Examine the consistency of the following system of equations: $5x - y + 4z = 5$ $2x + 3y + 5z = 2$ $5x - 2y + 6z = -1$

Examine the consistency of the following system of – Class 12 Mathematics

Exercise 4.5 · Q6

5x - y + 4z = 5

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Concept understanding — Matrix Equation Solving

Solving a System of Equations by the Matrix Method

A system of linear equations can be written as a single matrix equation and solved in one clean step using the inverse of a matrix. This is the Class-12 "matrix method" for simultaneous equations.

Writing the system as $AX=B$

Take the system

a_1x+b_1y+c_1z=d_1,\quad a_2x+b_2y+c_2z=d_2,\quad a_3x+b_3y+c_3z=d_3.

Collect the coefficients, the unknowns, and the constants into matrices:

A=\begin{pmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{pmatrix},\quad X=\begin{pmatrix}x\\y\\z\end{pmatrix},\quad B=\begin{pmatrix}d_1\\d_2\\d_3\end{pmatrix}.

Then the whole system is just

AX=B.

Solving when $A$ is invertible

If $\det(A)\neq0$ , then $A^{-1}$ exists, and multiplying both sides on the left by $A^{-1}$ gives

X=A^{-1}B,\qquad\text{where }\;A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A).

So you compute $\det(A)$ , then $\operatorname{adj}(A)$ , form $A^{-1}$ , and multiply by $B$ . The single column $X=A^{-1}B$ hands you $x$ , $y$ , $z$ at once, and because $A^{-1}$ is unique, the solution is unique.

Watch out

Multiply in the correct order: $X=A^{-1}B$ , not $BA^{-1}$ . Matrix multiplication is not commutative, and $BA^{-1}$ is not even defined here.

When $\det(A)=0$

If $\det(A)=0$ , $A^{-1}$ does not exist and the inverse method fails. The system is then either inconsistent (no solution) or has infinitely many solutions. Decide which by computing $(\operatorname{adj}A)\,B$ :

$(\operatorname{adj}A)\,B\neq O$ → no solution (inconsistent).
$(\operatorname{adj}A)\,B=O$ → infinitely many solutions (consistent, dependent). …

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Writing the system as $AX=B$ , we get $\det A=51\neq0$ , so $X=A^{-1}B=(3,\,2,\,-2)$ . Thus $x=3,\ y=2,\ z=-2$ .

The idea

The matrix method packs the whole system into one equation $AX=B$ , where $A$ holds the coefficients, $X$ the unknowns and $B$ the constants. If $\det A\neq0$ then $A^{-1}$ exists and multiplying on the left gives $X=A^{-1}B$ — the three values at once.

Set up

A=\begin{bmatrix}5&-1&4\\2&3&5\\5&-2&6\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}5\\2\\-1\end{bmatrix}.

Step 1 — Determinant

Expanding along the first row,

\det A=5\begin{vmatrix}3&5\\-2&6\end{vmatrix}-(-1)\begin{vmatrix}2&5\\5&6\end{vmatrix}+4\begin{vmatrix}2&3\\5&-2\end{vmatrix}=5(28)+1(-13)+4(-19)=51.

Since $\det A=51\neq0$ , $A$ is invertible and the solution is unique.

Step 2 — Cofactors and adjoint

Computing the nine cofactors gives the cofactor matrix

\begin{bmatrix}28&13&-19\\-2&10&5\\-17&-17&17\end{bmatrix},

and the adjoint is its transpose:

$\operatorname{adj}A=\begin{bmatrix}28&-2&-17\\13&10&-17\\-19&5&17\end{bmatrix}.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

JKBOSE Class 12 Annual Regular Examination 2022Set SZ2 marks
$x$
›Reveal solutionSolution

Equating corresponding entries of the two equal column matrices gives $x+y+z=9$ , $x+z=5$ , $y+z=7$ , which solve to $x=2,\ y=4,\ z=3$ .

Since the two $3\times1$ matrices are equal, their corresponding entries are equal:

$x+y+z=9 \quad\text{...(i)}$

$x+z=5 \quad\text{...(ii)}$

$y+z=7 \quad\text{...(iii)}$

Subtract (ii) from (i): $(x+y+z)-(x+z)=9-5 \Rightarrow y=4$ .

…

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JKBOSE Class 12 Annual Regular Examination 2021Set KD2 marks
$\begin{bmatrix} x+y+z \\ x+z \\ y+z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}$
›Reveal solutionSolution

Equate corresponding entries of the two equal column matrices to get three linear equations, then solve by elimination.

Since the matrices are equal, their corresponding entries are equal:

$x+y+z=9 \quad (1)$

$x+z=5 \quad (2)$

$y+z=7 \quad (3)$

Subtract (2) from (1): $(x+y+z)-(x+z)=9-5 \Rightarrow y=4$ .

…

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JKBOSE Class 12 Annual Regular Examination 2020Set SZ2 marks
$x$
›Reveal solutionSolution

Multiply the first matrix by 2, add the second, and match corresponding entries.

$2\begin{pmatrix}1&3\\0&x\end{pmatrix}+\begin{pmatrix}y&0\\1&2\end{pmatrix} = \begin{pmatrix}5&6\\1&8\end{pmatrix}$

…

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Q.Examine the consistency of the following system of equations: $5x - y + 4z = 5$ $2x + 3y + 5z = 2$ $5x - 2y + 6z = -1$

Concept understanding — Matrix Equation Solving

Solving a System of Equations by the Matrix Method

Writing the system as $AX=B$

Solving when $A$ is invertible

When $\det(A)=0$

The idea

Set up

Step 1 — Determinant

Step 2 — Cofactors and adjoint

Method: Solving a 3×3 System via the Inverse (Adjoint) when the Determinant is Non-zero

Steps

Common Mistakes

Q.Examine the consistency of the following system of equations: 5x−y+4z=55x - y + 4z = 55x−y+4z=5 2x+3y+5z=22x + 3y + 5z = 22x+3y+5z=2 5x−2y+6z=−15x - 2y + 6z = -15x−2y+6z=−1

The idea

Set up

Step 1 — Determinant

Step 2 — Cofactors and adjoint

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Q.Examine the consistency of the following system of equations: $5x - y + 4z = 5$ $2x + 3y + 5z = 2$ $5x - 2y + 6z = -1$

Unlock everything free for 14 days