Engineering Mathematics

Singular Matrix — GATE CE 2018

Question

Which one of the following matrices is singular?

Option A

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

Option B

B= \begin{bmatrix} 3 & 28\\ 2 & 3 \end{bmatrix}

Option C

C= \begin{bmatrix} 2 & 4\\ 3 & 6 \end{bmatrix}

Option D

D= \begin{bmatrix} 4 & 3\\ 6 & 2 \end{bmatrix}

Given

We have four \(2\times2\) matrices and need to determine which one is singular.

Concept Used

A square matrix is called singular if its determinant is zero.

\boxed{ \det(A)=0 \iff A\text{ is singular} }

If the determinant is non-zero, the matrix is non-singular:

\boxed{ \det(A)\neq0 \iff A\text{ is non-singular} }

Determinant of a \(2\times2\) matrix

For

A= \begin{bmatrix} a & b\\ c & d \end{bmatrix}

the determinant is

\boxed{ \det(A)=ad-bc }

Solution

Step 1: Check Option A

Consider

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

Using the formula

\det(A)=ad-bc

Here,

a=2,\qquad b=5,\qquad c=1,\qquad d=3

Therefore,

\det(A) =(2)(3)-(5)(1) \] \[ =6-5 \] \[ \boxed{\det(A)=1}

Since

\det(A)\neq0

Option A is non-singular.

Step 2: Check Option B

Consider

B= \begin{bmatrix} 3 & 28\\ 2 & 3 \end{bmatrix}

Using

\det(B)=ad-bc

We get

\det(B) =(3)(3)-(28)(2) \] \[ =9-56 \] \[ \boxed{\det(B)=-47}

Since

-47\neq0

Option B is non-singular.

Step 3: Check Option C

Consider

C= \begin{bmatrix} 2 & 4\\ 3 & 6 \end{bmatrix}

Using

\det(C)=ad-bc

Therefore,

\det(C) =(2)(6)-(4)(3) \] \[ =12-12 \] \[ \boxed{\det(C)=0}

Since

\det(C)=0

the matrix \(C\) is singular.

\boxed{ C\text{ is singular} }

Step 4: Check Option D

Consider

D= \begin{bmatrix} 4 & 3\\ 6 & 2 \end{bmatrix}

Using

\det(D)=ad-bc

Therefore,

\det(D) =(4)(2)-(3)(6) \] \[ =8-18 \] \[ \boxed{\det(D)=-10}

Since

-10\neq0

Option D is non-singular.

Step 5: Quick GATE Method

There is an even faster way to identify Option C.

Look at the rows of

C= \begin{bmatrix} 2 & 4\\ 3 & 6 \end{bmatrix}

The second row is proportional to the first row:

[3\quad6] = \frac{3}{2} [2\quad4]

Therefore, the two rows are linearly dependent.

A matrix having linearly dependent rows has determinant zero.

\boxed{ \text{Linearly dependent rows} \Rightarrow \det(A)=0 }

Hence, \(C\) is singular.

Final Answer

\boxed{\text{Option C}}

The singular matrix is

\boxed{ \begin{bmatrix} 2 & 4\\ 3 & 6 \end{bmatrix} }

because its determinant is zero:

\boxed{ \det(C)=12-12=0 }

Formulas & Concepts to Remember

1. Determinant of \(2\times2\) matrix

\boxed{ \begin{vmatrix} a & b\\ c & d \end{vmatrix} = ad-bc }

2. Singular matrix

\boxed{ \det(A)=0 \iff A\text{ is singular} }

3. Non-singular matrix

\boxed{ \det(A)\neq0 \iff A\text{ is non-singular} }

4. Linearly dependent rows or columns

\boxed{ \text{Rows/columns linearly dependent} \Rightarrow \det(A)=0 }

5. Inverse connection

A square matrix has an inverse if and only if its determinant is non-zero:

\boxed{ A^{-1}\text{ exists} \iff \det(A)\neq0 }

GATE Tip

Don't blindly calculate determinants for every option. First look for structural shortcuts.

If two rows or columns are proportional, you can immediately conclude:

\boxed{\det(A)=0}

In this question:

[3\quad6] = \frac32[2\quad4]

So the answer can be identified almost instantly as Option C.

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