Using properties of determinants solve for \(x\begin{vmatrix} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{vmatrix}=0\)

Using properties of determinants solve the following for X:
\(\begin{vmatrix} x-a & x & x \\ x & x+a & x \\ x & x & x+a \end{vmatrix}=0,a\neq 0\)

Use product \(\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}\begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}\) to solve the system of equations.

x - y + 2z = 1;
2y - 3z = 1;
3x - 2y + 4z = 2

solve for x,y,z
\({2\over x}+{3\over y}+{10\over z}=4\);
\({4\over x}-{6\over y}+{5\over z}=1\);
\({6\over x}+{9\over y}-{20\over z}=2\)

A school wants to award its student or the values of Honesty, Regularity and Hard Work with a total cash award of Rs.6,000. Three times the award money for Hard work added to that given for Honesty amounts to Rs.11,000. The award money given for Honesty and Hard work together is double the one given for regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.

If a \(\neq \)p b\(\neq \)q c\(\neq \)r  and \(\left| \begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix} \right| \) = 0 find the value of \(\frac { p }{ p-a } +\frac { q }{ q-b } +\frac { r }{ r-c } \)

For the matrix A = \(\left[ \begin{matrix} 3 & 2 \\ 1 & 1 \end{matrix} \right] \)find the numbers a and b such that A² +aA + bI = O. Hence find A^-1.

If A^-1 = \(\left[ \begin{matrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{matrix} \right] \) and B =\(\left[ \begin{matrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{matrix} \right] \) find (AB)^-1