Linear Algebra Set-1

[1994 : 1 M] The rank of (m x n) matrix (where m < n) cannot be more than

(a) m

(b) n

(c) mn

(d) none

Ans: (a)

Explanation:- Given that the rank of matrix is (m x n)

Now, If A is a matrix of order m x n (m < n) then rank of A is ≤ min {m, n}

\therefore e\left(A\right)\bcancel>m

So, option (a) is the correct answer.

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[1994 : 2 M] Solve the following system of equations

x₁ + x₂ + x₃ = 3

x₁ – x₂ = 0

x₁ – x₂ + x₃ = 1

(a) Unique solution

(b) No solution

(c) Infinite number of solutions

(d) Only one solution

Ans: (a)

Explanation:- The given system of equations are

x₁ + x₂ + x₃ = 3

x₁ – x₂ = 0

x₁ – x₂ + x₃ = 1

Now converting equations into matrix form, where

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

B=\begin{bmatrix}3\\0\\1\end{bmatrix}

Now, we know that C=AB, where C is the Augmented matrix

Augmented matrix (AB) =

\begin{bmatrix}1&1&1&3\\1&-1&0&0\\1&-1&1&1\end{bmatrix}\begin{array}{c}R_2-R_1\\R_3-R_1\end{array}\\ \begin{bmatrix}1&1&1&3\\0&-2&-1&-3\\0&-2&1&1\end{bmatrix}R_3-R_2\\ \begin{bmatrix}1&1&1&3\\0&-2&-1&-3\\0&0&2&4\end{bmatrix}\\ \therefore\rho(A)=\rho(AB)=3\\

⟹ Unique solution.

So, option (a) is the correct answer.

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[1998 : 1 M] The eigen values of matrix A=\begin{bmatrix}0&1\\1&0\end{bmatrix} are

(a) 1, 1

(b) -1, -1

(c) j, -j

(d) 1, -1

Ans: (d)

Explanation:- Given matrix is A=\begin{bmatrix}0&1\\1&0\end{bmatrix}

We know that ,

Characteristics equation is |A-λI| = 0

\begin{bmatrix}-\lambda&1\\-&-\lambda\end{bmatrix}=0

λ² – 1 = 0

λ = ± 1

So, option (d) is the correct answer.

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[2000 : 1 M] The eigen values of the matrix \begin{bmatrix}2&-1&0&0\\0&3&0&0\\0&0&-2&0\\0&0&-1&4\end{bmatrix} are

(a) 2, -2, 1, -1

(b) 2, 3, -2, 4

(c) 2, 3, 1, 4

(d) None

Ans: (b)

Explanation:- Given matrix is \begin{bmatrix}2&-1&0&0\\0&3&0&0\\0&0&-2&0\\0&0&-1&4\end{bmatrix}

The characteristics equation is given by,

Characteristic equation is | A – λI | = 0

\begin{vmatrix}2-\lambda&-1&0&0\\0&3-\lambda&0&0\\0&0&-2-\lambda&0\\0&0&-1&4-\lambda\end{vmatrix}=0 \begin{vmatrix}3-\lambda&0&0\\0&-2\lambda&0\\0&-1&4-\lambda\end{vmatrix}=0

(2 – λ)(3 – λ)[(-2-λ)(4-λ)] = 0

λ = 2, λ = 3, -λ = 2, 4-λ = 0

λ = -2 λ = 4

eigen values are 2, 3, -2, 4

So, option (b) is the correct answer.

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[2005 : 2 M] Given the matrix \begin{bmatrix}-4&2\\4&3\end{bmatrix}, the eigenvector is

(a) \begin{bmatrix}3\\2\end{bmatrix}

(b) \begin{bmatrix}4\\3\end{bmatrix}

(c) \begin{bmatrix}2\\-1\end{bmatrix}

(d) \begin{bmatrix}-1\\2\end{bmatrix}

Ans: (c)

Explanation :- Given matrix is \begin{bmatrix}-4&2\\4&3\end{bmatrix}

The characteristics equation is ,

A – λI = 0

\begin{bmatrix}-4&2\\4&3\end{bmatrix}-\begin{bmatrix}\lambda&0\\0&\lambda\end{bmatrix}=0 \Rightarrow\begin{bmatrix}-4-\lambda&2\\4&3-\lambda\end{bmatrix}=0

(-4 – λ)(3 – λ)-8 = 0

-12 + 4λ – 3λ + λ² – 8 = 0

λ² + λ – 20 = 0

(λ + 5)(λ – 4) = 0

∴ λ = -5, 4

Let, eigen vector be \begin{bmatrix}m_1\\m_2\end{bmatrix}; Putting, λ₁ = -5

\begin{bmatrix}1&2\\4&8\end{bmatrix}\begin{bmatrix}m_1\\m_2\end{bmatrix}=0 \begin{bmatrix}1&2\\1&2\end{bmatrix}\begin{bmatrix}m_1\\m_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix} \Rightarrow\begin{bmatrix}1&2\\0&0\end{bmatrix}\begin{bmatrix}m_1\\m_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}

m₁ + 2m₂ = 0

Taking the value, m₁ = 2 and m₂ = -1

Thus the eigen vector correspondence to eigen value \lambda_1=-5\;is\;X_1=\begin{bmatrix}2\\-1\end{bmatrix}

Option (c) is the correct answer.

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[2005 : 2 M] Let, A=\begin{bmatrix}2&-0.1\\0&3\end{bmatrix} and A^{-1}=\begin{bmatrix}\frac12&a\\0&b\end{bmatrix}. Then (a + b) =

(a) frac7{20}

(b) \frac3{20}

(c) \frac{19}{60}

(d) \frac{11}{20}

Ans: (a)

Explanation:- Given that A=\begin{bmatrix}2&-0.1\\0&3\end{bmatrix} ,

Now, we know thatthe identity matrix is given by,

\begin{bmatrix}AA^{-1}\end{bmatrix}=I \Rightarrow\begin{bmatrix}2&-0.1\\0&3\end{bmatrix}\begin{bmatrix}\frac12&a\\0&b\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix} \Rightarrow\begin{bmatrix}1&2a-0.1b\\0&3b\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}

⟹ 2a – 0.1b = 0

3b=1\Rightarrow b=\frac13 2a=\frac{0.1}3 \Rightarrow a=\frac{0.1}6=\frac1{60} a+b=\frac1{60}+\frac13=\frac{1+20}{60}=\frac{21}{60}=\frac7{20}

So, option (a) is the correct answer.

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[2005 : 2] Given an orthogonal matrix

A=\begin{bmatrix}1&1&1&1\\1&1&-1&-1\\1&-1&0&0\\0&0&1&-1\end{bmatrix}\;\begin{bmatrix}AA^T\end{bmatrix}^{-1} is

(a) \begin{bmatrix}\frac14&0&0&0\\0&\frac14&0&0\\0&0&\frac12&0\\0&0&0&\frac12\end{bmatrix}

(b) \begin{bmatrix}\frac12&0&0&0\\0&\frac12&0&0\\0&0&\frac12&0\\0&0&0&\frac12\end{bmatrix}

(c) \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}

(d) \begin{bmatrix}\frac14&0&0&0\\0&\frac14&0&0\\0&0&\frac14&0\\0&0&0&\frac14\end{bmatrix}

Ans: (c)

Explanation:- Given that A=\begin{bmatrix}1&1&1&1\\1&1&-1&-1\\1&-1&0&0\\0&0&1&-1\end{bmatrix}\;\begin{bmatrix}AA^T\end{bmatrix}^{-1}

\left[AA^T\right]^{-1}=I

For orthogonal matrix AA^T = I i.e. unity matrix.

Inverse of I = I

So, option (c) is the correct answer.

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[2006 : 1 M] The rank of the matrix \begin{bmatrix}1&1&1\\1&-1&0\\1&1&1\end{bmatrix} is

(a) 0

(b) 1

(c) 2

(d) 3

Ans: (c)

Explanation:- Given matrix is \begin{bmatrix}1&1&1\\1&-1&0\\1&1&1\end{bmatrix}

Let us apply row transformation for finding the rank of the matrix,

R₃ ⟶ R₁ – R₃

\begin{bmatrix}1&1&1\\1&-1&0\\0&0&0\end{bmatrix}

∴ Rank = 2

So, the rank is 2 and correct option is (c) .

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[2006 : 2 M] The eigenvalues and the corresponding eigenvectors of a 2 x 2 matrix are given by

Eigenvalue	Eigenvector
λ1 = 8	V_1=\begin{bmatrix}1\\1\end{bmatrix}
λ2 = 4	V_2=\begin{bmatrix}1\\-1\end{bmatrix}

The matrix is

(a) \begin{bmatrix}6&2\\2&6\end{bmatrix}

(b) \begin{bmatrix}4&6\\6&4\end{bmatrix}

(c) \begin{bmatrix}2&4\\4&2\end{bmatrix}

(d) \begin{bmatrix}4&8\\8&4\end{bmatrix}

Ans: (a)

Explanation:- Now, it is given that the value of eigen values are λ₁ = 8, λ₂ = 4,

So, for finding the matrix we will solve by taking \left[\lambda I-A\right]=0

\left[\lambda I-A\right]=0 \begin{bmatrix}\lambda&0\\0&\lambda\end{bmatrix}-\begin{bmatrix}6&2\\2&6\end{bmatrix}=1

(λ-6)² – 4 = 0

λ = 8, 4

∗ By property of eigen matrix sum of diagonal elements should be equal to sum of eigen values.

So, option (a) is the correct answer.

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[2006 : 2 M] For the matrix \begin{bmatrix}4&2\\2&4\end{bmatrix} the eigenvalue corresponding to the eigenvector \begin{bmatrix}101\\101\end{bmatrix} is

(a) 2

(b) 4

(c) 6

(d) 8

Ans: (c)

Explanation:- It is given that,

M=\begin{bmatrix} 4&2\\2&4\end{bmatrix} M-\lambda I=\begin{bmatrix}4-\lambda&2\\2&4-\lambda\end{bmatrix}

Given eigen vector

\therefore\begin{bmatrix}4-\lambda&2\\2&4-\lambda\end{bmatrix}\begin{bmatrix}101\\101\end{bmatrix}=0 \Rightarrow\left(4-\lambda\right)\left(101\right)+2\left(101\right)=0

⇒ 4 – λ + 2 = 0

λ = 6

So, option (c) is the correct answer.

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