If trace of the square matrix A-aij-n - n is zero- where aii-ii-3- then the order of the matrix is

Tr (A)=∑_i=1^na_ii=∑_i=1^ni(i 3)=∑_i=1^ni^2 3∑_i=1^ni =n(n+1)(2n+1)6 3n(n+1)2 =n(n+1)6[2n+1 9 ] nbsp; nbsp; =n(n+1)(n 4)3 nbsp; Given tr(A)=0 ∴ n=4 {∵ ...

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Standard XII

Mathematics

Second Fundamental Theorem of Calculus

If trace of t...

Question

If trace of the square matrix $A = [a_{i j}]_{n \times n}$ is zero, where $a_{i i} = i (i - 3),$ then the order of the matrix is

2 \times 2

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3 \times 3

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4 \times 4

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6 \times 6

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Solution

The correct option is C $4 \times 4$
$t r (A) = n \sum i = 1 a_{i i} = n \sum i = 1 i (i - 3) = n \sum i = 1 i^{2} - 3 n \sum i = 1 i = \frac{n (n + 1) (2 n + 1)}{6} - \frac{3 n (n + 1)}{2} = \frac{n (n + 1)}{6} [2 n + 1 - 9] = \frac{n (n + 1) (n - 4)}{3}$
Given $t r (A) = 0$
$∴ n = 4$ ${∵ n = 0, n = - 1 not possible}$

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If trace of the square matrix A=[aij]n×n is zero, where aii=i(i−3), then the order of the matrix is

The correct option is C 4×4tr(A)=n∑i=1aii=n∑i=1i(i−3)=n∑i=1i2−3n∑i=1i=n(n+1)(2n+1)6−3n(n+1)2=n(n+1)6[2n+1−9]=n(n+1)(n−4)3 Given tr(A)=0 ∴n=4 {∵n=0,n=−1 not possible}

If trace of the square matrix $A = [a_{i j}]_{n \times n}$ is zero, where $a_{i i} = i (i - 3),$ then the order of the matrix is