Q.Show that "if a matrix $A$ is invertible, then $A$ is non-singular".

We are given

and need to compute $(\operatorname{Adj}A)(\operatorname{Adj}(\operatorname{Adj}A))$.

Concept and Intuition

The adjugate (classical adjoint) of a matrix is intimately tied to its determinant and inverse. For an invertible $n \times n$ matrix $A$, we have the fundamental relation:

This means $\operatorname{Adj}(A) = |A| A^{-1}$. Applying the adjugate twice leads to a neat formula:

for $n \ge 2$. So instead of computing huge 3×3 adjugates directly, we can compute the determinant of $A$ once and plug into these formulas. The product then becomes a scalar multiple of the identity.

Step-by-step solution

1. Compute $|A|$ Expand along the first row:

2. Use the adjugate-inverse relation Since $|A| = 14 \neq 0$, $A$ is invertible and

3. Find $\operatorname{Adj}(\operatorname{Adj}(A))$ For a $3 \times 3$ matrix ($n=3$), the formula is: $$$$

We are given two matrices:

and the equation

We need $xyz - lmn - pqr$.

Concept and intuition:

Notice that $A$ is a permutation matrix that swaps the first and third rows/columns. Squaring it gives the identity: $A^2 = I$. That means $A^4 = (A^2)^2 = I^2 = I$. So $A^2B + A^4B = IB + IB = 2B$. The left-hand side simplifies dramatically. Then the equation becomes $(2B)^{-1} = \frac12 C$, where $C$ is the given $3\times3$ matrix. Inverting both sides lets us find $B$ exactly. The expression $xyz - lmn - pqr$ looks like the determinant of $B$ (expanded along the first row, with signs: $x(mr - nq) - y(lr - np) + z(lq - mp)$). But here it's $xyz - lmn - pqr$, which is not the full determinant — it’s only three terms. That suggests a special structure: if $B$ is a permutation of a diagonal matrix, many entries vanish. Let’s work it out.

Step-by-step solution:

1. Simplify the matrix power. Compute $A^2$:

Hence $A^2 = I$, so $A^4 = (A^2)^2 = I^2 = I$.

Therefore:

2. Rewrite the given equation. The equation becomes:

Since $(2B)^{-1} = \frac12 B^{-1}$, we have:

Multiply both sides by 2:

3. Find $B$ by inverting the given matrix. Let $C = \begin{bmatrix} 1 & 2 & -1 \\ 0 & 1 & 2 \\ 1 & 0 & 1 \end{bmatrix}$. Then $B = C^{-1}$. Compute $\det(C)$:

Now find the adjugate (or use row reduction). Compute cofactors:

• $C_{11} = +(1\cdot1 - 2\cdot0) = 1$
• $C_{12} = -(0\cdot1 - 2\cdot1) = -(-2) = 2$
• $C_{13} = +(0\cdot0 - 1\cdot1) = -1$
• $C_{21} = -(2\cdot1 - (-1)\cdot0) = -(2) = -2$
• $C_{22} = +(1\cdot1 - (-1)\cdot1) = 1 + 1 = 2$
• $C_{23} = -(1\cdot0 - 2\cdot1) = -(-2) = 2$
• $C_{31} = +(2\cdot2 - (-1)\cdot1) = 4 + 1 = 5$
• $C_{32} = -(1\cdot2 - (-1)\cdot0) = -(2) = -2$
• $C_{33} = +(1\cdot1 - 2\cdot0) = 1$

The cofactor matrix is:

Transpose to get the adjugate:

Then:

4. Identify entries of $B$. So: $$$$

Let $t = 10^{x}$ (so $10^{-x} = 1/t$). Rearranging the defining relation and isolating $t$ leads to

Taking base-10 logarithms of both sides: …

We are given three linear equations in $x, y, z$ and told the solution is unique. The question asks for $\alpha$, which is the $x$-coordinate of that unique solution. The most direct path is to eliminate $y$ and $z$ systematically to solve for $x$.

Concept & Intuition

When a system is linear and has a unique solution, we can combine the equations to cancel variables one at a time. Here, notice that the first two equations both contain $z$ with opposite signs ($+z$ and $-z$). Adding them will immediately eliminate $z$, giving a relation between $x$ and $y$. Then we can use the third equation to eliminate $y$ and solve for $x$.

Step-by-step solution

1. Write the equations in standard form

(We moved the constant terms to the right side.)

2. Eliminate $z$ using equations (1) and (2) Add (1) and (2):

The $z$ terms cancel:

3. Eliminate $z$ again, but using (1) and (3) Multiply (1) by 3 so that $z$ has coefficient $3$:

Now add this to (3):

The $z$ terms cancel:

Divide by 2:

4. Solve for $x$ using equations A and B We have: $\begin{cases} 5x - y = 3 \\ 5x - 7y = -9 \end{cases}$ …

The problem asks us to calculate a matrix expression involving $A^{-1}$ and $A^2$, given information about $A$ and $B$. The core idea is to first determine the matrix $A$ itself, or at least its inverse and square, using the given conditions. We'll leverage properties of determinants and the definition of a scalar matrix.

Here's how we approach this problem:

1. Determine the order of matrix A.

   Matrix $B$ is a $2 \times 2$ matrix. For the product $AB$ to be defined, matrix $A$ must have 2 columns. Let $A$ be an $m \times 2$ matrix. The product $AB$ will then be an $m \times 2$ matrix.

   We are told that $AB$ is a scalar matrix. A scalar matrix is a special type of diagonal matrix where all diagonal elements are equal, and it must be a square matrix. Therefore, $AB$ must be a $2 \times 2$ matrix. This implies that $m=2$, so $A$ is also a $2 \times 2$ matrix.

2. Find the determinant of A.

   We are given the condition $\det(3A) = 27$.

   For an $n \times n$ matrix $M$ and a scalar $k$, the determinant property is $\det(kM) = k^n \det(M)$.

   Since $A$ is a $2 \times 2$ matrix, $n=2$. Applying this property:

Equating this to the given value:

So, the determinant of matrix $A$ is $3$.

3. Determine the scalar value for the product AB.

We know that $AB$ is a scalar matrix. This means $AB$ can be written in the form $kI$, where $k$ is some scalar and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is the $2 \times 2$ identity matrix.

So, $AB = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}$.

Now, let's take the determinant of both sides:

We know that $\det(AB) = \det(A) \det(B)$.
From Step 2, we have $\det(A) = 3$.
For matrix $B = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$, its determinant is $\det(B) = (1)(3) - (2)(0) = 3$.
Therefore, $\det(AB) = (3)(3) = 9$.
Also, for a $2 \times 2$ scalar matrix $kI$, its determinant is $\det(kI) = k^2$.
Equating the two expressions for $\det(AB)$:

This gives us two possible values for $k$: $k = 3$ or $k = -3$. We will proceed with $k=3$ first, as it's a common convention to consider the positive root unless specified, and check if it leads to one of the given options. If not, we would try $k=-3$.

4. Find matrix A.

We have the relation $AB = kI$. To find $A$, we can multiply both sides by $B^{-1}$ on the right:

First, let's find $B^{-1}$:

The adjoint of $B$ is $\text{adj}(B) = \begin{bmatrix} 3 & -2 \\ -0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ 0 & 1 \end{bmatrix}$.

Now, substitute $B^{-1}$ and $k=3$ into the expression for $A$:

Let's quickly verify $\det(A) = (3)(1) - (-2)(0) = 3$, which matches our finding in Step 2.

5. Calculate $3A^{-1}$.

From the relation $AB = kI$, we can also find $A^{-1}$. Multiply by $A^{-1}$ on the left: …

Key identities ($n=3$).

For any invertible $3\times3$ matrix, $|\operatorname{Adj}A| = |A|^{\,n-1} = |A|^{2}$, and $|A^{-1}| = \dfrac{1}{|A|}$.

Evaluate the determinant.

With $|A| = \dfrac12$:

$\frac{1}{|A|^{3}} = \frac{1}{(1/2)^{3}} = 2^{3} = 8.$ …

Concept and Intuition

When faced with a messy expression involving inverses and products, the first instinct should be to simplify using the fundamental property of matrix inverses: $(XY)^{-1} = Y^{-1}X^{-1}$. This is the matrix analogue of "socks and shoes" — you reverse the order when undoing a composition. Here, we have two terms: $(A^{-1}B)^{-1}$ and $(AB^{-1})^{-1}$. Applying the property to each will turn them into products of $A$, $B$, and their inverses, which we can then compute directly since $A$ and $B$ are given explicitly. The matrices are small (3×3), so after simplification we can multiply them by hand.

Step-by-step solution

1. Simplify the first term Using $(XY)^{-1} = Y^{-1}X^{-1}$ with $X = A^{-1}$ and $Y = B$, we get

(Recall that $(A^{-1})^{-1} = A$.)

2. Simplify the second term Similarly, with $X = A$ and $Y = B^{-1}$,

3. Combine the two terms The whole expression becomes

4. Find $A^{-1}$ and $B^{-1}$ $A$ is a diagonal-like matrix (actually a permutation of signs):

Its inverse is obtained by taking reciprocals of the nonzero entries (since it's essentially a diagonal matrix after reordering rows/columns). Check: $A^2 = I$? Compute:

So $A^{-1} = A$ (it is its own inverse).

$B$ is a permutation matrix swapping rows 1 and 2:

Clearly $B^2 = I$ as well, so $B^{-1} = B$.

5. Substitute the inverses Since $A^{-1}=A$ and $B^{-1}=B$, we have

(Because $B^{-1}=B$ and $A^{-1}=A$, both terms are the same product $BA$.)

6. Compute $BA$ $$BA = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} …$$

The key idea is a known property of adjoints: for any square matrix $A$ of order $n$,

adj(adj $A$) = $|A|^{n-2} A$, provided $n \geq 2$.

This is not a random formula — it follows from the fact that $A \cdot \text{adj}(A) = |A| I$, and then applying the adjoint again.

Since $A$ here is diagonal, its determinant is trivial to compute, and the property gives the answer directly.

1. Find $|A|$. $A$ is diagonal with entries $1, 2, 3$, so

2. Apply the property. For an $n \times n$ matrix,

Here $n = 3$, so $n-2 = 1$, giving

$\text{adj}(\text{adj } A) = |A|^{1} A = 6A.$ …