Q.Let $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$, $C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$ Find each of the following:

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).

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:

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

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

   $= 2 \cdot (4 - 8) - 3 \cdot (6 - 4) + 4 \cdot (12 - 4)$

   $= 2 \cdot (-4) - 3 \cdot (2) + 4 \cdot (8) = -8 - 6 + 32 = 18$.

3. Find $|AB|$.

   Since $|AB| = |A| \cdot |B|$, we have $|AB| = 11 \times 18 = 198$.

4. Apply the adjugate property.

   For $n=3$, $|\text{Adj}(AB)| = |AB|^{2} = 198^2$.

   Then $\sqrt{|\text{Adj}(AB)|} = \sqrt{198^2} = 198$.

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:

So $|B| = 18$.

3. Find $|AB|$.

Since $|AB| = |A| \cdot |B|$, we have

4. Apply the adjugate determinant formula.

For a $3 \times 3$ matrix $M$, $|\operatorname{Adj}(M)| = |M|^{2}$.

Thus

Then

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