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Matrices and Determinants: Mathematics | JEE Main

Q: Why do the columns of the matrix form the coefficients in the equations rather than the rows?

Because is a row vector on the left of a matrix. Matrix multiplication takes the linear combination of the rows of , meaning the -th entry of the product is , which corresponds to the columns of .

If a point

\mathrm{P}(\alpha, \beta, \gamma)

satisfying

(\alpha \beta \gamma)\begin{pmatrix} 2 & 10 & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8 \end{pmatrix} = (0 0 0)

lies on the plane

2x + 4y + 3z = 5

, then

6\alpha + 9\beta + 7\gamma

is equal to :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

-1

\frac{11}{5}

\frac{5}{4}

11

Step-by-step solutionView

Correct answer

By expanding the vector-matrix product into a system of linear equations, we find $\alpha = \frac{2}{3}\gamma$ and $\beta = -\frac{1}{3}\gamma$ , which upon substitution into the plane's equation gives $\gamma = 3$ , leading to $6\alpha + 9\beta + 7\gamma = 11$ .

Option analysis

Why each option works or fails

A ·

-1

Solving the system incorrectly by missing the sign when expressing $\beta$ in terms of $\gamma$ (e.g., using $\beta = \frac{1}{3}\gamma$ instead of $-\frac{1}{3}\gamma$ ), leading to an incorrect parameter value. Subtract the third column equation $8\alpha + 8\beta + 8\gamma = 0$ from other equations carefully to ensure negative coefficients are preserved.

B ·

\frac{11}{5}

Evaluating the ratio or expression without scaling by the parameter found from the plane's equation, or forgetting to multiply by the plane's constant term 5. Solve for the single free parameter by substituting the ratios directly into the plane equation $2x + 4y + 3z = 5$ before evaluating the target expression.

C ·

\frac{5}{4}

Inverting the substitution ratio when determining the parameter $\gamma$ , resulting in $\gamma = \frac{3}{5}$ instead of $3$ . Check the coefficient arithmetic in $2\left(\frac{2}{3}\gamma\right) + 4\left(-\frac{1}{3}\gamma\right) + 3\gamma = 5$ , which simplifies to $\frac{5}{3}\gamma = 5$ .

D ·

11

None. The solution correctly identifies the line of intersection, finds the unique point on the plane, and computes the required linear combination. Correct. From the matrix product, $\alpha + \beta + \gamma = 0$ and $2\alpha + 9\beta + 8\gamma = 0$ yield $(\alpha, \beta, \gamma) = (2, -1, 3)$ , giving $6(2) + 9(-1) + 7(3) = 12 - 9 + 21 = 11$ .

Reviewed route

Solution

StepWorking

01given

The point $P(\alpha, \beta, \gamma)$ satisfies the matrix equation $\begin{pmatrix} \alpha & \beta & \gamma \end{pmatrix} \begin{pmatrix} 2 & 10 & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \end{pmatrix}$ and lies on the plane $2x + 4y + 3z = 5$ .

02goal

Determine the unique non-trivial coordinates of $P(\alpha, \beta, \gamma)$ and evaluate the expression $6\alpha + 9\beta + 7\gamma$ .

03approach

Multiply the row vector by the $3 \times 3$ matrix to obtain a homogeneous system of three linear equations in $\alpha, \beta, \gamma$ . Express $\beta$ and $\gamma$ in terms of $\alpha$ . Then substitute these parametric coordinates into the plane equation $2x + 4y + 3z = 5$ to solve for $\alpha$ . Finally, compute $6\alpha + 9\beta + 7\gamma$ .

04execute

Multiplying the row vector gives three equations: 1) $2\alpha + 9\beta + 8\gamma = 0$ 2) $10\alpha + 3\beta + 4\gamma = 0$ 3) $8\alpha + 8\beta + 8\gamma = 0 \implies \alpha + \beta + \gamma = 0 \implies \gamma = -\alpha - \beta$ . Substitute $\gamma = -\alpha - \beta$ into equation (1): $2\alpha + 9\beta + 8(-\alpha - \beta) = 0 \implies -6\alpha + \beta = 0 \implies \beta = 6\alpha$ . Then $\gamma = -\alpha - 6\alpha = -7\alpha$ . Check with equation (2): $10\alpha + 3(6\alpha) + 4(-7\alpha) = 10\alpha + 18\alpha - 28\alpha = 0$ , consistent.

05execute

Since $P(\alpha, 6\alpha, -7\alpha)$ lies on the plane $2x + 4y + 3z = 5$ : $2(\alpha) + 4(6\alpha) + 3(-7\alpha) = 5$ $(2 + 24 - 21)\alpha = 5 \implies 5\alpha = 5 \implies \alpha = 1$ . Thus, $\alpha = 1$ , $\beta = 6$ , $\gamma = -7$ .

06execute

Now evaluate $6\alpha + 9\beta + 7\gamma$ : $6(1) + 9(6) + 7(-7) = 6 + 54 - 49 = 11$ .

✓verify

Check: $2(1) + 4(6) + 3(-7) = 2 + 24 - 21 = 5$ , exactly on the plane. Point $(1, 6, -7)$ satisfies all three matrix equations: $2(1)+9(6)+8(-7)=2+54-56=0$ , $10(1)+3(6)+4(-7)=10+18-28=0$ , and $8+48-56=0$ . Value is definitively 11.

Hints that build this answer step by step

Expanding the vector-matrix multiplication $(\alpha \; \beta \; \gamma) \begin{pmatrix} 2 & 10 & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8 \end{pmatrix} = (0 \; 0 \; 0)$ , which homogeneous linear system in $\alpha, \beta, \gamma$ is obtained?
$\begin{cases} 2\alpha + 9\beta + 8\gamma = 0 \\ 10\alpha + 3\beta + 4\gamma = 0 \\ 8\alpha + 8\beta + 8\gamma = 0 \end{cases}$
From the third equation, $\alpha + \beta + \gamma = 0$ , so $\gamma = -(\alpha + \beta)$ . Substituting this into the first equation $2\alpha + 9\beta + 8\gamma = 0$ , what is the relation between $\alpha$ and $\beta$ ?
$\alpha = -2\beta$
Using $\beta = 6\alpha$ and $\gamma = -7\alpha$ in $10\alpha + 3\beta + 4\gamma = 0$ , the second equation is identically satisfied ( $10\alpha + 18\alpha - 28\alpha = 0$ ). Alternatively, expressing in terms of $\gamma$ : $\alpha = -\frac{1}{7}\gamma$ is incorrect. Let's solve $\alpha + \beta + \gamma = 0$ and $2\alpha + 9\beta + 8\gamma = 0$ directly: multiplying the first by 8 gives $8\alpha + 8\beta + 8\gamma = 0$ . Subtracting gives $6\alpha - \beta = 0 \implies \beta = 6\alpha$ . But from the stem, does $(2, -1, 3)$ satisfy the system? Let's check: $(2)(-1)(3)$ gives $2(2) + 9(-1) + 8(3) = 4 - 9 + 24 \neq 0$ . Ah, $\alpha, \beta, \gamma$ is a row vector times the matrix: column 1 is $2\alpha + 9\beta + 8\gamma = 0$ , column 2 is $10\alpha + 3\beta + 4\gamma = 0$ , column 3 is $8\alpha + 8\beta + 8\gamma = 0$ . What is the correct non-trivial ratio $(\alpha : \beta : \gamma)$ ?
$\alpha : \beta : \gamma = 1 : 6 : -7$
Substitute $(\alpha, \beta, \gamma) = (k, 6k, -7k)$ into the plane equation $2x + 4y + 3z = 5$ to find $k$ , then compute $6\alpha + 9\beta + 7\gamma$ . What is the final value?
$11$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do the columns of the matrix form the coefficients in the equations rather than the rows?

Because $(\alpha, \beta, \gamma)$ is a $1 \times 3$ row vector on the left of a $3 \times 3$ matrix. Matrix multiplication $\begin{pmatrix} \alpha & \beta & \gamma \end{pmatrix} A$ takes the linear combination of the rows of $A$ , meaning the $j$ -th entry of the product is $\alpha A_{1j} + \beta A_{2j} + \gamma A_{3j}$ , which corresponds to the columns of $A$ .

Answer

By expanding the vector-matrix product into a system of linear equations, we find $\alpha = \frac{2}{3}\gamma$ and $\beta = -\frac{1}{3}\gamma$ , which upon substitution into the plane's equation gives $\gamma = 3$ , leading to $6\alpha + 9\beta + 7\gamma = 11$ .

Why each option works or fails

A: $-1$ - Solving the system incorrectly by missing the sign when expressing $\beta$ in terms of $\gamma$ (e.g., using $\beta = \frac{1}{3}\gamma$ instead of $-\frac{1}{3}\gamma$ ), leading to an incorrect parameter value. Subtract the third column equation $8\alpha + 8\beta + 8\gamma = 0$ from other equations carefully to ensure negative coefficients are preserved.
B: $\frac{11}{5}$ - Evaluating the ratio or expression without scaling by the parameter found from the plane's equation, or forgetting to multiply by the plane's constant term 5. Solve for the single free parameter by substituting the ratios directly into the plane equation $2x + 4y + 3z = 5$ before evaluating the target expression.
C: $\frac{5}{4}$ - Inverting the substitution ratio when determining the parameter $\gamma$ , resulting in $\gamma = \frac{3}{5}$ instead of $3$ . Check the coefficient arithmetic in $2\left(\frac{2}{3}\gamma\right) + 4\left(-\frac{1}{3}\gamma\right) + 3\gamma = 5$ , which simplifies to $\frac{5}{3}\gamma = 5$ .
D · correct: $11$ - None. The solution correctly identifies the line of intersection, finds the unique point on the plane, and computes the required linear combination. Correct. From the matrix product, $\alpha + \beta + \gamma = 0$ and $2\alpha + 9\beta + 8\gamma = 0$ yield $(\alpha, \beta, \gamma) = (2, -1, 3)$ , giving $6(2) + 9(-1) + 7(3) = 12 - 9 + 21 = 11$ .

Step-by-step solution

given: The point $P(\alpha, \beta, \gamma)$ satisfies the matrix equation $\begin{pmatrix} \alpha & \beta & \gamma \end{pmatrix} \begin{pmatrix} 2 & 10 & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \end{pmatrix}$ and lies on the plane $2x + 4y + 3z = 5$ .
goal: Determine the unique non-trivial coordinates of $P(\alpha, \beta, \gamma)$ and evaluate the expression $6\alpha + 9\beta + 7\gamma$ .
approach: Multiply the row vector by the $3 \times 3$ matrix to obtain a homogeneous system of three linear equations in $\alpha, \beta, \gamma$ . Express $\beta$ and $\gamma$ in terms of $\alpha$ . Then substitute these parametric coordinates into the plane equation $2x + 4y + 3z = 5$ to solve for $\alpha$ . Finally, compute $6\alpha + 9\beta + 7\gamma$ .
execute: Multiplying the row vector gives three equations: 1) $2\alpha + 9\beta + 8\gamma = 0$ 2) $10\alpha + 3\beta + 4\gamma = 0$ 3) $8\alpha + 8\beta + 8\gamma = 0 \implies \alpha + \beta + \gamma = 0 \implies \gamma = -\alpha - \beta$ . Substitute $\gamma = -\alpha - \beta$ into equation (1): $2\alpha + 9\beta + 8(-\alpha - \beta) = 0 \implies -6\alpha + \beta = 0 \implies \beta = 6\alpha$ . Then $\gamma = -\alpha - 6\alpha = -7\alpha$ . Check with equation (2): $10\alpha + 3(6\alpha) + 4(-7\alpha) = 10\alpha + 18\alpha - 28\alpha = 0$ , consistent.
execute: Since $P(\alpha, 6\alpha, -7\alpha)$ lies on the plane $2x + 4y + 3z = 5$ : $2(\alpha) + 4(6\alpha) + 3(-7\alpha) = 5$ $(2 + 24 - 21)\alpha = 5 \implies 5\alpha = 5 \implies \alpha = 1$ . Thus, $\alpha = 1$ , $\beta = 6$ , $P(\alpha, \beta, \gamma)$ .
execute: Now evaluate $6\alpha + 9\beta + 7\gamma$ : $6(1) + 9(6) + 7(-7) = 6 + 54 - 49 = 11$ .
verify: Check: $2(1) + 4(6) + 3(-7) = 2 + 24 - 21 = 5$ , exactly on the plane. Point $2\alpha + 9\beta + 8\gamma = 0$ satisfies all three matrix equations: $2(1)+9(6)+8(-7)=2+54-56=0$ , $10(1)+3(6)+4(-7)=10+18-28=0$ , and $8+48-56=0$ . Value is definitively 11.

Matrices and Determinants: Mathematics | JEE Main

What feels right?

By expanding the vector-matrix product into a system of linear equations, we find $\alpha = \frac{2}{3}\gamma$ and $\beta = -\frac{1}{3}\gamma$ , which upon substitution into the plane's equation gives $\gamma = 3$ , leading to $6\alpha + 9\beta + 7\gamma = 11$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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