Question

Let B=(b_1,b_2,...,b_{r-1},b_r,b_{r+1},....,b_n) be a non-singular matrix. If column b_r is replaced by a and that the resulting matrix is called B_a along with a=\sum_{i=1}^{n}{y_ib_i}, then state the necessary and sufficient condition for B_a to be non-singular.

Properties for the Solution

Since B is a non-singular matrix composed of b_n column vectors, it satisfies the following properties:

1. B is a square matrix with size n \times n.
2. Transpose of B (B^T) will not change its non-singularity nature or rank.
3. B has full rank ⟹ rank(B)=rank(B^T)=r=n
4. B_T is a linearly independent system of rank n.

Solution

Now, we can say that BT is a matrix with row vectors B^T=(b_1,b_2,...,b_{r-1},b_r,b_{r+1},....,b_n) and say, column vectors B^T=(v_1,v_2,...,v_{r-1},v_r,v_{r+1},....,v_n).

Row vectors can be written as:

    \begin{align*} \left b_1=c_{11}.v_1 + c_{12}.v_2 +....+ c_{1r}.v_r +....+ c_{1n}.v_n   \end{align*}

    \begin{align*} \left b_2=c_{21}.v_1 + c_{22}.v_2 +....+ c_{2r}.v_r +....+ c_{2n}.v_n   \end{align*}

    \begin{align*} \left b_r=c_{r1}.v_1 + c_{r2}.v_2 +....+ c_{rr}.v_r +....+ c_{rn}.v_n   \end{align*}

    \begin{align*} \left b_n=c_{n1}.v_1 + c_{n2}.v_2 +....+ c_{nr}.v_r +....+ c_{nn}.v_n   \end{align*}

As per question, a=\sum_{i=1}^{n}{y_ib_i}. This means that,

    \begin{align*} \left a=y_1.b_1 + y_2.b_2 +....+ y_r.b_r +....+ y_n.b_n  \end{align*}

Assume that y_i is a scalar quantity. Our aim is to replace b_r with a. Let’s say we did that.

Using elementary row operations, if we subtract a with y_1b_1, y_2b_2,.,y_rb_r,..,y_nb_n etc., a will become a zero vector. If that happens, the rank of B^T will become n-1. This will cause the system to be linearly dependent & lose its non-singular nature.

This can be avoided if y_r is never equal to zero. In that case, the term y_rb_r will remain non-zero within the system after elementary row operations.

Thus, the necessary & sufficient condition for B_a to be non-singular is y_r \ne 0.

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