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**Solution of a system of linear equations using inverse of a matrix**

Consider the system of equations

a_{1}x + a_{2}y + a_{3}z = k_{1}

b_{1}x + b_{2}y + b_{3}z = k_{2}

c_{1}x + c_{2}y + c_{3}z = k_{3}

In matrix form, the system of equations can be expressed as

$\left[\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{a}}_{\text{2}}& {\text{a}}_{\text{3}}\\ {\text{b}}_{\text{1}}& {\text{b}}_{\text{2}}& {\text{b}}_{\text{3}}\\ {\text{c}}_{\text{1}}& {\text{c}}_{\text{2}}& {\text{c}}_{\text{3}}\end{array}\right]$ $\left[\begin{array}{c}\text{x}\\ \text{y}\\ \text{z}\end{array}\right]$ = $\left[\begin{array}{c}{\text{k}}_{\text{1}}\\ {\text{k}}_{\text{2}}\\ {\text{k}}_{\text{3}}\end{array}\right]$ â€¦(i)

Let A = $\left[\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{a}}_{\text{2}}& {\text{a}}_{\text{3}}\\ {\text{b}}_{\text{1}}& {\text{b}}_{\text{2}}& {\text{b}}_{\text{3}}\\ {\text{c}}_{\text{1}}& {\text{c}}_{\text{2}}& {\text{c}}_{\text{3}}\end{array}\right]$ ; X = $\left[\begin{array}{c}\text{x}\\ \text{y}\\ \text{z}\end{array}\right]$ and B = $\left[\begin{array}{c}{\text{k}}_{\text{1}}\\ {\text{k}}_{\text{2}}\\ {\text{k}}_{\text{3}}\end{array}\right]$

From (i), we have

AX = B

**Case (i):** A is a non-singular matrix. â‡’ A^{-1} exists.

We have AX = B

â‡’ A^{-1} (AX) = A^{-1}B

â‡’ (A^{-1} A)X = A^{-1}B (by associative property)

â‡’ (I)X = A^{-1}B ('.' A^{-1}A = I , inverse property)

â‡’ X = A^{-1}B â‡’ X = A^{-1}B ('.' IX = X , identity property)

This method of solving a system of linear equations is known as the matrix method.

**Case (ii):** A is a singular matrix.

â‡’ |A| = 0

We have

X = A^{-1}B â‡’ X = (adj A / Det A)B

**Sub-case (i):** **(Adj A)B ****is a non-zero matrix**.

In this case, the system of linear equations is inconsistent.

In other words, the system of linear equations has no solution.

**Sub-case (ii):** **(Adj A)B**** is a zero matrix.**

In this case, the system of linear equations has infinitely many solutions or no solution at all.

**Solution of a system of linear equations using inverse of a matrix**

Consider the system of equations

a_{1}x + a_{2}y + a_{3}z = k_{1}

b_{1}x + b_{2}y + b_{3}z = k_{2}

c_{1}x + c_{2}y + c_{3}z = k_{3}

In matrix form, the system of equations can be expressed as

$\left[\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{a}}_{\text{2}}& {\text{a}}_{\text{3}}\\ {\text{b}}_{\text{1}}& {\text{b}}_{\text{2}}& {\text{b}}_{\text{3}}\\ {\text{c}}_{\text{1}}& {\text{c}}_{\text{2}}& {\text{c}}_{\text{3}}\end{array}\right]$ $\left[\begin{array}{c}\text{x}\\ \text{y}\\ \text{z}\end{array}\right]$ = $\left[\begin{array}{c}{\text{k}}_{\text{1}}\\ {\text{k}}_{\text{2}}\\ {\text{k}}_{\text{3}}\end{array}\right]$ â€¦(i)

Let A = $\left[\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{a}}_{\text{2}}& {\text{a}}_{\text{3}}\\ {\text{b}}_{\text{1}}& {\text{b}}_{\text{2}}& {\text{b}}_{\text{3}}\\ {\text{c}}_{\text{1}}& {\text{c}}_{\text{2}}& {\text{c}}_{\text{3}}\end{array}\right]$ ; X = $\left[\begin{array}{c}\text{x}\\ \text{y}\\ \text{z}\end{array}\right]$ and B = $\left[\begin{array}{c}{\text{k}}_{\text{1}}\\ {\text{k}}_{\text{2}}\\ {\text{k}}_{\text{3}}\end{array}\right]$

From (i), we have

AX = B

**Case (i):** A is a non-singular matrix. â‡’ A^{-1} exists.

We have AX = B

â‡’ A^{-1} (AX) = A^{-1}B

â‡’ (A^{-1} A)X = A^{-1}B (by associative property)

â‡’ (I)X = A^{-1}B ('.' A^{-1}A = I , inverse property)

â‡’ X = A^{-1}B â‡’ X = A^{-1}B ('.' IX = X , identity property)

This method of solving a system of linear equations is known as the matrix method.

**Case (ii):** A is a singular matrix.

â‡’ |A| = 0

We have

X = A^{-1}B â‡’ X = (adj A / Det A)B

**Sub-case (i):** **(Adj A)B ****is a non-zero matrix**.

In this case, the system of linear equations is inconsistent.

In other words, the system of linear equations has no solution.

**Sub-case (ii):** **(Adj A)B**** is a zero matrix.**

In this case, the system of linear equations has infinitely many solutions or no solution at all.