Q.If $A = \begin{pmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{pmatrix}$, find $A^{-1}$. Using $A^{-1}$, solve the system of linear equations $x - 2y = 10$, $2x - y - z = 8$, $-2y + z = 7$.

Concept and Intuition

For any invertible square matrix, $A^{-1}=\dfrac{\mathrm{adj}(A)}{\det(A)}$. Here $\det(A)=2$, so $2A^{-1}=\mathrm{adj}(A)$ exactly — this avoids computing $A^{-1}$'s fractional entries and lets us add the whole-number adjugate matrix directly to $A$.

Step-by-Step Solution

1. $A=\begin{pmatrix}1&2&-2\\2&-1&2\\-1&1&-2\end{pmatrix}$. Expand along row 1: $\det(A)=1[(-1)(-2)-(2)(1)]-2[(2)(-2)-(2)(-1)]+(-2)[(2)(1)-(-1)(-1)]$ $=1(2-2)-2(-4+2)-2(2-1)=0+4-2=2$.
2. Compute all nine cofactors and transpose to get $\mathrm{adj}(A)$: $\mathrm{adj}(A)=\begin{pmatrix}0&2&2\\2&-4&-6\\1&-3&-5\end{pmatrix}$.
3. Since $\det(A)=2$, $A^{-1}=\tfrac12\mathrm{adj}(A)$, so $2A^{-1}=\mathrm{adj}(A)$ exactly (no fractions needed).
4. $A+2A^{-1}=A+\mathrm{adj}(A)$: Row1: $1+0=1,\ 2+2=4,\ -2+2=0$ Row2: $2+2=4,\ -1-4=-5,\ 2-6=-4$ Row3: $-1+1=0,\ 1-3=-2,\ -2-5=-7$ …

Concept and Intuition

By the Cayley–Hamilton theorem, every square matrix satisfies its own characteristic equation. For a $3\times3$ matrix with characteristic polynomial $\lambda^3-c_1\lambda^2+c_2\lambda-c_3=0$ (where $c_3=\det A$), substituting $A$ gives $A^3-c_1A^2+c_2A-c_3I=0$. Multiplying through by $A^{-1}$ (valid since $\det A\ne0$) expresses $A^{-1}$ as a polynomial in $A$ — this avoids computing the full inverse via cofactors.

Step-by-Step Solution

1. $A=\begin{bmatrix}1&0&1\\0&1&1\\0&1&0\end{bmatrix}$. Compute $\det A = 1(1\cdot0-1\cdot1)-0+1(0\cdot1-1\cdot0) = -1$.
2. Compute $A^2$: multiplying $A$ by itself row by row gives $A^2=\begin{bmatrix}1&1&1\\0&2&1\\0&1&1\end{bmatrix}$. …

Concept and Intuition

$A^{-1}=\dfrac{1}{\det A}\,\text{adj}(A)$. If $\det A$ happens to equal 10, then $10A^{-1}=\text{adj}(A)$ exactly — so instead of inverting $A$, just build its adjugate (transpose of the cofactor matrix) and read off the unknown entry.

Step-by-Step Solution

1. $A=\begin{bmatrix}1&-1&1\\2&1&-3\\1&1&1\end{bmatrix}$. Expand along row 1: $\det A = 1(1\cdot1-(-3)\cdot1) -(-1)(2\cdot1-(-3)\cdot1) + 1(2\cdot1-1\cdot1) = 1(4)+1(5)+1(1)=10$.
2. Since $\det A=10$, $A^{-1}=\frac{1}{10}\text{adj}(A) \Rightarrow 10A^{-1}=\text{adj}(A)$, and since $B=A^{-1}$, $10B=\text{adj}(A)$.
3. Compute cofactors of $A$: $C_{11}=4,\;C_{12}=-5,\;C_{13}=1$ $C_{21}=2,\;C_{22}=0,\;C_{23}=-2$ $C_{31}=2,\;C_{32}=5,\;C_{33}=3$
4. $\text{adj}(A)$ = transpose of the cofactor matrix: …

Concept and Intuition

A square matrix is called orthogonal precisely when its rows (equivalently columns) form an orthonormal set — each row has unit length and distinct rows are perpendicular. For any orthogonal matrix, $AA^T=I$, which directly means $A^{-1}=A^T$ (since the inverse is whatever matrix multiplies A to give the identity).

Step-by-Step Solution

1. Let the rows (before the $\frac17$ factor) be $R_1=(3,-2,6)$, $R_2=(-6,-3,2)$, $R_3=(-2,6,3)$.
2. Check lengths: $|R_1|^2 = 9+4+36=49$, $|R_2|^2=36+9+4=49$, $|R_3|^2=4+36+9=49$ — each scaled row (divided by 7) has unit length.
3. Check orthogonality: $R_1\cdot R_2 = (3)(-6)+(-2)(-3)+(6)(2) = -18+6+12=0$; $R_1\cdot R_3 = (3)(-2)+(-2)(6)+(6)(3)=-6-12+18=0$; $R_2\cdot R_3=(-6)(-2)+(-3)(6)+(2)(3)=12-18+6=0$.
4. All rows orthonormal $\Rightarrow$ $AA^T = I \Rightarrow A^{-1}=A^T$. …

We want $\operatorname{tr}(A^{-1})$, the sum of the diagonal entries of $A^{-1}$. Computing $A^{-1}$ directly is possible but tedious. Instead, we use a clever relationship:

For any invertible matrix $A$,

where $\operatorname{adj}(A)$ is the adjugate (transpose of the cofactor matrix).

Then

The trace of the adjugate is simply the sum of the cofactors of the diagonal entries of $A$ (since the adjugate’s diagonal entries are exactly the cofactors of the corresponding diagonal entries of $A$). So we only need $\det(A)$ and the sum of the three cofactors $C_{11}, C_{22}, C_{33}$.

1. Compute $\det(A)$

Using the first row:

So $\det(A) = -18$.

2. Find the cofactors of the diagonal entries
   • $C_{11}$: minor is $\begin{bmatrix}1&2\\2&3\end{bmatrix}$, determinant $= 1\cdot3 - 2\cdot2 = -1$, so $C_{11} = (-1)^{1+1}(-1) = -1$. …

Concept and Intuition

With three linear equations in three unknowns, eliminate one variable at a time to reduce to two equations in two unknowns, then back-substitute. Once $\alpha,\beta,\gamma$ are known, simply test each answer option numerically — far faster than trying to derive the relation symbolically.

Step-by-Step Solution

1. Equations: (1) $2x-3y+2z=-15$, (2) $3x+y-z=-2$, (3) $x-3y-3z=-8$.
2. From (2): $y = -3x+z-2$.
3. Substitute into (1): $2x-3(-3x+z-2)+2z=-15 \Rightarrow 2x+9x-3z+6+2z=-15 \Rightarrow 11x-z=-21 \Rightarrow z=11x+21$.
4. Substitute $y=-3x+z-2 = -3x+(11x+21)-2 = 8x+19$ and $z=11x+21$ into (3):

5. Then $z = 11(-2)+21 = -1$, and $y = 8(-2)+19 = 3$. …

Concept and Intuition

The scalar polynomial $x^3-6x^2+11x-6$ factors neatly:

(verify: $x=1: 1-6+11-6=0$; $x=2: 8-24+22-6=0$; $x=3: 27-54+33-6=0$ — all check out). So the matrix equation $A^3-6A^2+11A-6I=0$ is satisfied when $A$'s eigenvalues are drawn from $\{1,2,3\}$; in particular, a scalar matrix $A=kI$ satisfies the equation exactly when $k \in \{1,2,3\}$ (substituting $A=kI$ turns the matrix equation into the same scalar cubic in $k$).

Since $6 \ne 0$ is the (nonzero) product of the roots $1\times2\times3$, $A$ is guaranteed invertible, and we can find $A^{-1}$ for each candidate scalar case:

• $A=I \Rightarrow A^{-1}=I$
• $A=2I \Rightarrow A^{-1}=\tfrac12I$
• $A=3I \Rightarrow A^{-1}=\tfrac13I$

Checking the answer choices against which scalar $A$ they'd imply:

• (A) $A^{-1}=\tfrac14I \Rightarrow A=4I$; but $4$ is not a root of the cubic ($64-96+44-6=6\ne0$) — invalid.
• (B) $A^{-1}=4I \Rightarrow A=\tfrac14I$; $\tfrac14$ is not a root — invalid.
• (C) $A^{-1}=3I \Rightarrow A=\tfrac13I$; $\tfrac13$ is not a root — invalid.
• (D) $A^{-1}=\tfrac13I \Rightarrow A=3I$; $3$ is a root — valid.

Step-by-Step Solution

1. Factor the scalar cubic: $x^3-6x^2+11x-6=(x-1)(x-2)(x-3)$, roots $1,2,3$. …

Concept and Intuition

A matrix of the form $\begin{bmatrix}1&-\tan\theta\\ \tan\theta&1\end{bmatrix}$ is exactly

$\sec\theta$ times the standard rotation matrix $R(\theta)$ (since $1=\cos\theta\sec\theta$ and

$\tan\theta=\sin\theta\sec\theta$). Rotation matrices are orthogonal, so their inverse equals

their transpose, i.e. $R(\theta)^{-1}=R(-\theta)$ — this makes inverting $N$ effortless and

turns the whole problem into "rotate by $\theta$, then rotate by $\theta$ again", i.e. rotate by

$2\theta$.

Step-by-Step Solution

1. Write $M = \sec\theta\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix} = \sec\theta\, R(\theta)$.
2. Similarly $N = \begin{bmatrix}1&\tan\theta\\-\tan\theta&1\end{bmatrix} = \sec\theta\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix} = \sec\theta\, R(-\theta)$.
3. Since $R(-\theta)^{-1}=R(\theta)$ (rotation matrices are orthogonal, inverse = transpose), $N^{-1} = \dfrac{1}{\sec\theta}R(\theta) = \cos\theta\, R(\theta)$.
4. Then $MN^{-1} = \sec\theta\,R(\theta)\cdot\cos\theta\,R(\theta) = R(\theta)R(\theta)=R(2\theta)$. …

Concept and Intuition

For a $2\times2$ matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$, the inverse is $\frac{1}{\det}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$. Using $\cot^2\theta-\csc^2\theta=-1$ (identity), $\det A=-1$ always, so inverting is just a sign flip combined with swapping the diagonal — this makes the algebra very light.

Step-by-Step Solution

1. $\det A=(-\cot\theta)(-\cot\theta)-(\csc\theta)(\csc\theta)=\cot^2\theta-\csc^2\theta=-1$ (identity $\csc^2\theta-\cot^2\theta=1$).
2. $A^{-1}=\dfrac{1}{-1}\begin{pmatrix}-\cot\theta&-\csc\theta\\-\csc\theta&-\cot\theta\end{pmatrix}=\begin{pmatrix}\cot\theta&\csc\theta\\ \csc\theta&\cot\theta\end{pmatrix}$.
3. $A^{-1}=A$: compare diagonal entries — $\cot\theta=-\cot\theta\Rightarrow2\cot\theta=0\Rightarrow\cot\theta=0\Rightarrow\theta_1=\pi/2$ (off-diagonal $\csc\theta=\csc\theta$ is automatic). …

Concept and Intuition

For a $2\times2$ matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ with determinant $\Delta$, the inverse is $\dfrac{1}{\Delta}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$. Here the matrix has the structure of an orthogonal (rotation-like) matrix, so its determinant works out to exactly 1 via the Pythagorean identity, making the inverse easy to write down directly by swapping/negating entries.

Step-by-Step Solution

1. Compute $\det A = \sin\alpha \cdot \sin\alpha - (-\cos\alpha)\cdot\cos\alpha = \sin^2\alpha + \cos^2\alpha = 1$.
2. Since $\det A = 1$, $A^{-1} = \begin{bmatrix}\sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha\end{bmatrix}$ (swap diagonal entries — same here since both are $\sin\alpha$ — negate off-diagonal, then divide by $\Delta=1$).
3. Add: $A+A^{-1} = \begin{bmatrix}\sin\alpha+\sin\alpha & -\cos\alpha+\cos\alpha \\ \cos\alpha-\cos\alpha & \sin\alpha+\sin\alpha\end{bmatrix} = \begin{bmatrix}2\sin\alpha & 0\\0 & 2\sin\alpha\end{bmatrix}$. …