}$ expressed as a combination of $A^2$, $A$, and $I$. This is a standard technique for problems where the inverse is given as a polynomial in the matrix.

Let’s work through it.

1. Find the characteristic polynomial of $A$. For a $3 \times 3$ matrix, the characteristic polynomial is

Here $\text{tr }A = 2 + 1 + 2 = 5$.

The sum of principal minors of order 2:

• Minor from rows 1,2 and cols 1,2: $\det\begin{pmatrix}2 & 1 \\ 0 & 1\end{pmatrix} = 2$
• Minor from rows 1,3 and cols 1,3: $\det\begin{pmatrix}2 & 1 \\ 1 & 2\end{pmatrix} = 4 - 1 = 3$
• Minor from rows 2,3 and cols 2,3: $\det\begin{pmatrix}1 & 0 \\ 1 & 2\end{pmatrix} = 2$ Sum = $2 + 3 + 2 = 7$.

Now $\det A$: expanding along the second row (which has a zero),

So the characteristic polynomial is

2. Apply Cayley–Hamilton. Since $A$ satisfies its own characteristic equation,

Rearranging,

3. Express $A^{-1}$ in terms of $A^2$ and $A$. …

We are given a matrix $A$ with unknown entries $\alpha, \beta, \gamma$, and a known vector equation:

We need to compute $(\operatorname{adj} A)^{-1} + \operatorname{adj} A^{-1}$.

Concept and intuition

For any invertible square matrix $A$, two important identities are:

• $\operatorname{adj} A = (\det A) A^{-1}$, so $(\operatorname{adj} A)^{-1} = \frac{A}{\det A}$.
• $\operatorname{adj}(A^{-1}) = (\det A^{-1}) A = \frac{A}{\det A}$.

Thus, both terms are actually the same: each equals $\frac{A}{\det A}$. Their sum is therefore $2 \cdot \frac{A}{\det A}$.

So the problem reduces to finding $\det A$. We can find it using the given vector equation without fully determining $\alpha, \beta, \gamma$.

Step-by-step reasoning

1. Express the given equation as a system

   Let $v = (5, -13, 11)^T$ and $b = (-290, -119, 210)^T$. We have $A v = b$.

   This gives three linear equations in $\alpha, \beta, \gamma$:

   • Row 1: $7 \cdot 5 + 3 \cdot (-13) + \alpha \cdot 11 = -290$

     $35 - 39 + 11\alpha = -290 \implies -4 + 11\alpha = -290 \implies 11\alpha = -286 \implies \alpha = -26$.

   • Row 2: $\beta \cdot 5 + 1 \cdot (-13) + (-11) \cdot 11 = -119$

     $5\beta - 13 - 121 = -119 \implies 5\beta - 134 = -119 \implies 5\beta = 15 \implies \beta = 3$.

   • Row 3: $(-5) \cdot 5 + \gamma \cdot (-13) + 19 \cdot 11 = 210$

     $-25 - 13\gamma + 209 = 210 \implies 184 - 13\gamma = 210 \implies -13\gamma = 26 \implies \gamma = -2$.

   So $A = \begin{pmatrix} 7 & 3 & -26 \\ 3 & 1 & -11 \\ -5 & -2 & 19 \end{pmatrix}$.

2. Compute $\det A$

   Expand along the first row:

   $$\det A = 7 \cdot \det\begin{pmatrix} 1 & -11 \\ -2 & 19 \end{pmatrix} - 3 \cdot \det\begin{pmatrix} 3 & -11 \\ -5 & 19 \end{pmatrix} + (-26) \cdot \det\begin{pmatrix} 3 & 1 \\ -5 & -2 \end{pmatrix}. …$$

The problem asks us to evaluate a sum of determinants. The first step is to understand how to calculate the determinant of the given $2 \times 2$ matrix $\Delta_r$. Once we have a simplified expression for $\Delta_r$, we will look for a pattern that allows us to efficiently sum the terms from $r=1$ to $r=33$. Often, such sums involve telescoping series, which can be achieved by expressing each term as a difference of two consecutive terms.

1. Calculate the determinant $\Delta_r$. For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is given by $ad - bc$. In our case, $\Delta_r = \begin{vmatrix} \frac{1}{3r-2} & \frac{2}{3r-5} \\ 0 & \frac{3}{3r+1} \end{vmatrix}$. Applying the formula:

Notice that this is a lower triangular matrix (all elements above the main diagonal are zero). For such matrices, the determinant is simply the product of the diagonal elements.

2. Decompose $\Delta_r$ using partial fractions.

To make the sum $\sum \Delta_r$ easier to evaluate, we aim to express $\Delta_r$ as a difference of two terms. This is a common technique for telescoping sums. We use partial fraction decomposition for the expression $\frac{3}{(3r-2)(3r+1)}$.

Let

Multiplying both sides by $(3r-2)(3r+1)$, we get:

To find $A$, set $3r-2=0$, which means $r = \frac{2}{3}$:

To find $B$, set $3r+1=0$, which means $r = -\frac{1}{3}$:

So, the partial fraction decomposition is:

3. Evaluate the sum $\sum_{r=1}^{33} \Delta_r$. Now we need to sum this expression from $r=1$ to $r=33$: …

We are asked for $\sqrt{|\operatorname{Adj}(AB)|}$. The notation $|\cdot|$ means determinant. So we need the determinant of the adjugate of $AB$, then its square root.

Concept and intuition

For any square matrix $M$ of order $n$, the adjugate satisfies

If $M$ is invertible, then $\operatorname{Adj}(M) = |M| \, M^{-1}$. Taking determinants of both sides gives a powerful formula:

Here $n=3$, so $|\operatorname{Adj}(AB)| = |AB|^{2}$. Then $\sqrt{|\operatorname{Adj}(AB)|} = |AB|$ (since determinants are positive here). So the problem reduces to finding $|AB| = |A| \cdot |B|$.

Step-by-step solution

1. Compute $|A|$.

Use expansion or row operations. Let’s do cofactor expansion along the first row:

So $|A| = 11$.

2. Compute $|B|$.

Expand along the first row:

We are asked for $\sqrt{|\text{Adj}(AB)|}$. The matrices $A$ and $B$ are $3 \times 3$, so $n=3$. A classic result: for any square matrix $M$ of order $n$, the determinant of its adjugate is $|M|^{n-1}$. This saves us from actually computing the adjugate — we just need the determinant of $AB$, then raise it to the appropriate power and take the square root.

Why this works: The adjugate is the transpose of the cofactor matrix, and it satisfies $M \cdot \text{Adj}(M) = |M| I$. Taking determinants of both sides gives $|M| \cdot |\text{Adj}(M)| = |M|^n$, so if $|M| \neq 0$, we get $|\text{Adj}(M)| = |M|^{n-1}$. Here $M = AB$, so $|\text{Adj}(AB)| = |AB|^{3-1} = |AB|^2$. Then $\sqrt{|\text{Adj}(AB)|} = \sqrt{|AB|^2} = |AB|$ (since determinants are real numbers, and we take the principal square root). So the problem reduces to finding $|AB| = |A| \cdot |B|$.

Now we compute step by step.

1. Compute $|A|$.

   $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{bmatrix}$.

   Expand along the first row:

   $|A| = 1 \cdot \begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix}$

   $= 1 \cdot (1\cdot1 - 1\cdot3) - 2 \cdot (2\cdot1 - 1\cdot1) + 3 \cdot (2\cdot3 - 1\cdot1)$

   $= 1 \cdot (1 - 3) - 2 \cdot (2 - 1) + 3 \cdot (6 - 1)$

   $= 1 \cdot (-2) - 2 \cdot (1) + 3 \cdot (5) = -2 - 2 + 15 = 11$.

2. Compute $|B|$.

   $B = \begin{bmatrix} 2 & 3 & 4 \\ 3 & 2 & 2 \\ 2 & 4 & 2 \end{bmatrix}$.

   Expand along the first row:

   $|B| = 2 \cdot \begin{vmatrix} 2 & 2 \\ 4 & 2 \end{vmatrix} - 3 \cdot \begin{vmatrix} 3 & 2 \\ 2 & 2 \end{vmatrix} + 4 \cdot \begin{vmatrix} 3 & 2 \\ 2 & 4 \end{vmatrix}$

   $= 2 \cdot (2\cdot2 - 2\cdot4) - 3 \cdot (3\cdot2 - 2\cdot2) + 4 \cdot (3\cdot4 - 2\cdot2)$ …

The core idea behind this problem is the consistency condition for a system of linear equations. When we have a system of linear equations represented as $AX = B$, where $A$ is the coefficient matrix, $X$ is the column vector of variables, and $B$ is the column vector of constants, the system has a solution if and only if the rank of the coefficient matrix $A$ is equal to the rank of the augmented matrix $[A : B]$.

The augmented matrix $[A : B]$ is formed by appending the column vector $B$ to the matrix $A$. When we perform elementary row operations on $[A : B]$ to reduce $A$ to its row echelon form, any row that becomes entirely zero in the $A$ part must also have a zero in the corresponding position in the $B$ part. If a row in the $A$ part becomes all zeros, but the corresponding entry in the $B$ part is non-zero, it implies an equation of the form $0 = k$ (where $k \neq 0$), which is a contradiction. In such a case, the system would be inconsistent, and the rank of $[A : B]$ would be greater than the rank of $A$.

Therefore, to find the condition for $\text{rank}(A) = \text{rank}([A : B])$, we will reduce the augmented matrix $[A : B]$ to its row echelon form and ensure that any row of zeros in the $A$ portion also results in a zero in the $B$ portion.

1. Form the augmented matrix $[A : B]$. Given $A = \begin{bmatrix} 1 & -4 & 7 \\ 0 & 3 & -5 \\ -2 & 5 & -9 \end{bmatrix}$ and $B = \begin{bmatrix} a \\ -b \\ -c \end{bmatrix}$, the augmented matrix is:

2. Perform elementary row operations to reduce the matrix $A$ to row echelon form. Our goal is to create zeros below the leading entries. First, we make the entry in the third row, first column zero using $R_1$:

The new third row will be:
$( -2 + 2(1) \quad 5 + 2(-4) \quad -9 + 2(7) \quad : \quad -c + 2(a) )$
$( 0 \quad 5 - 8 \quad -9 + 14 \quad : \quad 2a - c )$
$( 0 \quad -3 \quad 5 \quad : \quad 2a - c )$
The augmented matrix becomes:

3. Continue row operations to further reduce $A$. Next, we make the entry in the third row, second column zero using $R_2$:

The new third row will be: …

The key idea is that for any $2 \times 2$ matrix, its characteristic polynomial is $t^2 - (\operatorname{tr} A) t + \det A = 0$. By the Cayley–Hamilton theorem, the matrix itself satisfies this polynomial: $A^2 - (\operatorname{tr} A) A + (\det A) I = 0$. This lets us reduce higher powers of $A$ to linear combinations of $A$ and $I$, making the trace of $A^3$ computable from just the trace and determinant of $A$.

1. Write the Cayley–Hamilton relation. For a $2 \times 2$ matrix $A$, we have

Rearranging:

2. Find $A^3$ in terms of $A$ and $I$. Multiply both sides by $A$:

Now substitute the expression for $A^2$ from step 1:

Simplify:

3. Take the trace of both sides. Recall that $\operatorname{tr}(cA) = c \operatorname{tr} A$ and $\operatorname{tr}(cI) = 2c$ for a $2 \times 2$ identity. So:

Simplify:

4. Plug in the given values. …