HOW TO CHECK WHETHER THE MATRIX IS INVERTIBLE OR NOT

A square matrix whose determinant is 0 is called singular matrix.

|A| ≠ 0

To check if the function is invertible or not, we have to follow the steps.

i) Let us consider the given matrix as A.

ii) Finding (|A|) determinant of A.

iii)  If |A| ≠ 0, then the given matrix is non singular and it is not invertible. Inverse does not exists.

iv)  If |A| = 0, then the given matrix is singular matrix and it is invertible.

Identify the singular and non-singular matrix.

Problem 1 :

149491691625

Solution:

In order to check if the given matrix is singular or non singular, we have to find the determinant of the given matrix.

Let A=149491691625=149491691625=1[225-256]-4[100-144]+9[64-81]=1[-31]-4[-44]+9[-17]=-31+176-153=-184+176=-8

It is not equal to zero. Hence it is non singular matrix.

Problem 2 :

123456-2-4-6

Solution:

Let A=123456-2-4-6=123456-2-4-6=1[-30+24]-2[-24+12]+3[-16+10]=1[-6]-2[-12]+3[-6]=-6+24-18=24-24=0

Hence the matrix is singular matrix.

Problem 3 :

Solve:

2x4321123=-3

Solution:

2x4321123=-32[6-2]-x[9-1]+4[6-2]=-32[4]-x[8]+4[4]=-38-8x+16=-3-8x+24=-3-8x=-3-24-8x=-27x=278

Problem 4 :

4393-2744x=-1

Solution:

4393-2744x=-14[-2x-28]-3[3x-28]+9[12+8]=-14[-2x-28]-9x+84+9[20]=-1-8x-112-9x+84+180=-1-17x+152=-117x=153x=15317x=9

Problem 5 :

Find the value of the determinant 264-5-15-10132 without usual expansion.

Solution:

264-5-15-10132

Taking out 2 common from C3.

=2262-5-15-5131

If any two row or column are identical then value of determinant is zero.

Here C1 and C3 are identical.

= 2 × 0

= 0

Hence the determinant of given is zero by calculating without usual expansion.

Problem 6 :

Solve:

x-1xx-20x-2x-300x-3=0

Solution:

Since, all the entries below the principal diagonal are zero, the value of the determinant is (x - 1)(x - 2)(x -3) = 0 which gives x = 1, 2, 3.

Problem 7 :

Solve for x if 010x2x13x=0

Solution:

010x2x13x=00(2x-3x)-1x2-x+0(3x-2)=00-1x2-x+0=0-x2+x=0x(1-x)=0x=0 and x=1

Problem 8 :

Solve for x if x57x+1-2-11=0

Solution:

x57x+1-2-11=0x2-35+(1-2)=0x2-35-1=0x2-36=0x2=36x=±6

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