Matrices- PYQs | Engineering Mathematics

In examinations, questions on matrices are considered very important. Most PYQs are based on definitions, basic operations, and special properties of matrices. To solve such questions, one must have a clear understanding of the fundamentals of matrices, their types, and standard results.

Short Question on Matrices

Question 1: Find the value of k such that the rank of A = \begin{bmatrix}1 & 1 & 1 & 1 \\1 & -1 & k & 1 \\3 & 1 & 0 & 1 \\\end{bmatrix}

Question 2: Find the Inverse of the matrix A = \begin{bmatrix}3 + 2i & i \\- i & 3 - 2i \\\end{bmatrix}

Question 3: If A = \begin{bmatrix}3 & 1 & 4 \\0 & 2 & 6 \\0 & 0 & 5\end{bmatrix} then find the eigenvalues of A ^-1.

Question 4: Determine if A is diagonalizable and, if so, find a diagonal matrix

A = \begin{bmatrix}2 & 0 & 3 & 0 \\0 & -3 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 5 & 0 & 2\end{bmatrix}

Question 5: Find the characteristic equation of the matrix A = \begin{bmatrix}2 & 0 & 1 \\0 & 2 & 0 \\1 & 0 & 2\end{bmatrix}

Question 6: Solve the system of equations by the Gauss-Elimination Method: 4x - y = 3, 2x + y = 5

Question 7: Show that the matrix A = \begin{bmatrix}1 & -2 \\2 & 1\\\end{bmatrix}satisfies its own characteristic equation.

Question 8: Determine the null space of A and verify the Rank-Nullity Theorem. A = \begin{bmatrix}2 & -1 \\-4 & 2 \\\end{bmatrix}

Question 9: Solve the following equation by using Cramer's Rule: 2x + 3y = 7, 3x + 5y = 9

Question 10: Find the LU Decomposition of A = \begin{bmatrix}3 & 1 \\- 6 & -4\\\end{bmatrix}

Long Question on Matrices

Question 1: Find the eigen values A = \begin{bmatrix}8 & -6 & 2 \\-6 & 7 & -4 \\2 & -4 & 3\end{bmatrix}.

Question 2: Solve the following question by gauss jordan method:

\begin{cases}10x + y + z = 12, \\2x + 10y + z = 13, \\x + y + 5z = 7\end{cases}

Question 3: Obtain the matrix A⁶ −25A ² + 122A where A = \begin{bmatrix}0 & 0 & 2 \\2 & 1 & 0 \\-1 & -1 & 3\end{bmatrix}.

Question 4: Verify Cayley- Hamilton for the matrix A = \begin{bmatrix}1 & 3 & 7 \\4 & 2 & 3 \\1 & 2 & 1\end{bmatrix}.

Question 5: Solve the following equations by using Cramer’s rule: x + 4 y + 3z = 2, 2x − 6 y + 6z = −3, 5x − 2 y + 3z = −5

Check if you were right - full answer with solution below.

Short Question on Matrices: Answers

Solution 1:

A = \begin{bmatrix}1 & 1 & 1 & 1 \\1 & -1 & k & 1 \\3 & 1 & 0 & 1 \\\end{bmatrix}

To find rank,

det(M₁​)=2k+4.

• If k ≠ −2 then det⁡(M₁) ≠ 0. rank = 3.
• If k = −2, then det⁡(M₁) = 0

M_2 = \begin{bmatrix}1 & 1 & 1 \\ 1 & -2 & 1 \\ 3 & 0 & 1\end{bmatrix} , det(M₂) = 6 ≠ 0.

so, rank = 3 for all values of k.

Solution 2:

A = \begin{bmatrix}3 + 2i & i \\- i & 3 - 2i \\\end{bmatrix}

det(A)=(3+2i)(3−2i) − (i)(−i).

= (3+2i)(3−2i)=9 − (2i)² = 9 + 4= 13

(i)(−i) =− i² = 1.

Therefore,

det⁡(A)= 13 − 1 = 12.

Adjugate matrix

adj⁡(A) = \begin{bmatrix}3 - 2i & i \\- i & 3 + 2i \\\end{bmatrix}

Finding inverse

A ^-1 = 1/12 \begin{bmatrix}3 - 2i & i \\- i & 3 + 2i \\\end{bmatrix}

Simplifying :

A ^-1 = \begin{bmatrix}\frac{1}{4} - \frac{i}{6} & -\frac{i}{12} \\\frac{i}{12} & \frac{1}{4} + \frac{i}{6}\end{bmatrix}

Solution 3:

If 𝜆 is an eigen values of A with eigenvectors v , then for A^-1 :

Av = 𝜆u
A ^-1 v = 1/ 𝜆 v

so matrix A is upper triangular matrix, so eigen values are the diagonal elements:

λ₁ = 3, λ₂ = 2, λ₃ = 5

Eigenvalues of A^-1

u₁ = 1/3 , u₂ = 1/2, u₃ = 1/5

Solution 4:

We solve det(A - 𝜆) = 0

A - 𝜆 = \begin{bmatrix}2-\lambda & 0 & 3 & 0 \\0 & -3-\lambda & 0 & 0 \\0 & 0 & -1-\lambda & 0 \\0 & 5 & 0 & 2-\lambda\end{bmatrix}

So it is upper traingular matrix

Block 1: \begin{bmatrix}2 - λ & 2\\0 & 1 - λ \\\end{bmatrix} = det = ( 2 - λ) (-1 - λ)

Block 2: \begin{bmatrix}-3 - λ & 3\\0 & 2 - λ \\\end{bmatrix} = det = (-3 - λ)(2 - λ)

So characteristic polynomial:

( 2 - λ) (-1 - λ)(-3 - λ)(2 - λ)

EigenValues:

λ₁ = 2, λ₂ = -1, λ₃ = -3

Solution 5:

The characteristic equation is

𝝀 2 − s₁𝝀 + s₂ = 0

s₁ = sum of the main diagonal element = 1 + 2 = 3

s₂ = |A| = \begin{bmatrix}1 & 2\\0 & 2 \\\end{bmatrix} = 2

Characteristic equation is 𝝀² 2 − 3𝝀 + 2 = 0

Solution 6:

Write in augmented matrix form:

\left[\begin{array}{cc|c}2 & 1 & 5 \\4 & -1 & 3\end{array}\right]

Elimination : R₂ = R₂ - 2R₁

\left[\begin{array}{cc|c}2 & 1 & 5 \\0 & -3 & -7\end{array}\right]

Back Subsitution :

y = 7/3 , x 5 - y /2 = 4 /3

so x = 4/3 and y = 7/3

Solution 7:

A = \begin{bmatrix}1 & -2 \\2 & 1\end{bmatrix}

The characteristic equation of the given matrix is

|A - λI |= 0

A -λI = \begin{bmatrix}1-\lambda & -2 \\2 & 1-\lambda\end{bmatrix}|A - \lambda I| = (1 - λ)(1 - λ) - (-2)(2) = λ² - 2λ + 5

Hence, the characteristic equation is

λ² - 2λ + 5 = 0

A² - 2A + 5 = 0

A² = \begin{bmatrix}1 & -2 \\2 & 1\end{bmatrix}\begin{bmatrix}1 & -2 \\2 & 1\end{bmatrix}

2A = \begin{bmatrix}-2 & 4 \\-4 & -2\end{bmatrix}

5I = \begin{bmatrix}5 & 0 \\0 & 5\end{bmatrix}

\begin{bmatrix}-3 & -4 \\4 & -3\end{bmatrix}+\begin{bmatrix}-2 & 4 \\-4 & -2\end{bmatrix}+\begin{bmatrix}5 & 0 \\0 & 5\end{bmatrix}=\begin{bmatrix}0 & 0 \\0 & 0\end{bmatrix}.

Solution 8:

A = \begin{bmatrix}2 & -1 \\-4 & 2 \\\end{bmatrix}

• Solve Ax = 0  ⟹  x₂ = 2x₁
  Null space: N(A) = span{(1,2)^T} Nullity = 1.
• Rows are dependent ⇒ Rank = 1.
• Rank–Nullity: 1 + 1 = 2 = number of columns.

Solution 9:

2x + 3y = 7, \quad 3x + 5y = 9

Determinats:

\begin{vmatrix}2 & 3 \\[4pt] 3 & 5 \end{vmatrix} = 2(5) - 3(3) = 1

D_x = \begin{vmatrix}7 & 3 \\[4pt] 9 & 5 \end{vmatrix} = 7(9) - 3(5) = 8

D_y = \begin{vmatrix}2 & 7 \\[4pt] 3 & 9 \end{vmatrix} = 2(9) - 7(3) = -3

x = D_x/D = 8/1 = 8

y= D_y/D = -3/1 = -3

(x, y) = ( 8, -3)

Solution 10:

A = \begin{bmatrix}3 & 1 \\[4pt] -6 & -4\end{bmatrix}

Let

L = \begin{bmatrix}1 & 0 \\[4pt] \ell_{21} & 1 \end{bmatrix}, \quad U = \begin{bmatrix}u_{11} & u_{12} \\[4pt] 0 & u_{22} \end{bmatrix}.

From calculations:

u_{11} = 3,\quad u_{12} = 1,\quad \ell_{21} = \frac{-6}{3} = -2,

\quad u_{22} = -4 - (-2)(1) = -2.

Thus

L = \begin{bmatrix}1 & 0 \\[4pt] -2 & 1\end{bmatrix},

\quad

U = \begin{bmatrix}3 & 1 \\[4pt] 0 & -2\end{bmatrix},

\boxed{A = LU}

Long Question on Matrices: Answers

Solution 1:

A = \begin{bmatrix}8 & -6 & 2 \\-6 & 7 & -4 \\2 & -4 & 3\end{bmatrix}

Characteristic polynomial:

\det(A-\lambda I)=\det\begin{bmatrix}8-\lambda & -6 & 2\\[4pt]-6 & 7-\lambda & -4\\[4pt]2 & -4 & 3-\lambda\end{bmatrix}

Expanding along the first row:

(8 - λ)\begin{vmatrix}7-\lambda & -4\\[4pt]-4 & 3-\lambda\end{vmatrix}- (-6)\begin{vmatrix}-6 & -4\\[4pt]2 & 3-\lambda\end{vmatrix}+2\begin{vmatrix}-6 & 7-\lambda\\[4pt]2 & -4\end{vmatrix}\\

= ( 8 - λ)((7 - λ)(3 - λ) -16) + 6 ((-6)(3 - λ) - (-8)) + 2(24 - (14 - 2λ))

Simplifying the minors:

( 7 - λ)(3 - λ) - 16 = λ² - 10λ + 5

(- 6)(3 - λ) - (- 8) = 6λ - 10

24 - (14 - 2λ) = (8 - λ)(λ² - 10λ + 5) + 6(6λ - 10) + 2(10 + 2λ)

Expand and combine terms:

(8−λ)(λ²−10λ+5) = −λ³ + 18λ² − 85λ + 40 + 40λ − 40
= −λ³+ 18λ² − 45

Factor: −λ(λ² − 18λ + 45)
λ² − 18λ + 45=(λ−3)(λ−15)

Hence eigenvalues: λ=0, 3, 15.

Solution 2:

\begin{cases}10x + y + z = 12, \\ 2x + 10y + z = 13, \\ x + y + 5z = 7\end{cases}

Step 1: Augmented matrix:

\left[\begin{array}{ccc|c}10 & 1 & 1 & 12 \\ 2 & 10 & 1 & 13 \\ 1 & 1 & 5 & 7\end{array}\right]

Step 2: Make the first pivot 1 R ₁ → R₁/10

\left[\begin{array}{ccc|c}10 & 1 & 1 & 12 \\ 2 & 10 & 1 & 13 \\ 1 & 1 & 5 & 7\end{array}\right]

R_2​ → R_{2 ​}− 2R₁​, R₃ ​→ R₃ ​− R_1​

\left[\begin{array}{ccc|c}1 & 0.1 & 0.1 & 1.2 \\ 0 & 9.8 & 0.8 & 10.6 \\ 0 & 0.9 & 4.9 & 5.8\end{array}\right]

R_2​ → 9.8R_2​​

\left[\begin{array}{ccc|c}1 & 0.1 & 0.1 & 1.2 \\ 0 & 1 & 0.0816 & 1.0816 \\ 0 & 0.9 & 4.9 & 5.8\end{array}\right]

R_1​ → R₁​ − 0.1R₂​, R_3​ → R₃​ − 0.9R₂​

\left[\begin{array}{ccc|c}1 & 0 & 0.0918 & 1.0918 \\ 0 & 1 & 0.0816 & 1.0816 \\ 0 & 0 & 4.826 & 4.826\end{array}\right]

Step 7: Eliminate third column in R₁ and R₂

\left[\begin{array}{ccc|c}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1\end{array}\right]

R₃​ → 4.826/R_3​​

\left[\begin{array}{ccc|c}1 & 0 & 0.0918 & 1.0918 \\ 0 & 1 & 0.0816 & 1.0816 \\ 0 & 0 & 1 & 1\end{array}\right]

R_1​ →R₁​ − 0.0918R₃ ​, R₂​ → R₂​ − 0.0816R₃​

\left[\begin{array}{ccc|c}1 & 0 & 0 & 1 \\0 & 1 & 0 & 1 \\0 & 0 & 1 & 1\end{array}\right]

x = 1, y = 1, z =1

Solution 3:

Find characteristic polynomial

det⁡(A−λI) = 0

A - λ = \begin{bmatrix} -\lambda & 0 & 2 \\ 2 & 1-\lambda & 0 \\ -1 & -1 & 3-\lambda \end{bmatrix}

Compute the determinant:

det⁡(A−λI) = -\lambda \begin{vmatrix}1-\lambda & 0 \\ -1 & 3-\lambda\end{vmatrix} + 2 \begin{vmatrix}2 & 1-\lambda \\ -1 & -1\end{vmatrix}

Compute first minor:

\begin{vmatrix}1-\lambda & 0 \\ -1 & 3-\lambda\end{vmatrix} ​​= (1 − λ)(3 − λ) − (0)(−1) = (1 − λ)(3 − λ)

Compute second minor:

\begin{vmatrix}2 & 1-\lambda \\ -1 & -1\end{vmatrix} = -2 + 1 - λ = -1 - λ

Now, determinant:

det⁡(A − λI) = −λ(λ² − 4λ + 3) − 2 − 2λ =− λ³ + 4λ² − 3λ − 2 − 2λ = −λ³ + 4λ² − 5λ − 2

Multiply by -1 for simplicity:

λ³ − 4λ² + 5λ + 2 = 0

So the characteristic equation is:

λ³ − 4λ² + 5λ + 2=0

Characterstic Equation :

A³− 4A² + 5A + 2I = 0  ⟹  A3 = 4A² − 5A −2I= 0

A⁴ = A⋅A³ = A(4A² − 5A − 2I) = 4A³ −5A² − 2A = 4(4A² − 5A −2I) − 5A² − 2A = 11A² − 22A − 8I

A⁶ = (A³)² = (4A² − 5A − 2I)²

Compute step by step:

16A⁴− 40A³+ 9A² + 20A + 4I

Substitute A⁴ and A³:

16(11A² − 22A − 8I) − 40(4A² − 5A − 2I) + 9A² + 20A + 4I

Simplify:

A⁶ = 25A² − 132A − 44I

A⁶ − 25A² + 122A = (25A²− 132A −44I) − 25A² + 122A = −10A−44I

−10A − 44I = -10 \begin{bmatrix}0 & 0 & 2 \\2 & 1 & 0 \\-1 & -1 & 3\end{bmatrix}-44\begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{bmatrix}=

\begin{bmatrix}-44 & 0 & -20 \\-20 & -54 & 0 \\10 & 10 & -74\end{bmatrix}

Solution 4:

Characteristic equation: ∣A−λI∣ = 0

λ³− S₁λ² + S₂λ − S₃ = 0

Where:

• S₁ = sum of diagonal elements= 1+ 2 + 1 = 4
• S₂ = sum of minors of leading diagonal elements.

Compute S₂ :

• Minor of a₁₁ = \begin{vmatrix}2 & 3 \\ 2 & 1\end{vmatrix} = 2.1 - 3.2 = 2 - 6 = -4
• Minor of a₂₂ = \begin{vmatrix}1 & 7 \\ 1 & 1\end{vmatrix} = 1.1 -7.1 = 1 - 7 = -6
• Minor of a₃₃ = \begin{vmatrix}1 & 3 \\ 4 & 2\end{vmatrix} = 1.2 -3.4 = 2 -12 = -10​

S₂ = −4 − 6 − 10= −20

S₃ = \begin{bmatrix}1 & 4 & 1 \\3 & 2 & 2 \\7 & 3 & 1\end{bmatrix} = 1(2.1 − 3.2) − 3(4.1 − 3∗.1) + 7(4.2 − 2.1) = 1(−4) − 3(1) + 7(6) = −4 −3 + 42 = 35

λ³ − 4λ² − 20λ − 35 = 0

By Cayley-Hamilton:

A³ − 4A² − 20A − 35I = 0  

A = \begin{bmatrix}1 & 3 & 7 \\4 & 2 & 3 \\1 & 2 & 1\end{bmatrix}

A² = A . A

= \begin{bmatrix}1 & 3 & 7 \\4 & 2 & 3 \\1 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 3 & 7 \\4 & 2 & 3 \\1 & 2 & 1\end{bmatrix}

=\begin{bmatrix}1*1 + 3*4 + 7*1 & 1*3 + 3*2 + 7*2 & 1*7 + 3*3 + 7*1 \\4*1 + 2*4 + 3*1 & 4*3 + 2*2 + 3*2 & 4*7 + 2*3 + 3*1 \\1*1 + 2*4 + 1*1 & 1*3 + 2*2 + 1*2 & 1*7 + 2*3 + 1*1\end{bmatrix}

= \begin{bmatrix}20 & 23 & 23\\15 & 22 & 37 \\10 & 9 & 14\end{bmatrix}

A³ - 4A² - 20A - 35I = 0

A^3 =\begin{bmatrix}135 & 152 & 232 \\140 & 163 & 208 \\60 & 76 & 111\end{bmatrix}

4A^2 =\begin{bmatrix}80 & 92 & 92 \\60 & 88 & 148 \\40 & 36 & 56\end{bmatrix}

20A =\begin{bmatrix}20 & 60 & 140 \\80 & 40 & 60 \\20 & 40 & 20\end{bmatrix}

35I = 35I =\begin{bmatrix}35 & 0 & 0 \\0 & 35 & 0 \\0 & 0 & 35\end{bmatrix}

Perform the subtraction element-wise:

A³ - 4A² - 20A - 35I

\begin{bmatrix}135 - 80 - 20 - 35 & 152 - 92 - 60 - 0 & 232 - 92 - 140 - 0 \\140 - 60 - 80 - 0 & 163 - 88 - 40 - 35 & 208 - 148 - 60 - 0 \\60 - 40 - 20 - 0 & 76 - 36 - 40 - 0 & 111 - 56 - 20 - 35\end{bmatrix}

Calculate each element:

= \begin{bmatrix}0 & 0 & 0 \\0 & 0 & 0 \\0 & 0 & 0\end{bmatrix}

Thus, we verify that:

A³ - 4A² - 20A - 35I = 0

Solution 5:

Solve the system using Cramer's Rule:

\begin{cases}x + 4y + 3z = 2 \\2x - 6y + 6z = -3 \\5x - 2y + 3z = -5\end{cases}

Step 1: Write coefficient matrix and constants vector

A = \begin{bmatrix}1 & 4 & 3 \\2 & -6 & 6 \\5 & -2 & 3\end{bmatrix}, \quad

\mathbf{B} = \begin{bmatrix}2 \\-3 \\-5\end{bmatrix}\]

Step 2: Compute det(A_x)

\det(A) = \begin{vmatrix}1 & 4 & 3 \\2 & -6 & 6 \\5 & -2 & 3\end{vmatrix}

1\cdot\begin{vmatrix}-6 & 6 \\-2 & 3\end{vmatrix}- 4 \cdot \begin{vmatrix}2 & 6 \\5 & 3\end{vmatrix}+ 3 \cdot \begin{vmatrix}2 & -6 \\5 & -2\end{vmatrix}

Calculate minors:

M₁₁ = (-6)(3) - (6)(-2) = -18 + 12 = - 6

M₁₂ = 2(3) - 6(5) = 6 - 30 = - 24

M₁₃ = 2(-2) - (-6)5 = -4 + 30 = 26

\begin{aligned}M_{11} &= (-6)(3) - (6)(-2) = -18 + 12 = -6 \\M_{12} &= 2 \cdot 3 - 6 \cdot 5 = 6 - 30 = -24 \\M_{13} &= 2 \cdot (-2) - (-6) \cdot 5 = -4 + 30 = 26\end{aligned}

Substitute back:

det(A) = 1(-6) - 4(-24) + 3(26) = -6 + 96 + 78 = 168

A_x = \begin{bmatrix}2 & 4 & 3 \\-3 & -6 & 6 \\-5 & -2 & 3\end{bmatrix}

\det(A_x) = 2 \cdot\begin{vmatrix}-6 & 6 \\-2 & 3\end{vmatrix}- 4 \cdot\begin{vmatrix}-3 & 6 \\-5 & 3\end{vmatrix}+ 3 \cdot\begin{vmatrix}-3 & -6 \\-5 & -2\end{vmatrix}\]

Calculate minors:

M₁₁ = (-6)(3) - (6)(-2) = -18 + 12 = -6

M₁₂ = (-3)(3) - 6(-5) = -9 + 30 = 21

M₁₃ = (-3)(-2) - (-6)(-5) = 6 - 30 = -24

Substitute:

det(A_x) = 2(-6) - 4(21) + 3(-24) = -12 - 84 - 72 = -168

A_y = \begin{bmatrix}1 & 2 & 3 \\2 & -3 & 6 \\5 & -5 & 3\end{bmatrix}

\det(A_y) = 1 \cdot\begin{vmatrix}-3 & 6 \\-5 & 3\end{vmatrix}- 2 \cdot\begin{vmatrix}2 & 6 \\5 & 3\end{vmatrix}+ 3 \cdot\begin{vmatrix}2 & -3 \\5 & -5\end{vmatrix}

Calculate minors:

M₁₁ = (-3)(3) - 6(-5) = -9 + 30 = 21

M₁₂ = 2(3) - 6(5) = 6 - 30 = -24

M₁₃ = 2(-5) - (-3)5 = -10 + 15 = 5

det(A_y) = 1(21) - 2(-24) + 3(5) = 21 + 48 + 15 = 34

\det(A_y) = 1 \times 21 - 2 \times (-24) + 3 \times 5 = 21 + 48 + 15 = 84

A_z = \begin{bmatrix}1 & 4 & 2 \\2 & -6 & -3 \\5 & -2 & -5\end{bmatrix}

det(A_z) = 1 \cdot\begin{vmatrix}-6 & -3 \\-2 & -5\end{vmatrix}- 4 \cdot\begin{vmatrix}2 & -3 \\5 & -5\end{vmatrix}+ 2 \cdot\begin{vmatrix}2 & -6 \\5 & -2\end{vmatrix}

Calculate minors:

M₁₁ = (-6)(-5) - (-3)(-2) = 30 - 6 = 24

M₁₂= 2(-5) - (-3)5 = -10 + 15 = 5

M₁₃ = 2(-2) - (-6)5 = -4 + 30 = 26

Substitute:

det(A_z) = 1(24) - 4(5) + 2(26) = 24 -20 + 52 = 56

x = \frac{\det(A_x)}{\det(A)} = \frac{-168}{168} = -1

y = \frac{\det(A_y)}{\det(A)} = \frac{84}{168} = \frac{1}{2}

z = \frac{\det(A_z)}{\det(A)} = \frac{56}{168} = \frac{1}{3}

x = - 1, y = 1/2 , z = 1/3