Q.Compute the following:

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:

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

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

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