How to solve AX=B when adj(A) is given?

Use A⁻¹ = adj(A)/|A|. So X = A⁻¹B = adj(A)·B / |A|. First find |A| from det(adj A) = |A|^(n-1).

What is the relation between det(adj A) and det(A)?

For n×n matrix: det(adj A) = |A|^(n-1). For n=3: det(adj A) = |A|². So |A| = ±√(det(adj A)).

How to compute det(adj A) for the given matrix?

det([[4,2,2],[-5,0,5],[1,-2,3]]) = 4(10) - 2(-20) + 2(10) = 40+40+20 = 100. So |A|²=100, |A|=±10.

What is X = adj(A)·B / |A|?

With |A|=10: X = [20,-10,10]/10 = [2,-1,1]. So x+y+z = 2-1+1 = 2. |x+y+z|=2.

Why does |A| = ±10 not matter for the final answer?

If |A|=-10: X=[-2,1,-1], sum=-2. |sum|=2. Either way |x+y+z|=2.

What is the adjugate (adjoint) matrix?

adj(A) is the transpose of the cofactor matrix of A. Key property: A·adj(A) = |A|·I, so A⁻¹ = adj(A)/|A| when |A|≠0.

A·adj(A) = adj(A)·A = |A|·I (identity scaled by determinant). This is the key identity for matrix inverse.

How to verify the solution x=2, y=-1, z=1?

We need AX=B. Since adj(A)·X = adj(A)·A⁻¹·B... Actually, verify by checking |x+y+z|=|2-1+1|=|2|=2 ✓

What if |A|=0 (singular matrix)?

If |A|=0, A⁻¹ doesn't exist and the system either has no solution or infinitely many. Here |A|=±10≠0, so unique solution exists.

If X=[x y z]^T is a solution of AX=B where adj A=[[4,2,2],[-5,0,5],[1,-2,3]] and B=[4,0,2]^T then |x+y+z| is equal to

Q: How to compute adj(A)·B?

[[4,2,2],[-5,0,5],[1,-2,3]]·[4,0,2]^T = [16+0+4, -20+0+10, 4+0+6] = [20,-10,10].

If X=[x y z]^T is a solution of AX=B where adj A=[[4,2,2],[-5,0,5],[1,-2,3]] and B=[4,0,2]^T then |x+y+z| is equal to | JEE Main Mathematics

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Math Matrices

Q MCQ Matrices

If $X = \begin{bmatrix}x\\y\\z\end{bmatrix}$ is a solution of the system of equations $AX = B$, where $$\text{adj}\,A = \begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}, \quad B = \begin{bmatrix}4\\0\\2\end{bmatrix}$$ then $|x+y+z|$ is equal to:

A) 3

B) 2 ✓

C) $\dfrac{3}{2}$

D) 1

✅ Correct Answer

B) 2

✓

Solution Steps

Key identity: use A·adj(A) = |A|·I

From $AX = B$, multiply both sides by $\text{adj}(A)$:

$$\text{adj}(A)\cdot AX = \text{adj}(A)\cdot B$$

Since $\text{adj}(A)\cdot A = |A|\cdot I$:

$$|A|\cdot X = \text{adj}(A)\cdot B$$

Find |A| using det(adj A) = |A|²

For a $3\times3$ matrix: $\det(\text{adj}\,A) = |A|^{n-1} = |A|^2$

Expand $\det(\text{adj}\,A)$ along Row 1:

$= 4\bigl[(0)(3)-(5)(-2)\bigr] – 2\bigl[(-5)(3)-(5)(1)\bigr] + 2\bigl[(-5)(-2)-(0)(1)\bigr]$

$= 4(10) – 2(-20) + 2(10)$

$= 40 + 40 + 20 = 100$

$$|A|^2 = 100 \implies |A| = \pm 10$$

Compute adj(A) · B

$$\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}\begin{bmatrix}4\\0\\2\end{bmatrix}$$

Row 1: $4(4)+2(0)+2(2) = 16+0+4 = 20$

Row 2: $(-5)(4)+0(0)+5(2) = -20+0+10 = -10$

Row 3: $1(4)+(-2)(0)+3(2) = 4+0+6 = 10$

$$\text{adj}(A)\cdot B = \begin{bmatrix}20\\-10\\10\end{bmatrix}$$

Find X = adj(A)·B / |A|

Case |A| = +10:

$X = \dfrac{1}{10}\begin{bmatrix}20\\-10\\10\end{bmatrix} = \begin{bmatrix}2\\-1\\1\end{bmatrix}$

$x+y+z = 2+(-1)+1 = 2$

Case |A| = −10:

$X = \dfrac{1}{-10}\begin{bmatrix}20\\-10\\10\end{bmatrix} = \begin{bmatrix}-2\\1\\-1\end{bmatrix}$

$x+y+z = -2+1+(-1) = -2$

Final answer

In both cases: $|x+y+z| = |2| = |-2| = \boxed{2}$

✓ The absolute value ensures a unique answer regardless of the sign of |A|.

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Key Insight

Multiply AX=B by adj(A) to get |A|·X = adj(A)·B. Find |A|=±10 from det(adj A)=|A|². Then |x+y+z|=2 regardless of sign of |A|.

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Theory

1. Adjugate Matrix and Inverse

For a square matrix $A$: $A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = |A| \cdot I$. Therefore $A^{-1} = \dfrac{\text{adj}(A)}{|A|}$ when $|A| \neq 0$. The adjugate $\text{adj}(A)$ is the transpose of the cofactor matrix. This identity is the backbone of Cramer’s rule and solving $AX=B$ without explicitly inverting $A$.

2. Determinant of Adjugate Matrix

For an $n \times n$ matrix: $\det(\text{adj}\,A) = |A|^{n-1}$. For $n=3$: $\det(\text{adj}\,A) = |A|^2$. So $|A| = \pm\sqrt{\det(\text{adj}\,A)}$. This allows finding $|A|$ directly from the given $\text{adj}(A)$ without recovering $A$ itself — a powerful shortcut in JEE problems.

3. Solving AX = B via Adjugate

From $AX = B$: multiply both sides on the left by $\text{adj}(A)$: $\text{adj}(A) \cdot AX = \text{adj}(A) \cdot B$, giving $|A| \cdot X = \text{adj}(A) \cdot B$. So $X = \dfrac{\text{adj}(A) \cdot B}{|A|}$. This method avoids finding $A$ itself when only $\text{adj}(A)$ and $B$ are given.

4. Why the Absolute Value Matters

$|A| = \pm 10$ gives two cases for $X$. With $|A|=+10$: $X=(2,-1,1)$, sum $=2$. With $|A|=-10$: $X=(-2,1,-1)$, sum $=-2$. The question asks for $|x+y+z|$ (absolute value), giving 2 in both cases. In JEE, when $|A|$ sign is ambiguous, asking for the absolute value of the answer is the standard trick to make the answer unique.

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FAQs

What is adj(A)·A equal to?

⌄

adj(A)·A = A·adj(A) = |A|·I. This is the key identity used in this solution.

How is det(adj A) = |A|² derived?

⌄

For n×n: det(adj A) = |A|^(n-1). For n=3: det(adj A) = |A|². Taking determinant of both sides of A·adj(A)=|A|·I: |A|·det(adj A) = |A|^3 → det(adj A) = |A|².

Compute det(adj A) step by step.

⌄

Expand along row 1: 4[(0)(3)-(5)(-2)] – 2[(-5)(3)-(5)(1)] + 2[(-5)(-2)-(0)(1)] = 4(10)-2(-20)+2(10) = 40+40+20 = 100.

How to compute adj(A)·B?

⌄

Row1: 4(4)+2(0)+2(2)=20. Row2: -5(4)+0(0)+5(2)=-10. Row3: 1(4)-2(0)+3(2)=10. Result: [20,-10,10].

Why does |A| = ±10?

⌄

det(adj A)=100=|A|². Taking square root: |A|=±10. Both signs are valid since we only know det(adj A).

What is X when |A|=10?

⌄

X = [20,-10,10]/10 = [2,-1,1]. So x=2, y=-1, z=1. x+y+z = 2.

What is X when |A|=-10?

⌄

X = [20,-10,10]/(-10) = [-2,1,-1]. x+y+z = -2+1-1 = -2. |x+y+z| = 2. Same final answer.

What is the adjugate matrix?

⌄

adj(A) = transpose of cofactor matrix. Each element C_ij = (-1)^(i+j) × minor M_ij. Then adj(A)_ij = C_ji (transposed).

Can we solve this without finding A?

⌄

Yes! That’s the beauty: use |A|·X = adj(A)·B directly. No need to recover A from adj(A).

What if the matrix were 2×2?

⌄

For 2×2: det(adj A) = |A|^1 = |A|. For n×n: det(adj A) = |A|^(n-1). The formula changes with matrix size.

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