Q.If $A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}$, then verify that $A\,\operatorname{adj} A = |A|\,I$. Also find $A^{-1}$.

Concept and Intuition

For an n x n matrix, |adj B| = |B|^(n-1); here n = 3 so it is |B|^2. Also |3A| = 3^3 |A| = |B|. The ratio collapses to |B|.

Step-by-Step Solution

1. |B| = |3A| = 3^3 * |A| = 27 * 5 = 135.
2. |adj B| = |B|^(3-1) = 135^2.
3. |adj B| / |3A| = 135^2 / 135 = 135.

Common Mistakes

• Using |adj B| = |B|^n instead of |B|^(n-1), or forgetting |3A| = 3^3|A|.

ANSWER: D

For an $n\times n$ matrix, $|\text{adj}\,A|=|A|^{n-1}$. Applying twice,

With $n=3$ and $|A|=9$:

$A$ is upper triangular, so $|A|=1\cdot2\cdot(-2)=-4$.

Using $|\mathrm{adj}\,A|=|A|^{n-1}$ twice, for $n=3$:

With $n=3$ and $|A|=7$: $|B|=|3A|=3^3|A|=27\cdot7=189$, and $|\mathrm{adj}\,A|=|A|^{n-1}=7^2=49$. Therefore $\dfrac{|\mathrm{adj}\,A|}{|B|}=\dfrac{49}{189}=\dfrac{7}{27}$.

For $A=\begin{bmatrix}3 & 7\\ 2 & 5\end{bmatrix}$, $\det A=15-14=1$.

The fundamental identity is $A\,(\mathrm{adj}\,A)=(\det A)\,I=1\cdot I=I$.

Therefore

For an $n\times n$ matrix, $|\text{adj}A|=|A|^{n-1}$. With $n=3$ and $|A|=4$, $|B|=4^2=16$.

Compute $\det B$ for $B=\begin{pmatrix}1&\alpha&3\\1&3&3\\2&4&4\end{pmatrix}$:

Set $2\alpha-6=16\Rightarrow 2\alpha=22\Rightarrow\alpha=11$.

Concept and Intuition

For an n×n matrix, det(adj A) = (det A)^(n−1). With complex entries det(A) may be any complex number satisfying the resulting equation.

Step-by-Step Solution

1. n = 4, so det(adj A) = det(A)^(4−1) = det(A)³.
2. det(A)³ = 5.
3. Over ℂ, z³ = 5 has exactly 3 distinct cube roots.
4. So det(A) has 3 possible values.

Common Mistakes

• Using exponent n instead of n−1.

ANSWER: D

Concept and Intuition

A·adj(A) = det(A)·I always, so the condition A·adj A = AA^T means AA^T = det(A)·I. This requires the off-diagonal of AA^T to vanish and the diagonal to equal det(A).

Step-by-Step Solution

1. A = [[5a, −b],[3, 2]]. AA^T = [[25a²+b², 15a−2b],[15a−2b, 13]].
2. det(A)·I = [[10a+3b, 0],[0, 10a+3b]].
3. Off-diagonal: 15a − 2b = 0 ⇒ b = 7.5a.
4. Lower-right: 13 = det(A) = 10a + 3b ⇒ 32.5a = 13 ⇒ a = 0.4, b = 3.
5. Check 25a²+b² = 4 + 9 = 13 ✓. So det(A) = 13 ≥ 0.

Common Mistakes

• Testing options like 5a−b or 5a+b without solving; only det(A) ≥ 0 holds (13).

ANSWER: E

Concept and Intuition

Compute the adjoint of the given matrix (transpose of the cofactor matrix); then read off the required diagonal sum and specific off-diagonal entries.

Step-by-Step Solution

1. For M = [[1,3,3],[1,4,3],[1,3,4]], det(M) = 1.
2. Cofactors give adj(M) = [[7,−3,−3],[−1,1,0],[−1,0,1]].
3. a1 = 7, b2 = 1, c3 = 1 ⇒ a1+b2+c3 = 9 (trace).
4. b1 = −3, a2 = −1 ⇒ b1·a2 = 3.
5. Ratio = 9/3 = 3.

Common Mistakes

• Forgetting adjoint is the transpose of the cofactor matrix.

ANSWER: B

Concept and Intuition

For an $n\times n$ matrix, $|kM|=k^n|M|$ and $|\mathrm{adj}(A)|=|A|^{n-1}$.

Step-by-Step Solution

1. Compute $|A|=2(1\cdot3-(-1)\cdot0)-1(1\cdot3-(-1)\cdot1)+(-2)(1\cdot0-1\cdot1)$.
2. $=2(3)-1(4)+(-2)(-1)=6-4+2=4$.
3. $B=|A|\,\mathrm{adj}(A)=4\,\mathrm{adj}(A)$, so $|B|=4^3\,|\mathrm{adj}(A)|$ (since $n=3$).
4. $|\mathrm{adj}(A)|=|A|^{n-1}=4^2=16$.
5. $|B|=64\times16=1024$.

Common Mistakes

• Forgetting the scalar factor $k^n$ with $n=3$.
• Using $|\mathrm{adj}(A)|=|A|$ instead of $|A|^{n-1}$.

ANSWER: D

Concept and Intuition

Since $A^{-1}=\frac{\mathrm{adj}(A)}{|A|}$, we have $\mathrm{adj}(A)=|A|\,A^{-1}$; find $|A|$ from $|A^{-1}|=1/|A|$.

Step-by-Step Solution

1. $A^{-1}=\frac12\begin{bmatrix}-10&-4\\2&1\end{bmatrix}$.
2. $|A^{-1}|=\left(\frac12\right)^2\big((-10)(1)-(-4)(2)\big)=\frac14(-10+8)=-\frac12$.
3. $|A|=\frac{1}{|A^{-1}|}=-2$.
4. $\mathrm{adj}(A)=|A|\,A^{-1}=-2\cdot\frac12\begin{bmatrix}-10&-4\\2&1\end{bmatrix}=-\begin{bmatrix}-10&-4\\2&1\end{bmatrix}=\begin{bmatrix}10&4\\-2&-1\end{bmatrix}$.

Common Mistakes

• Forgetting the $(1/2)^2$ factor when computing $|A^{-1}|$.
• Sign error making $|A|=+2$.

ANSWER: B