JEE Main 2025 Matrices PYQ with solution

JEE Main 2025 Matrices PYQ with solution helps you score easy marks in JEE Main and advance. These questions are based on simple matrices formulas. Solving more question helps student to improve score in exam.

Question :

Let A be a 3×3 real matrix such that A²[A – 2I] – 4[A – I] = 0 Where I and 0 are the identity and null matrices respectively If A⁵ = $\alpha$A² + $\beta$A + $\gamma$I Where $\alpha$, $\beta$ and $\gamma$ are real constants, then $\alpha$ + $\beta$ + $\gamma$

Options :

(A) 12

(B) 20

(C) 76

(D) 4

JEE Main 2025 Matrices PYQ with solution

Solution :

A²[A – 2I] – 4[A – I] = 0

A³ – 2A² – 4A + 4I = 0

A³ = 2A² + 4A – 4I

Multiply by A on both sides

A⁴ = 2A³ + 4A² – 4A

A⁴ = 2[ 2A² + 4A – 4I] + 4A² – 4A

A⁴ = 4A² +8A – 8I + 4A² – 4A

A⁴ = 8A² + 4A – 8I

Multiply by A on both sides

A⁵ = 8A³ + 4A² – 8A

A⁵ = 8(2A² + 4A – 4I) + 4A² – 8A

A⁵ = 16A² + 32A – 32I + 4A² – 8A

A⁵ = 20A² + 24A – 32I

By camparing with

A⁵ = $\alpha$A² + $\beta$A + $\gamma$I

$\alpha$ = 20, $\beta$ = 24, $\gamma$ = -32

$\alpha$ + $\beta$ + $\gamma$ = 20 + 24 – 32 = 12

Practice Question :

1. If A + B = $\begin{bmatrix} 12 & 6 &2 \\ 8 & 4 &0\end{bmatrix}$ and A – B = $\begin{bmatrix}2 & 0 & 4 \\4 & -2& 6\end{bmatrix}$ Find Matrices A and B

Solution :

Given

A + B = $\begin{bmatrix} 12 & 6 &2 \\ 8 & 4 &0\end{bmatrix}$ ——————————- 1

A – B = $\begin{bmatrix}2 & 0 & 4 \\4 & -2& 6\end{bmatrix}$ ——————————–2

Adding 1 and 2 equation

2A = $\begin{bmatrix}14 & 6 & 6 \\12 & 2 & 6 \end{bmatrix}$

A = $\frac{1}{2}\begin{bmatrix}14 & 6 & 6 \\12 & 2 & 6 \end{bmatrix}$

A = $\begin{bmatrix} 7 & 3 & 3 \\6 & 1 & 3\end{bmatrix}$

Subtracting 1 and 2 equation

2B = $\begin{bmatrix}10 & 6 & -2 \\4 & -2 & 6 \end{bmatrix}$

B = $\begin{bmatrix} 5 & 3 & -1 \\ 2 & 3 & -3\end{bmatrix}$

2. If $\begin{bmatrix} x + 3 & 2y + x \\ z – 1 & 4w – 8 \end{bmatrix}$ = $\begin{bmatrix} -x -1 & 0 \\ 3 & 2w \end{bmatrix}$ Find the value of | x + y| + | z + y |

Solution :

x + 3 = -x – 1

2x = -4

x = -2

2y + x = 0

2y – 2 = 0

2y = 2

y = 1

z – 1 = 3

z = 4

4w – 8 = 2w

2w = 8

w = 4

Substitude x, y , z, w in | x + y| + | z + y |

| -2 + 1| + |4 + 4| = 1 + 8 = 9

3. Let A be a square matrix such that AA^T = I Then $\frac{1}{2}$A[ ( A + A^T)² + [A – A^T]² is equal to

Solution :

Given

$\frac{1}{2}$A[ A + A^T]² + [A – A^T]²]

$\frac{1}{2}$A [A² + (A^T)² + 2AA^T + A² + (A^T)² – 2AA^T]

$\frac{1}{2}$A [ 2(A² + (A^T)²]

A³ + A.A^T.A^T

A³ + A^T

4. If |A| = 2 where A is a square matrix of order n = 3 the find

(a) |adj A|

Formula : |adjA| = |A|^{n – 1}

|adj A|= |A|²

2² = 4

(b) |adj(adjA)|

Formula : |adj(adjA)|= $|A|^{(n-2)^2}$

= 2⁴

= 16

(c) |adj (A^T)|

Formula : |adjA| = |A|^{n – 1}

= |A|² = 4

(d) |adj(5A)|

Formula : |adjkA| = K^{n(n – 1)}.|adjA|

= 5³⁽²⁾.2²

= 5⁶.4

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