[Solved] The Number Of Values Of K, For Which The System Of Equations

Concept:

Let \(A = \left[ {\begin{array}{*{20}{c}} a&b\\ c&d \end{array}} \right]\;then\;adj\;\left( A \right) = \left[ {\begin{array}{*{20}{c}} d&{ - b}\\ { - c}&a \end{array}} \right]\)

Matrix Method:

Let us consider the system of linear equations:

a11 × x + a12 × y = b1

a21 × x + a22 × y = b2

We can write these equations in matrix form as: A X = B, where \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right],\;X = \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right]\;and\;B = \left[ {\begin{array}{*{20}{c}} {{b_1}}\\ {{b_2}} \end{array}} \right]\)

Determinant of A	Solution of system of equations
Case – 1	|A| ≠ 0	The system of equations is consistent and has unique solution which is given by: X = A-1 B, where \({A^{ - 1}} = \frac{{adj\;\left( A \right)}}{{\left| A \right|}}\)
Case –2	|A| = 0 and adj(A) × B = O, where O is the null matrix	The system of equations is consistent and has infinitely many solutions
Case -3	|A| = 0 and adj(A) × B ≠ O, where O is the null matrix	The system of equations is inconsistent and has no solution

Calculation:

Given: (k + 1) x + 8y = 4k and kx + (k + 3)y = 3k - 1

Here, \(A = \left[ {\begin{array}{*{20}{c}} {{k+1}}&{{8}}\\ {{k}}&{{k + 3}} \end{array}} \right],\;X = \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right]\;and\;B = \left[ {\begin{array}{*{20}{c}} {{4k}}\\ {{3k - 1}} \end{array}} \right]\)

As we know that, for infinite solution |A| = 0 and (adj (A)) × B = O, where O is the null matrix then the system of equations is consistent and has infinitely many solutions.

⇒ (k + 1)(k + 3) - 8k = 0

⇒ k2 - 4k + 3 = 0

⇒ k = 1 or k = 3

Now let's calculate (adj (A)) × B

\(⇒ \;adj\;\left( A \right) = \left[ {\begin{array}{*{20}{c}} k + 3&{ - 8}\\ { - k}&k + 1 \end{array}} \right]\)

\(⇒ adj\left( A \right) \cdot B = \left[ {\begin{array}{*{20}{c}} {4{k^2} - 12k + 8}\\ { - {k^2} + 2k - 1} \end{array}} \right]\)

Now by using (adj (A)) × B = O, where O is the null matrix, we get

⇒ 4k2 - 12k + 8 = 0 and - k2 + 2k - 1 = 0

⇒ k2 - 3k + 2 = 0 and k2 - 2k + 1 = 0

⇒ k = 1 or 2 and k = 1

So, for k = 1 given system will have infinitely many solutions.

Hence, there are 1 value of k for which the given system will have infinitely many solutions.

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