evaluation of determinant without expanding

Question

If $\;\det \begin{pmatrix} a & x & x & x \\ x & b & x & x \\ x & x & c & x \\ x & x & x & d \end{pmatrix} =f(x)-xf’(x)$

where $f'(x)$ denotes the derivative of $f(x)$ w.r.t. $x$, then $f(x)$ is equal to

$(a) \quad (x-a)(x-b)(x-c)(x-d)$

$(b) \quad (x+a)(x+b)(x+c)(x+d)$

$(c) \quad 2(x-a)(x-b)(x-c)(x-d)$

$(d) \quad 2(x+a)(x+b)(x+c)(x+d)$

I have tried using the properties of determinant to get the determinant in the form of $$\det \begin{pmatrix} a & x & x & x \\ 0 & b-x & x-c & 0 \\ 0 & 0 & c-x & x-d \\ x-a & 0 & 0 & d-x \end{pmatrix}$$

which is equal to $a \det \begin{pmatrix} x & x & x \\ b-x & x-c & 0 \\ 0 & c-x & x-d \end{pmatrix} +(a-x)\det \begin{pmatrix} x & x & x \\ b-x & x-c & 0 \\ 0 & c-x & x-d \end{pmatrix}$

which seems difficult to evaluate knowing that it is a objective problem there must be some way out. Please help me with this one.

Every time you format a question like this, a kitten sheds a tear. Please see the help center for proper formatting. — , Jul 23 '16 at 06:16
You applied wrong Laplace expansion. You must eliminate the corresponding row. In fact, the first $(3\times 3)$ matrix should be upper triangular. — , Jul 23 '16 at 07:38

score 0 · Answer 1 · edited Jun 12 '20 at 10:38

0

This is not a rigorous proof:

Substitute $x=0$, we have

$$f(0)=abcd$$

Substitute $x=a$, we have

$$f(a)-af'(a)=-a(a-b)(a-c)(a-d)$$

We'll have similar expressions for $x=b$, etc.

Considering

$$f(x)=(x-a)(x-b)(x-c)(x-d)$$

which gives $f(a)=0$ and $f'(a)=(a-b)(a-c)(a-d)$ and matches all conditions.

P.S.:
The general solution is $$f(x)=(x-a)(x-b)(x-c)(x-d)+k x$$

edited Jun 12 '20 at 10:38

Community

1

answered Jul 23 '16 at 10:37

Ng Chung Tak

18,990

evaluation of determinant without expanding

1 Answers1

This is not a rigorous proof: