Q.Find $X$ and $Y$, if $X + Y = \begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix}$ and $X - Y = \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}$.

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:

Compute each minor:

• First: $1 \cdot 19 - (-11)(-2) = 19 - 22 = -3$.
• Second: $3 \cdot 19 - (-11)(-5) = 57 - 55 = 2$.
• Third: $3 \cdot (-2) - 1 \cdot (-5) = -6 + 5 = -1$.

So:

3. Use the adjugate identities Since $\det A = -1$, we have:

Therefore:

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:
$( 0 + 0 \quad -3 + 3 \quad 5 + (-5) \quad : \quad (2a - c) + (-b) )$
$( 0 \quad 0 \quad 0 \quad : \quad 2a - b - c )$
The augmented matrix in row echelon form is:

4. Apply the rank condition. From the row echelon form, we can see that the rank of matrix $A$ (the $3 \times 3$ block on the left) is 2, because there are two non-zero rows. For the rank of the augmented matrix $[A : B]$ to also be 2, the third row of the augmented matrix must be entirely zero. This means the entry in the $B$ part of the third row must be zero. Therefore, we must have:

5. Simplify the condition and match with options.

   The condition $2a - b - c = 0$ can be rewritten as $2a = b + c$.

   Dividing by 2, we get $a = \frac{b+c}{2}$.

   Comparing this with the given options:

   (A) $2a + b + c = 0$

   (B) $a = \dfrac{b + c}{2}$

   (C) $b = \dfrac{a + c}{2}$

   (D) $c = \dfrac{a + b}{2}$

   Our derived condition matches option (B).

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 $A$. Multiply the equation $A^3 - 5A^2 + 7A - 3I = 0$ by $A^{-1}$ (which exists since $\det A = 3 \neq 0$):

Hence

4. Match coefficients. The problem states $A^{-1} = \alpha A^2 + \beta A + \gamma I$. Comparing,

5. Compute $17\alpha + 5\beta + \gamma$.

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

So the determinant equals $-1$ for every $k$, i.e. $Ak^2 + Bk + C = -1$, giving $A=0$, $B=0$, $C=-1$.

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.

Expanding along the third column (only the middle entry $-1$ survives): $\det = -(-1)\cdot\big[\,-\det\begin{pmatrix}2&-1\\5&-1\end{pmatrix}\big]\cdot(-1) = (2\cdot(-1)-(-1)\cdot 5) = 3 \neq 0.$

The three images are linearly independent.

Step 4 — conclude. Since $(A-I)$ maps a basis to three independent vectors, $\operatorname{rank}(A-I) = 3$. (Consistency check: $Av_2 = 0$ means $0$ is an eigenvalue of $A$, i.e. $1$ is not forced to be an eigenvalue of $A$ — nothing requires $A - I$ to be singular, and indeed it isn't.)

ANSWER: A

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 $r=33$:

Let's write out the first few terms and the last term to observe the telescoping pattern:
For $r=1$: $\frac{1}{3(1)-2} - \frac{1}{3(1)+1} = \frac{1}{1} - \frac{1}{4}$
For $r=2$: $\frac{1}{3(2)-2} - \frac{1}{3(2)+1} = \frac{1}{4} - \frac{1}{7}$
For $r=3$: $\frac{1}{3(3)-2} - \frac{1}{3(3)+1} = \frac{1}{7} - \frac{1}{10}$
...
For $r=33$: $\frac{1}{3(33)-2} - \frac{1}{3(33)+1} = \frac{1}{99-2} - \frac{1}{99+1} = \frac{1}{97} - \frac{1}{100}$

When we sum these terms, the intermediate terms cancel out:

The sum simplifies to the first part of the first term and the second part of the last term:

4. Calculate the final value.