Let A = [aij]2×2, where aij ≠ 0 for all i, j and A² = I. Let a be the sum of all diagonal elements o

Q: Wait, why do we reject the possibility that $p+s \neq 0$?

Great question! We found that the off-diagonal elements of $A^2$ are $q(p+s)$ and $r(p+s)$. For $A^2$ to be the identity matrix, these must be zero. Since the problem explicitly states that all elements $a_{ij}$ are non-zero, $q$ and $r$ cannot be zero. Therefore, the only way for the product to be zero is if $p+s$ itself is zero.

Q: Why is the determinant $b = -1$ and not $1$?

Excellent observation. We know $|A^2| = |A|^2 = |I| = 1$, so $|A|$ could be $1$ or $-1$. However, we derived $b = -(p^2+qr)$. From the top-left element of $A^2=I$, we have $p^2+qr=1$. Substituting this, we get $b = -(1) = -1$. The constraint $a_{ij} \neq 0$ forces this specific value.

Q: Can we use the Cayley-Hamilton theorem here?

Absolutely! The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation. For a $2 \times 2$ matrix, this is $A^2 - \text{trace}(A)A + \det(A)I = 0$. Given $A^2 = I$, we have $I - aA + bI = 0$, or $aA = (1+b)I$. Since $A$ has non-zero elements, $a$ cannot be zero unless $1+b=0$. This is a powerful alternative path!

Question

Let $A = [a_{ij}]_{2 	imes 2}$, where $a_{ij} 
eq 0$ for all $i, j$ and $A^2 = I$. Let $a$ be the sum of all diagonal elements of $A$ and $b = |A|$. Then $3a^2 + 4b^2$ is equal to

Options:

(A) 4

(B) 14

(C) 7

(D) 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the precipice of a matrix transformation, looking at a $2 	imes 2$ matrix $A = \begin{bmatrix} p & q \ r & s \end{bmatrix}$.

The problem presents a beautiful, almost poetic constraint: $A^2 = I$, and every single element is non-zero. We are told that $a$ is the trace (the sum of the diagonal elements) and $b$ is the determinant.

Our mission is to find the value of $3a^2 + 4b^2$.

### The Power of the Identity

The condition $A^2 = I$ is our North Star. When we multiply $A$ by itself, we obtain:

$$A^2 = \begin{bmatrix} p^2+qr & q(p+s) \ r(p+s) & qr+s^2 \end{bmatrix}$$

Setting this equal to the identity matrix $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$, we unlock the secrets of the matrix. The off-diagonal elements must vanish, giving us $q(p+s) = 0$ and $r(p+s) = 0$.

Because the problem guarantees that all elements are non-zero, $q$ and $r$ cannot be zero. Therefore, the only logical conclusion is that $p+s = 0$.

This is a profound realization: the trace $a = p+s$ is exactly $0$.

### Unveiling the Determinant

Now, let us turn our attention to the determinant $b = |A| = ps - qr$. We know that $p+s = 0$, which implies $s = -p$.

Substituting this into our determinant formula, we get:

$$b = p(-p) - qr = -(p^2+qr)$$

Look back at the top-left element of our matrix multiplication: $p^2+qr = 1$. This is the key that unlocks the final door.

If $p^2+qr = 1$, then our determinant $b$ must be $-1$.

### The Final Synthesis

We have arrived at the summit. We found that $a = 0$ and $b = -1$.

The expression we need to evaluate is $3a^2 + 4b^2$. Substituting our values, we get:

$$3(0)^2 + 4(-1)^2$$

This simplifies beautifully to $0 + 4(1) = 4$.

It is a testament to the harmony of linear algebra that such a complex-looking constraint collapses into such a clean, integer result. The final answer is **4**.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = [a_{ij}]_{2 \times 2}$ , where $a_{ij} \neq = 0$ for all $i, j$ and $A^{2} = I$ . Let $a$ be the sum of all diagonal elements of $A$ and $b = ∣ A ∣$ . Then $3 a^{2} + 4 b^{2}$ is equal to

Select Answer:

Visualized Solution

Defining Matrix $A$

The Condition $A^{2} = I$

Computing $A^{2}$

Equating to $I$

Comparing Off-Diagonal Elements

Applying the Constraint

Finding the Value of $a$

Comparing Diagonal Elements

Finding the Value of $b$

Simplifying $b$

Calculating $b$

Final Calculation

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Power of the Identity

Unveiling the Determinant

The Final Synthesis

Similar Questions

Let $A = [21 - 1 1]$ . If the sum of the diagonal elements of $A^{13}$ is $3^{n}$ , then $n$ is equal to_________

If $A = [a b b a]$ and $A^{2} = [α β β α]$ , then

Let M be a $3 \times 3$ matrix satisfying $M 010 = - 1 23, M 1 - 1 0 = 11 - 1$ , and $M 111 = 0012$ . Then the sum of the diagonal entries of M is

Let $P$ be a square matrix such that $P^{2} = I - P$ . For $α, β, γ, δ \in N$ , if $P^{α} + P^{β} = γ I - 29 P$ and $P^{α} - P^{β} = δ I - 13 P$ , then $α + β + γ - δ$ is equal to

Suppose A is any $3 \times 3$ non-singular matrix and $(A - 3 I) (A - 5 I) = 0$ , where $I = I_{3}$ and $O = O_{3}$ . If $α A + β A^{- 1} = 4 I$ , then $α + β$ is equal to :-

Let $A = [a_{ij}]$ be $3 \times 3$ matrix such that $A 010 = 001$ , $A 413 = 010$ and $A 212 = 100$ , then $a_{23}$ equals:

Let $A$ be a $2 \times 2$ matrix with real entries such that $A^{T} = α A + I$ , where $α \in R - {- 1, 1}$ . If $det (A^{2} - A) = 4$ , the sum of all possible values of $α$ is equal to

Let A be a matrix such that $A \cdot [1023]$ is a scalar matrix and $∣3 A ∣ = 108$ . Then $A^{2}$ equals :

Let $A = 1000412 0 - 1 - 3$ . Then the sum of the diagonal elements of the matrix $(A + I)^{11}$ is equal to:

Let $A = [a_{ij}]$ be a matrix of order $3 \times 3$ , with $a_{ij} = (2)^{i + j}$ . If the sum of all the elements in the third row of $A^{2}$ is $α + β 2, α, β \in Z$ , then $α + β$ is equal to :

Select Answer:

Visualized Solution

Defining Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

The Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2=I

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2

Equating to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I

Comparing Off-Diagonal Elements

Applying the Constraint

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Comparing Diagonal Elements

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b