1.

If A = \(\begin{bmatrix}1 & 0 & 1 \\[0.3em]0&1 & 2 \\[0.3em]0 & 0 & 4\end{bmatrix}\), then show that |3A| = 27|A|.

Answer»

|A| = \(\begin{vmatrix} 1 & 0 & 1 \\[0.3em] 0&1 & 2 \\[0.3em] 0 & 0 & 4 \end{vmatrix}\) 

Expanding along the first row,

|A| = 1\(\begin{vmatrix} 1 & 2 \\[0.3em] 0&4 \\[0.3em] \end{vmatrix}\) - 0\(\begin{vmatrix} 0 & 2 \\[0.3em] 0&4 \\[0.3em] \end{vmatrix}\) + 1\(\begin{vmatrix} 0 & 1 \\[0.3em] 0&0 \\[0.3em] \end{vmatrix}\)

= 1(1×4 – 2×0) – 0(0×4 – 0×2) + 1(0×0 – 0×1) 

= 1(4 – 0) + 0 + 1(0 + 0) 

= 1×4 

= 4 

Now,

|3A| = \(\begin{vmatrix} 3 & 0 & 3 \\[0.3em] 0&3 & 6 \\[0.3em] 0 & 0 & 12 \end{vmatrix}\) 

Expanding along the first row,

 |3A| = 3\(\begin{vmatrix} 3 & 6 \\[0.3em] 0&12 \\[0.3em] \end{vmatrix}\) - 0\(\begin{vmatrix} 0 & 6 \\[0.3em] 0&12 \\[0.3em] \end{vmatrix}\) + 3\(\begin{vmatrix} 0 & 3 \\[0.3em] 0&0 \\[0.3em] \end{vmatrix}\)

= 3(3×12 – 6×0) – 0(0×12 – 0×6) + 3(0×0 – 0×3) 

= 3(36 – 0) + 0 + 3(0 + 0) 

= 3×36 = 108 

= 27 × 4 

= 27 |A| 

Hence, 

|3A|= 27 |A|


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