CAT 2003 Slot 1 — QA Question 8

On substituting the values of p, q and r in the options we see that the values of p, q and r satisfy only the equation 5p − 2q − r = 0.

Hence, option (a).

Alternatively,

In order to understand this solution we need to have an understanding of a few concepts of higher mathematics like “Rank of a matrix”.

A number p is said to be the rank of a matrix A if

(i) A possesses at least one p-rowed determinant whose value is not zero

(ii) A does not possess any non zero (p + 1) rowed determinant.

In other words, the number of non-zero rows in the row-reduced form of a matrix is called the rank of a matrix.

For example,

Consider the matrix, 

$A=\left(\begin{array}{ccc}4& 5& 6\\ 1& 2& 3\\ 3& 4& 5\end{array}\right)$

Matrix A has only one 3 rowed determinant, namely

$\left|\begin{array}{ccc}4& 5& 6\\ 1& 2& 3\\ 3& 4& 5\end{array}\right|$

The value of this determinant is zero. Hence, the rank of A will be less than 3.

Now we will try to find a 2 rowed non zero determinant.

$\left|\begin{array}{cc}4& 5\\ 1& 2\end{array}\right|$ = 42 - 5.1 = 3 ≠ 0

The fact that every 3 rowed determinant of A is zero and there is at least one 2 rowed determinant of A which is not zero, is generally expressed by saying that the rank of A is 2.

Consider the system of equations,

x + 2y − 3z = p

2x + 6y − 11z = q

x − 2y + 7z = r

The coefficient matrix (whose elements are coefficients of the three unknowns x, y and z) of the system is

$\left(\begin{array}{ccc}1& 2& -3\\ 2& 6& -11\\ 1& -2& 7\end{array}\right)$

And the Augmented matrix is

$\left(\begin{array}{cccc}1& 2& -3& p\\ 2& 6& -11& à\\ 1& -2& 7& r\end{array}\right)$

For the system of equations to possess solutions it is necessary that the rank of the augmented matrix and the rank of the coefficient matrix should be equal.

Let us first find the rank of the coefficient matrix.

We will first find the value of the only 3 rowed determinant of A,

$\left|\begin{array}{ccc}1& 2& -3\\ 2& 6& -11\\ 1& -2& 7\end{array}\right|$ = 1(6 × 7 - 11 × (-2)) - 2(2 × 7 - (-11)1) - 3(2 × (-2) - 6 × 1) = 0

Hence, the rank of the coefficient matrix will be less than 3.

The following is a two-rowed non-zero determinant in the coefficient matrix

$\left|\begin{array}{cc}1& 2\\ 2& 6\end{array}\right|$ = 6 - 4 = 2 ≠ 0

Hence, the rank of the coefficient matrix is 2.

For the system of equations to possess solutions the rank of the augmented matrix should also be 2.

We will first try to simplify the augmented matrix using row operations.

$\left(\begin{array}{cccc}1& 2& -3& p\\ 2& 6& -11& q\\ 1& -2& 7& r\end{array}\right)$

By the row operation R₃ → R₃ - R₁

we reduce the augmented matrix to the equivalent matrix,

$~\left(\begin{array}{cccc}1& 2& 3& p\\ 2& 6& -11& q\\ 0& -4& 10& r-p\end{array}\right)$

$~\left(\begin{array}{cccc}1& 2& -3& p\\ 0& 2& -5& q-2p\\ 0& -4& 10& r-p\end{array}\right)$ (By the row operation R₂ → R₂ - 2R₁)

$~\left(\begin{array}{cccc}1& 2& -3& p\\ 0& 2& -5& q-2p\\ 0& 0& 0& -5p+2q+r\end{array}\right)$ (By the row operation R₃ → R₃ - 2R₂)

For the augmented matrix to have rank 2 every 3 rowed determinant should be zero.

Hence,

5p − 2q − r = 0

Hence, option (a).