What is the value of p for which A = [[p, 3, 3], [3, p, 3], [3, 3, p]] is of rank 1?

Steps to Solve

1. Set up the determinant To find the value of $p$ for which the matrix $A$ has a rank of 1, we need to calculate its determinant and set it to zero.

The matrix ( A ) is given as: $$ A = \begin{bmatrix} p & 3 & 3 \ 3 & p & 3 \ 3 & 3 & p \end{bmatrix} $$

2. Calculate the determinant We will use the formula for the determinant of a 3x3 matrix: $$ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) $$

For our matrix, we have ( a = p, b = 3, c = 3, d = 3, e = p, f = 3, g = 3, h = 3, i = p ).

The determinant becomes: $$ |A| = p(p \cdot p - 3 \cdot 3) - 3(3 \cdot p - 3 \cdot 3) + 3(3 \cdot 3 - 3 \cdot p) $$

3. Simplify the determinant expression Expanding this gives: $$ |A| = p(p^2 - 9) - 3(3p - 9) + 3(9 - 3p) $$

Further simplifying:

• The first part: ( p(p^2 - 9) = p^3 - 9p )
• The second part: ( -3(3p - 9) = -9p + 27 )
• The third part: ( 3(9 - 3p) = 27 - 9p )

Combining these: $$ |A| = p^3 - 9p - 9p + 27 + 27 - 9p = p^3 - 27p + 54 $$

4. Set the determinant to zero To find the values of ( p ): $$ p^3 - 27p + 54 = 0 $$

5. Solve the polynomial equation We can use techniques such as factoring or the Rational Root Theorem to find possible rational roots. Testing for rational roots, we can find that ( p = 3 ) is a root.

6. Factor the polynomial Using synthetic division or polynomial long division, we can factor the cubic polynomial: $$ p^3 - 27p + 54 = (p - 3)(p^2 + 3p - 18) $$

7. Find other roots Next, we solve the quadratic equation: $$ p^2 + 3p - 18 = 0 $$

Using the quadratic formula: $$ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-18)}}{2 \cdot 1} $$

Calculate the discriminant: $$ \sqrt{9 + 72} = \sqrt{81} = 9 $$

Thus: $$ p = \frac{-3 \pm 9}{2} $$

Giving the roots:

• ( p = \frac{6}{2} = 3 )
• ( p = \frac{-12}{2} = -6 )

So the values of ( p ) for which the rank of ( A ) is 1 are ( p = 3 ) and ( p = -6 ).