Identify the singular and non-singular matrices: (i) 1 2 3 \4 5 6 \7 8 9 (ii) 2 -3 5 \6 0 4 \1 5 -7 (iii) 0 a-

Exercise 7.4 · Q3

Q. Identify the singular and non-singular matrices:

(i)

\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}

(ii)

\begin{bmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{bmatrix}

(iii)

\begin{bmatrix} 0 & a-b & k \\ b-a & 0 & 5 \\ -k & -5 & 0 \end{bmatrix}

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Test each matrix by its determinant: (i) singular ( $|A|=0$ ),

(ii) non-singular ( $|A|=-28$ ),

(iii) singular (skew-symmetric of odd order, so $|A|=0$ ).

A square matrix $A$ is singular if $|A|=0$ and non-singular if $|A|\ne0$ — so the whole problem reduces to evaluating three determinants.

Step 1 — matrix (i). For $\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}$ , expand along row 1:

|A| = 1(5\cdot9-6\cdot8) - 2(4\cdot9-6\cdot7) + 3(4\cdot8-5\cdot7) = 1(-3) - 2(-6) + 3(-3) = -3+12-9 = 0.

Since $|A|=0$ , matrix (i) is singular.

Step 2 — matrix (ii). For $\begin{bmatrix}2&-3&5\\6&0&4\\1&5&-7\end{bmatrix}$ :

|A| = 2(0\cdot(-7)-4\cdot5) - (-3)(6\cdot(-7)-4\cdot1) + 5(6\cdot5-0\cdot1).

= 2(-20) + 3(-46) + 5(30) = -40 - 138 + 150 = -28.

Since $|A|=-28\ne0$ , matrix (ii) is non-singular. …

Q.Identify the singular and non-singular matrices: